Click vào số 0 bên phải cùng để vote cho mình nếu lời giải đúng nhé :DNeu $A>120^0$, thi $\angle EAD=\angle EDA >60^0$ nen $\angle AED<60^0$. Do vay $ADChung minh tuong tu$AD $$\frac{AD}{AB}+\frac{AD}{AC} <\frac{AF}{AB}+\frac{AE}{AC}=\frac{DC}{CB}+\frac{BD}{CB}=1$$ Neu $A<120^0$ thi $\angle EAD=\angle EDA <60^0$ nen $\angle AED > 60^0$. Nhu vay $AD>AE$, tuong tu $AD>AF$. Nhu vay $$\frac{AD}{AB}+\frac{AD}{Ac}>\frac{AF}{AB}+\frac{AE}{AC}=1$$ Nhu vay $\angle A=120^0$

Click vào số 0 bên phải cùng để vote cho mình nếu lời giải đúng nhé :Db) Neu $A>120^0$, thi $\angle EAD=\angle EDA >60^0$ nen $\angle AED<60^0$. Do vay $AD<AE$.Chung minh tuong tu $AD<AF$. Nhu vay $$\frac{AD}{AB}+\frac{AD}{AC} <\frac{AF}{AB}+\frac{AE}{AC}=\frac{DC}{CB}+\frac{BD}{CB}=1$$ Neu $A<120^0$ thi $\angle EAD=\angle EDA <60^0$ nen $\angle AED > 60^0$. Nhu vay $AD>AE$, tuong tu $AD>AF$. Nhu vay $$\frac{AD}{AB}+\frac{AD}{Ac}>\frac{AF}{AB}+\frac{AE}{AC}=1$$ Nhu vay $\angle A=120^0$

Click vào số 0 bên phải cùng để vote cho mình nếu lời giải đúng nhé :Db) Neu $A>120^0$, thi $\angle EAD=\angle EDA >60^0$ nen $\angle AED<60^0$. Do vay $ADChung minh tuong tu$AD $$\frac{AD}{AB}+\frac{AD}{AC} <\frac{AF}{AB}+\frac{AE}{AC}=\frac{DC}{CB}+\frac{BD}{CB}=1$$ Neu $A<120^0$ thi $\angle EAD=\angle EDA <60^0$ nen $\angle AED > 60^0$. Nhu vay $AD>AE$, tuong tu $AD>AF$. Nhu vay $$\frac{AD}{AB}+\frac{AD}{Ac}>\frac{AF}{AB}+\frac{AE}{AC}=1$$ Nhu vay $\angle A=120^0$

Click vào số 0 bên phải cùng để vote cho mình nếu lời giải đúng nhé :DDat $S=x+y, P=xy$, ta co $$3=x^2+y^2+xy =S^2-P$$ Do vay $$(x^2-xy-3y^2+3)^2=(x^2-xy-3y^2+x^2+y^2+xy)^2$$ $$=4(x^2-y^2)^2=4(x-y)^2(x+y)^2=4S^2(S^2-4P)=12(4-S^2)\geq 48$$Do vay $x^2-xy-3y^2+3\in [-4\sqrt{3}, 4\sqrt{3}]$. Do vay ta co dpcm

Click vào số 0 bên phải cùng để vote cho mình nếu lời giải đúng nhé :DDat $S=x+y, P=xy$, ta co $$3=x^2+y^2+xy =S^2-P$$ Do vay $$(x^2-xy-3y^2+3)^2=(x^2-xy-3y^2+x^2+y^2+xy)^2$$ $$=4(x^2-y^2)^2=4(x-y)^2(x+y)^2=4S^2(S^2-4P)=12(4-S^2)\leq 48$$Do vay $x^2-xy-3y^2+3\in [-4\sqrt{3}, 4\sqrt{3}]$. Do vay ta co dpcm

Click vào số 0 bên phải cùng để vote cho mình nếu lời giải đúng nhé :DBan tu ve hinh nhe. Qua $D$ ta ke $DE//AB, DF//AC$ voi $E\in AC, F\in AB$. Do $\angle A=120^0$ nen ban se chung to duoc rang $\triangle ADE,ADF$ la cac tam giac deu. Ta co $$\frac{AD}{AC}+\frac{AD}{AB}=\frac{AF}{AB}+\frac{AE}{AC}=\frac{DC}{CB}+\frac{BD}{CB}=1$$ Nen $$\frac{1}{AC}+\frac{1}{AB}=\frac{1}{AD}$$

