|
|
giải đáp
|
Chứng minh rằng dãy số bị chặn!
|
|
|
|
Dễ thấy: $x_n>0,\forall n$
Ta sẽ chứng minh: $x_n<2,\forall n$.
Thật vậy,
Với $n=1$, ta có: $x_1=\sqrt2<2$, đúng.
Giả sử BĐT đúng với $n=k,k\ge1$, hay $x_k<2$
Ta sẽ chứng minh: $x_{k+1}<2$
Ta có: $x_{k+1}=\sqrt{2+x_k}<\sqrt{2+2}=2$, đúng.
Vậy theo nguyên lý quy nạp, ta có: $x_h<2,\forall n$.
Vậy $(x_n)$ bị chặn.
|
|
|
|
giải đáp
|
Bất đẳng thức trong tam giác(tt).
|
|
|
|
Ta có:
$\sin^2A+\sin^2B+\sin^2C$
$=\dfrac{1-\cos2A}{2}+\dfrac{1-\cos2B}{2}+1-\cos^2C$
$=2-\dfrac{1}{2}(\cos2A+\cos2B)-\cos^2C$
$=2-\cos(A+B)\cos(A-B)-\cos^2C$
$=2+\cos C[\cos(A-B)-\cos C]$
$=2+\cos C[\cos(A-B)+\cos(A+B)]$
$=2+\cos A\cos B\cos C$
Khi đó:
$\sin^2A+\sin^2B+\sin^2C>2$
$\Leftrightarrow \cos A\cos B\cos C>0$
$\Leftrightarrow \cos A;\cos B;\cos C>0$
$\Leftrightarrow \Delta ABC$ nhọn.
|
|
|
|
giải đáp
|
Đơn điệu.
|
|
|
|
a.
Ta có:
$U_n=\sqrt{2n+1}-\sqrt{2n}$
$=\dfrac{1}{\sqrt{2n+1}+\sqrt{2n}}$
$>\dfrac{1}{\sqrt{2n+3}+\sqrt{2n+2}}$
$=\sqrt{2n+3}-\sqrt{2n+2}=U_{n+1}$
Suy ra: $U_n$ là dãy giảm.
b.
Dễ thấy: $U_n>0,\forall n\ge1$
Mà $U_n$ là dãy giảm nên $U_n\le U_1=\sqrt3-\sqrt2,\forall n\ge1$
Vậy: $U_n$ bị chặn.
|
|
|
|
sửa đổi
|
Bất đẳng thức trong tam giác.
|
|
|
|
c.Ta có:$\cos\frac{A}{2}+\cos\frac{B}{2}=2\cos(\frac{A}{4}+\frac{B}{4})\cos(\frac{A}{4}-\frac{B}{4})\le2\cos(\frac{A}{4}+\frac{B}{4})$$\cos\frac{C}{2}+\cos\frac{\pi}{6}=2\cos(\frac{C}{4}+\frac{\pi}{12})\cos(\frac{C}{4}-\frac{\pi}{12})\le2\cos(\frac{C}{4}+\frac{\pi}{12})$
$\cos(\frac{A}{4}+\frac{B}{4})+\cos(\frac{C}{4}+\frac{\pi}{12})=2\cos\frac{\pi}{6}\cos(\frac{A}{8}+\frac{B}{8}-\frac{C}{8}-\frac{\pi}{24})\le2\cos\frac{\pi}{6}$Suy ra: $\cos\frac{A}{2}+\cos\frac{B}{2}+\cos\frac{C}{2}\le3\cos\frac{\pi}{6}=\frac{3\sqrt3}{2}$Ta có:
$\frac{\displaystyle\cos\frac{A}{2}}{1+\cos
A}+\frac{\displaystyle\cos\frac{B}{2}}{1+\cos
B}+\frac{\displaystyle\cos\frac{C}{2}}{1+\cos C}$$=\frac{1}{\displaystyle 2\cos\frac{A}{2}}+\frac{1}{\displaystyle 2\cos\frac{B}{2}}+\frac{1}{\displaystyle 2\cos\frac{C}{2}}$$\ge\frac{9}{\displaystyle 2\cos\frac{A}{2}+2\cos\frac{B}{2}+2\cos\frac{C}{2}}\ge\sqrt3$Dấu bằng xảy ra khi: $\Delta ABC$ đều.