Click vào số 0 bên phải cùng để vote cho mình nếu lời giải đúng nhé :DBan tu ve hinh nhe. Qua $D$ ta ke $DE//AB, DF//AC$ voi $E\in AC, F\in AB$. a) Neu $\angle A=120^0$ ta se chung to duoc rang $\triangle ADE,ADF$ la cac tam giac deu. Ta co $$\frac{AD}{AC}+\frac{AD}{AB}=\frac{AF}{AB}+\frac{AE}{AC}=\frac{DC}{CB}+\frac{BD}{CB}=1$$ Nen $$\frac{1}{AC}+\frac{1}{AB}=\frac{1}{AD}$$

Click vào số 0 bên phải cùng để vote cho mình nếu lời giải đúng nhé :DTa co $y'=3x^2-12x+9=3(x-1)(x-3)$. Ta co bang bien thien ( tu ve)$y$ tang tu $-\infty$ den $d+4$ tren $(-\infty, 1]$, giam tu $d+4$ den $d$ tren $[1,3]$, tang tu $d$ den $d+4$ tren khoang $[3,+\infty)$. Nhu vay $y=0$ co ba nghiem phan biet $x_1<x_2<x_3$ neu $d<0<d+4$. Khi do $y(0)y(1)=y(1)y(3)=y(3)y(4)=d(d+4)<0$. Do vay $0<x_1<1<x_2<3<x_3<4$

Click vào số 0 bên phải cùng để vote cho mình nếu lời giải đúng nhé :DBan tu ve hinh nhe. Ta co $MN//AH\perp BC$ nen $$CN^2=MN^2+MC^2=(\frac{\sqrt{2}}{2})^2+(\frac{3}{4}\sqrt{3})^2=\frac{35}{16}$$ va $$AM^2=AH^2+MH^2=(\sqrt{2})^2+(\frac{\sqrt{3}}{4})^2=\frac{35}{16}$$ Nhu vay $AM=CN=\frac{\sqrt{35}}{4}\quad (1)$. Dung dinh ly Melenauyt ta co $$\frac{NA}{NB}\frac{KM}{KA}\frac{CB}{CM}=1\Rightarrow \frac{KM}{KA}=\frac{NB}{NA}\frac{CM}{CB}=\frac{3}{4}\Rightarrow \fac{KM}{AM}=\frac{3}{7} \quad (2)$$ Ta cung co $$\frac{KN}{KC}\frac{MC}{MB}\frac{AB}{AN}=1\Rightarrow \frac{KN}{KC}=\frac{MB}{MC}\frac{AN}{AB}=\frac{1}{6}\Rightarrow \frac{KC}{CN}=\frac{6}{7} \quad (3)$$ Tu $(1)(2)(3)$ suy ra $\frac{KM}{KC}=\frac{1}{2}=\frac{HM}{HC}$. Do vay $KH$ la tia phan giac cua $\angle CKM$

Click vào số 0 bên phải cùng để vote cho mình nếu lời giải đúng nhé :DBan tu ve hinh nhe. Ta co $MN//AH\perp BC$ nen $$CN^2=MN^2+MC^2=(\frac{\sqrt{2}}{2})^2+(\frac{3}{4}\sqrt{3})^2=\frac{35}{16}$$ va $$AM^2=AH^2+MH^2=(\sqrt{2})^2+(\frac{\sqrt{3}}{4})^2=\frac{35}{16}$$ Nhu vay $AM=CN=\frac{\sqrt{35}}{4}\quad (1)$. Dung dinh ly Melenauyt ta co $$\frac{NA}{NB}\frac{KM}{KA}\frac{CB}{CM}=1\Rightarrow \frac{KM}{KA}=\frac{NB}{NA}\frac{CM}{CB}=\frac{3}{4}\Rightarrow \frac{KM}{AM}=\frac{3}{7} \quad (2) $$Ta cung co$$ \frac{KN}{KC}\frac{MC}{MB}\frac{AB}{AN}=1\Rightarrow \frac{KN}{KC}=\frac{MB}{MC}\frac{AN}{AB}=\frac{1}{6}\Rightarrow \frac{KC}{CN}=\frac{6}{7} \quad (3)$$Tu $(1)(2)(3)$ suy ra $\frac{KM}{KC}=\frac{1}{2}=\frac{HM}{HC}$. Do vay $KH$ la tia phan giac cua $\angle CKM$

Click vào số 0 bên phải cùng để vote cho mình nếu lời giải đúng nhé :DBan tu ve hinh nhe. Ta co $MN//AH\perp BC$ nen $$CN^2=MN^2+MC^2=(\frac{\sqrt{2}}{2})^2+(\frac{3}{4}\sqrt{3})^2=\frac{35}{16}$$ va $$AM^2=AH^2+MH^2=(\sqrt{2})^2+(\frac{\sqrt{3}}{4})^2=\frac{35}{16}$$ Nhu vay $AM=CN=\frac{\sqrt{35}}{4}\quad (1)$. Dung dinh ly Melenauyt ta co $$\frac{NA}{NB}\frac{KM}{KA}\frac{CB}{CM}=1\Rightarrow \frac{KM}{KA}=\frac{NB}{NA}\frac{CM}{CB}=\frac{3}{4}\Rightarrow KM=\frac{3}{7}AM \quad (2) $$Ta cung co$$ \frac{KN}{KC}\frac{MC}{MB}\frac{AB}{AN}=1\Rightarrow \frac{KN}{KC}=\frac{MB}{MC}\frac{AN}{AB}=\frac{1}{6}\Rightarrow \frac{KC}{CN}=\frac{6}{7} \quad (3)$$Tu $(1)(2)(3)$ suy ra $\frac{KM}{KN}=\frac{1}{2}=\frac{HM}{HC}$. Do vay$KH$la tia phan giac cua$\angle CKM$

Click vào số 0 bên phải cùng để vote cho mình nếu lời giải đúng nhé :DBan tu ve hinh nhe. Ta co $MN//AH\perp BC$ nen $$CN^2=MN^2+MC^2=(\frac{\sqrt{2}}{2})^2+(\frac{3}{4}\sqrt{3})^2=\frac{35}{16}$$ va $$AM^2=AH^2+MH^2=(\sqrt{2})^2+(\frac{\sqrt{3}}{4})^2=\frac{35}{16}$$ Nhu vay $AM=CN=\frac{\sqrt{35}}{4}\quad (1)$. Dung dinh ly Melenauyt ta co $$\frac{NA}{NB}\frac{KM}{KA}\frac{CB}{CM}=1\Rightarrow \frac{KM}{KA}=\frac{NB}{NA}\frac{CM}{CB}=\frac{3}{4}\Rightarrow \fac{KM}{AM}=\frac{3}{7} \quad (2) $$Ta cung co$$ \frac{KN}{KC}\frac{MC}{MB}\frac{AB}{AN}=1\Rightarrow \frac{KN}{KC}=\frac{MB}{MC}\frac{AN}{AB}=\frac{1}{6}\Rightarrow \frac{KC}{CN}=\frac{6}{7} \quad (3)$$Tu $(1)(2)(3)$ suy ra $\frac{KM}{KC}=\frac{1}{2}=\frac{HM}{HC}$. Do vay$KH$la tia phan giac cua$\angle CKM$

Click vào số $0$ bên phải cùng để vote cho mình nếu lời giải đúng nhé :DBainay giai theo hinh hoc, tuong tuong, khong nen dung phuong trinh de tinh toan nhieu. De thay $A,B,C,D,0$ la cac dinh cua hinh lap phuong don vi nen duong tron $(S)$ chinh la duong tron tam $I_1=(1/2,1/2,1/2)$, ban kinh $\sqrt{3}/2$ va $D'\in (S)$.De thay $I_1,D', A', B'$ cung nam tren mat phang $(P):y-z=0$. Hon nua $\vec{C'I_1}\perp (P)$ do $\vec{C'I_1}=(-1/2,-1/2,1/2)$ nen $I_1$ chinh la tam cua duong tron giao tuyen.Vi $D'\in (S')\cap (S)$ nen $I_1D'$ chinh la ban kinh duong tron giao tuyen va bang $\sqrt{3}/2$.