c.Ta có:$\cos\dfrac{A}{2}+\cos\dfrac{B}{2}=2\cos(\dfrac{A}{4}+\dfrac{B}{4})\cos(\dfrac{A}{4}-\dfrac{B}{4})\le2\cos(\dfrac{A}{4}+\dfrac{B}{4})$$\cos\dfrac{C}{2}+\cos\dfrac{\pi}{6}=2\cos(\dfrac{C}{4}+\dfrac{\pi}{12})\cos(\dfrac{C}{4}-\dfrac{\pi}{12})\le2\cos(\dfrac{C}{4}+\dfrac{\pi}{12})$$\cos(\dfrac{A}{4}+\dfrac{B}{4})+\cos(\dfrac{C}{4}+\dfrac{\pi}{12})=2\cos\dfrac{\pi}{6}\cos(\dfrac{A}{8}+\dfrac{B}{8}-\dfrac{C}{8}-\dfrac{\pi}{24})\le2\cos\dfrac{\pi}{6}$Suy ra: $\cos\dfrac{A}{2}+\cos\dfrac{B}{2}+\cos\dfrac{C}{2}\le3\cos\dfrac{\pi}{6}=\dfrac{3\sqrt3}{2}$Ta có:
$\dfrac{\cos\dfrac{A}{2}}{1+\cos
A}+\dfrac{\cos\dfrac{B}{2}}{1+\cos
B}+\dfrac{\cos\dfrac{C}{2}}{1+\cos C}$$=\dfrac{1}{2\cos\dfrac{A}{2}}+\dfrac{1}{2\cos\dfrac{B}{2}}+\dfrac{1}{2\cos\dfrac{C}{2}}$$\ge\dfrac{9}{2\cos\dfrac{A}{2}+2\cos\dfrac{B}{2}+2\cos\dfrac{C}{2}}\ge\sqrt3$Dấu bằng xảy ra khi: $\Delta ABC$ đều.
|
|
|
|
sửa đổi
|
Bất đẳng thức trong tam giác.
|
|
|
|
c.Ta có:$\sin\dfrac{A}{2}+\sin\dfrac{B}{2}=2\sin(\dfrac{A}{4}+\dfrac{B}{4})\cos(\dfrac{A}{4}-\dfrac{B}{4})\le2\sin(\dfrac{A}{4}+\dfrac{B}{4})$$\sin\dfrac{C}{2}+\sin\dfrac{\pi}{6}=2\sin(\dfrac{C}{4}+\dfrac{\pi}{12})\cos(\dfrac{C}{4}-\dfrac{\pi}{12})\le2\sin(\dfrac{C}{4}+\dfrac{\pi}{12})$$\sin(\dfrac{A}{4}+\dfrac{B}{4})+\sin(\dfrac{C}{4}+\dfrac{\pi}{12})=2\sin\dfrac{\pi}{6}\cos(\dfrac{A}{8}+\dfrac{B}{8}-\dfrac{C}{8}-\dfrac{\pi}{24})\le2\sin\dfrac{\pi}{6}$Suy ra: $\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}\le3\sin\dfrac{\pi}{6}=\dfrac{3}{2}$Suy ra: $\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}\le\left(\dfrac{\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}}{3}\right)^3\le\dfrac{1}{8}$Ta có:
$\left(1+\dfrac{1}{\sin\dfrac{A}{2}}\right)\left(1+\dfrac{1}{\sin\dfrac{B}{2}}\right)\left(1+\dfrac{1}{\sin\dfrac{C}{2}}\right)$$=1+\left(\dfrac{1}{\sin\dfrac{A}{2}}+\dfrac{1}{\sin\dfrac{B}{2}}+\dfrac{1}{\sin\dfrac{C}{2}}\right)+\left(\dfrac{1}{\sin\dfrac{A}{2}\sin\dfrac{B}{2}}+\dfrac{1}{\sin\dfrac{B}{2}\sin\dfrac{C}{2}}+\dfrac{1}{\sin\dfrac{C}{2}\sin\dfrac{A}{2}}\right)+\dfrac{1}{\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}}$$\ge1+\dfrac{3}{\sqrt[3]{\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}}}+\dfrac{3}{\sqrt[3]{\sin^2\dfrac{A}{2}\sin^2\dfrac{B}{2}\sin^2\dfrac{C}{2}}}+8\ge27$Dấu bằng xảy ra khi: $\Delta ABC$ đều.