Click vào số $0$ bên phải cùng để vote cho mình nếu lời giải đúng nhé :DBainay giai theo hinh hoc, tưởng tượng, khong nen dung phuong trinh de tinh toan nhieu. De thay $A,B,C,D,0$ la cac dinh cua hinh lap phuong don vi nen duong tron $(S)$ chinh la duong tron tam $I_1=(1/2,1/2,1/2)$, ban kinh $\sqrt{3}/2$ va $D'\in (S)$.De thay $I_1,D', A', B'$ cung nam tren mat phang $(P):y-z=0$. Hon nua $\vec{C'I_1}\perp (P)$ do $\vec{C'I_1}=(-1/2,-1/2,1/2)$ nen $I_1$ chinh la tam cua duong tron giao tuyen.Vi $D'\in (S')\cap (S)$ nen $I_1D'$ chinh la ban kinh duong tron giao tuyen va bang $\sqrt{3}/2$.

Dat $x^2+x+3=a$ va $2x^2+4x+5=b$, ta co $$log_3^{\frac{a}{b}}=b-a$$ . Hay $$log_2^a-loog_3^b=b-a$$ Dieu nay tuong duong voi $$log_3^a+a=log_3^b+b\quad (*)$$ .Vi $log_3^t+t$ la ham dong bien theo $t$ tren $[0,+\infty)$ nen $(*)$ dung neu va chi neu $a=b$. Nhu vay $$b-a=x^2+3x+2=(x+1)(x+2)=0$$ De thay tai $x=1,x=0$thi$b\ne 0$. Do vay nghiem cua phuong trinh da cho la$x=1,x=0$

Dat $x^2+x+3=a$ va $2x^2+4x+5=b$, ta co $$log_3^{\frac{a}{b}}=b-a$$ . Hay $$log_2^a-loog_3^b=b-a$$ Dieu nay tuong duong voi $$log_3^a+a=log_3^b+b\quad (*)$$ .Vi $log_3^t+t$ la ham dong bien theo $t$ tren $[0,+\infty)$ nen $(*)$ dung neu va chi neu $a=b$. Nhu vay $$b-a=x^2+3x+2=(x+1)(x+2)=0$$ De thay tai $x=-1,x=-2$thi$b\ne 0$. Do vay nghiem cua phuong trinh da cho la$x=-1,x=-2$

Gia su $x+y+z$ khong chia het cho $3$ thi $(x-y)(y-z)(z-x)$ khong chia het cho $3$. Do vay $x,y,z$ khong co cung so du khi chia cho $3$ nen co the gia su rang $x\equiv 0 (mod 3), y\equiv 1(od 3), z\equiv 2 (mod 3)$. Tuy nhien, khi do $x+y+z\equiv 0(mod 3)$ (Mau thuan). Vay nen $x+y+z=(x-y)(y-z)(z-x)$ chia het cho $3$. Suy ra ton tai hai trong ba so $x,y,z$ chia het cho $3$. Khong mat tinh tong quat, gia su $x\equiv y \equiv r(mod 3)$.Neu $r=1$, do $x+y+z$ chia het cho $3$ nen $z\equiv 1 (mod 3)$. Neu $r=2$, tuong tu ta co $z\equiv 2 (mod 3)$. Neu $r=0$, ta co $z\equiv 0 (mod 3)$.Tom lai $x\equiv y\equiv z (mod 3)$. Do vay $(x-y)(y-z)(z-x)$ chia het cho $27$. Nen $x+y+z$ chia het cho $27$.

Gia su $x+y+z$ khong chia het cho $3$ thi $(x-y)(y-z)(z-x)$ khong chia het cho $3$. Do vay $x,y,z$ khong co cung so du khi chia cho $3$ nen co the gia su rang $x\equiv 0 (mod 3), y\equiv 1(mod 3), z\equiv 2 (mod 3) $. Tuy nhien, khi do$x+y+z\equiv 0(mod 3)$(Mau thuan). Vay nen$x+y+z=(x-y)(y-z)(z-x)$chia het cho$3$. Suy ra ton tai hai trong ba so$x,y,z$chia het cho$3$. Khong mat tinh tong quat, gia su$x\equiv y \equiv r(mod 3)$.Neu$r=1$, do$x+y+z$chia het cho$3$nen$z\equiv 1 (mod 3)$. Neu$r=2$, tuong tu ta co$z\equiv 2 (mod 3)$. Neu$r=0$, ta co$z\equiv 0 (mod 3)$.Tom lai$x\equiv y\equiv z (mod 3)$. Do vay$(x-y)(y-z)(z-x)$chia het cho$27$. Nen$x+y+z$chia het cho$27$.

Gia su $x+y+z$ khong chia het cho $3$ thi $(x-y)(y-z)(z-x)$ khong chia het cho $3$. Do vay $x,y,z$ khong co cung so du khi chia cho $3$ nen co the gia su rang $x\equiv 0 (mod 3), y\equiv 1(od 3), z\equiv 2 (mod 3)$. Tuy nhien, khi do $x+y+z\equiv 0(mod 3)$ (Mau thuan). Vay nen $x+y+z=(x-y)(y-z)(z-x)$ chia het cho $3$. Suy ra ton tai hai trong ba so $x,y,z$ chia het cho $3$. Khong mat tinh tong quat, gia su $x\equiv y \equiv r(mod 3)$.Neu $r=1$, do $x+y+z$ chia het cho $3$ nen $z\equiv 1 (mod 3)$. Neu $r=2$, tuong tu ta co $z\equiv 2 (mod 3)$. Neu $r=0$, ta co $z\equiv 0 (mod 3)$.Tom lai $x\equiv y\equiv z (mod 3)$. Do vay $(x-y)(y-z)(z-x)$ chia het cho $27$. Nen $x+y+z$ chia het cho $27$.

Gia su $x+y+z$ khong chia het cho $3$ thi $(x-y)(y-z)(z-x)$ khong chia het cho $3$. Do vay $x,y,z$ khong co cung so du khi chia cho $3$ nen co the gia su rang $x\equiv 0 (mod 3), y\equiv 1(od 3), z\equiv 2 (mod 3)$. Tuy nhien, khi do $x+y+z\equiv 0(mod 3)$ (Mau thuan). Vay nen $x+y+z=(x-y)(y-z)(z-x)$ chia het cho $3$. Suy ra ton tai hai trong ba so $x,y,z$ chia het cho $3$. Khong mat tinh tong quat, gia su $x\equiv y \equiv r(mod 3)$.Neu $r=1$, do $x+y+z$ chia het cho $3$ nen $z\equiv 1 (mod 3)$. Neu $r=2$, tuong tu ta co $z\equiv 2 (mod 3)$. Neu $r=0$, ta co $z\equiv 0 (mod 3)$.Tom lai $x\equiv y\equiv z (mod 3)$. Do vay $(x-y)(y-z)(z-x)$ chia het cho $27$. Nen $x+y+z$ chia het cho $27$.

Dat $AB=c, AC=b$ thi $BC=\sqrt{a^2+b^2}$. Khi do $$HD/AC=HB/BC=AB^2/BC^2=c^2/(b^2+c^2)$$ tuong tu $HE/AB=b^2/(b^2+c^2)$, Ta co $$\frac{S_{ADHE}}{S_{ABC}}= 2 \frac{HD. HE}{AB.AC} = 2\frac{b^2c^2}{(b^2+c^2)^2} =\frac{1}{2}$$ Suy ra $4b^2c^2=(b^2+c^2)$ nen $b=c$. Suy ra dpcm

Dat $AB=c, AC=b$ thi $BC=\sqrt{a^2+b^2}$. Khi do $$HD/AC=HB/BC=AB^2/BC^2=c^2/(b^2+c^2)$$ tuong tu $HE/AB=b^2/(b^2+c^2)$, Ta co $$\frac{S_{ADHE}}{S_{ABC}}= 2 \frac{HD. HE}{AB.AC} = 2\frac{b^2c^2}{(b^2+c^2)^2} =\frac{1}{2}$$ Suy ra $4b^2c^2=(b^2+c^2)^2$nen$b=c$. Suy ra dpcm

$A=(-\infty, 1]$ va $B=(m,2]$. Neu $m\geq 1$thi$A\cup B=(-\infty, 1]\cup (m,2]\ne (-\infty, 2]$. Neu$m\leq 1$thi$A\cup B=(-\infty, 2]$. Nhu vay$m\leq 1$

$A=(-\infty, 1]$ va $B=(m,2]$. Neu $m&gt; 1$thi$A\cup B=(-\infty, 1]\cup (m,2]\ne (-\infty, 2]$. Neu$m\leq 1$thi$A\cup B=(-\infty, 2]$. Nhu vay$m\leq 1$

Ban giai sai roi, khong can dieu kien $\triangle ABC$ deu. Loi giai nhu sau: Goi diem doi xung cua $O$ qua $AB, BC, CA$ lan luot la $C',A',B'$. Vi $O$ la tam duong tron ngoai tiep $\triangle ABC$ nen $OA=OB=OC$ nen cacs hinh sau la hinh thoi : OBA'C, OAC'B, OCB'A. Do vay $$\vec{OA}+\vec{OB}=\vec{OC'} \\\vec{OA}+\vec{OC}=\vec{OB'}\\ \vec{OC}+\vec{OB}=\vec{OA'}$$ Tu do say ra $M=C', N=A', P=B'$

Ban giai sai roi, khong can dieu kien $\triangle ABC$ deu. Loi giai nhu sau: Goi diem doi xung cua $O$ qua $AB, BC, CA$ lan luot la $C',A',B'$. Vi $O$ la tam duong tron ngoai tiep $\triangle ABC$ nen $OA=OB=OC$ nen cac hinh sau la hinh thoi : $OBA'C, OAC'B, OCB'A.$ Do vay $$\vec{OA}+\vec{OB}=\vec{OC'} \\\vec{OA}+\vec{OC}=\vec{OB'}\\ \vec{OC}+\vec{OB}=\vec{OA'}$$ Tu do say ra $M=C', N=A', P=B'$

Giai tong quat luon cho cac cau: Voi $a+b+c+d\ne 0$, ta luon xac dinh duoc diem $Q$ duy nhat sao cho $$a\vec{OA}+ b\vec{OB}+ c\vec{OC} +d\vec{OD}=(a+b+c+d)\vec{OQ}$$ Ta co $$a\vec{MA}+ b\vec{MB}+ c\vec{MC}+d\vec{MD}\\= a(\vec{MO}+\vec{OA})+b(\vec{MO}+\vec{BO})+c(\vec{MO}+\vec{OC})+d(\vec{MO}+\vec{OD})\\= (a+b+c+d)(\vec{MO}+\vec{OQ})=(a+b+c+d)vec{MQ}$$ Ta co $a\vec{OA}+b\vec{OB}+c\vec{OC}+d\vec{OD}=k\vec{MI}$ voi moi $M$ nen $k\vec{MI}=(a+b+c+d)\vec{MQ}$ voi moi $M$. Do vay $k=a+b+c+d$ va $I =Q$. Voi cac cau a)b)c)d) ta chi can thay so la OK.

Giai tong quat luon cho cac cau: Voi $a+b+c+d\ne 0$, ta luon xac dinh duoc diem $Q$ duy nhat sao cho $$a\vec{OA}+ b\vec{OB}+ c\vec{OC} +d\vec{OD}=(a+b+c+d)\vec{OQ}$$ Ta co $$a\vec{MA}+ b\vec{MB}+ c\vec{MC}+d\vec{MD}\\= a(\vec{MO}+\vec{OA})+b(\vec{MO}+\vec{BO})+c(\vec{MO}+\vec{OC})+d(\vec{MO}+\vec{OD})\\= (a+b+c+d)(\vec{MO}+\vec{OQ})=(a+b+c+d)\vec{MQ}$$Ta co $a\vec{OA}+b\vec{OB}+c\vec{OC}+d\vec{OD}=k\vec{MI}$ voi moi $M$ nen $k\vec{MI}=(a+b+c+d)\vec{MQ}$ voi moi $M$. Do vay $k=a+b+c+d$ va $I =Q $. Voi cac cau a)b)c)d) ta chi can thay so la OK.