e.Ta có:$\sin\dfrac{A}{2}+\sin\dfrac{B}{2}=2\sin(\dfrac{A}{4}+\dfrac{B}{4})\cos(\dfrac{A}{4}-\dfrac{B}{4})\le2\sin(\dfrac{A}{4}+\dfrac{B}{4})$$\sin\dfrac{C}{2}+\sin\dfrac{\pi}{6}=2\sin(\dfrac{C}{4}+\dfrac{\pi}{12})\cos(\dfrac{C}{4}-\dfrac{\pi}{12})\le2\sin(\dfrac{C}{4}+\dfrac{\pi}{12})$$\sin(\dfrac{A}{4}+\dfrac{B}{4})+\sin(\dfrac{C}{4}+\dfrac{\pi}{12})=2\sin\dfrac{\pi}{6}\cos(\dfrac{A}{8}+\dfrac{B}{8}-\dfrac{C}{8}-\dfrac{\pi}{24})\le2\sin\dfrac{\pi}{6}$Suy ra: $\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}\le3\sin\dfrac{\pi}{6}=\dfrac{3}{2}$Suy ra: $\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}\le\left(\dfrac{\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}}{3}\right)^3\le\dfrac{1}{8}$Ta có:
$\left(1+\dfrac{1}{\sin\dfrac{A}{2}}\right)\left(1+\dfrac{1}{\sin\dfrac{B}{2}}\right)\left(1+\dfrac{1}{\sin\dfrac{C}{2}}\right)$$=1+\left(\dfrac{1}{\sin\dfrac{A}{2}}+\dfrac{1}{\sin\dfrac{B}{2}}+\dfrac{1}{\sin\dfrac{C}{2}}\right)+\left(\dfrac{1}{\sin\dfrac{A}{2}\sin\dfrac{B}{2}}+\dfrac{1}{\sin\dfrac{B}{2}\sin\dfrac{C}{2}}+\dfrac{1}{\sin\dfrac{C}{2}\sin\dfrac{A}{2}}\right)+\dfrac{1}{\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}}$$\ge1+\dfrac{3}{\sqrt[3]{\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}}}+\dfrac{3}{\sqrt[3]{\sin^2\dfrac{A}{2}\sin^2\dfrac{B}{2}\sin^2\dfrac{C}{2}}}+8\ge27$Dấu bằng xảy ra khi: $\Delta ABC$ đều.
|
|
|
|
giải đáp
|
Bất đẳng thức trong tam giác.
|
|
|
|
e.
Ta có:
$\sin\dfrac{A}{2}+\sin\dfrac{B}{2}=2\sin(\dfrac{A}{4}+\dfrac{B}{4})\cos(\dfrac{A}{4}-\dfrac{B}{4})\le2\sin(\dfrac{A}{4}+\dfrac{B}{4})$
$\sin\dfrac{C}{2}+\sin\dfrac{\pi}{6}=2\sin(\dfrac{C}{4}+\dfrac{\pi}{12})\cos(\dfrac{C}{4}-\dfrac{\pi}{12})\le2\sin(\dfrac{C}{4}+\dfrac{\pi}{12})$
$\sin(\dfrac{A}{4}+\dfrac{B}{4})+\sin(\dfrac{C}{4}+\dfrac{\pi}{12})=2\sin\dfrac{\pi}{6}\cos(\dfrac{A}{8}+\dfrac{B}{8}-\dfrac{C}{8}-\dfrac{\pi}{24})\le2\sin\dfrac{\pi}{6}$
Suy ra: $\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}\le3\sin\dfrac{\pi}{6}=\dfrac{3}{2}$
Suy ra: $\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}\le\left(\dfrac{\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}}{3}\right)^3\le\dfrac{1}{8}$
Ta có:
$\left(1+\dfrac{1}{\sin\dfrac{A}{2}}\right)\left(1+\dfrac{1}{\sin\dfrac{B}{2}}\right)\left(1+\dfrac{1}{\sin\dfrac{C}{2}}\right)$
$=1+\left(\dfrac{1}{\sin\dfrac{A}{2}}+\dfrac{1}{\sin\dfrac{B}{2}}+\dfrac{1}{\sin\dfrac{C}{2}}\right)+\left(\dfrac{1}{\sin\dfrac{A}{2}\sin\dfrac{B}{2}}+\dfrac{1}{\sin\dfrac{B}{2}\sin\dfrac{C}{2}}+\dfrac{1}{\sin\dfrac{C}{2}\sin\dfrac{A}{2}}\right)+\dfrac{1}{\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}}$
$\ge1+\dfrac{3}{\sqrt[3]{\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}}}+\dfrac{3}{\sqrt[3]{\sin^2\dfrac{A}{2}\sin^2\dfrac{B}{2}\sin^2\dfrac{C}{2}}}+8\ge27$
Dấu bằng xảy ra khi: $\Delta ABC$ đều.
|
|
|
|
giải đáp
|
Bất đẳng thức trong tam giác.
|
|
|
|
d.
Ta có:
$\dfrac{\cos\dfrac{B-C}{2}}{\sin\dfrac{A}{2}}+\dfrac{\cos\dfrac{C-A}{2}}{\sin\dfrac{B}{2}}+\dfrac{\cos\dfrac{A-B}{2}}{\sin\dfrac{C}{2}}$
$=\dfrac{2\sin\dfrac{B+C}{2}\cos\dfrac{B-C}{2}}{2\sin\dfrac{A}{2}\cos\dfrac{A}{2}}+\dfrac{2\sin\dfrac{C+A}{2}\cos\dfrac{C-A}{2}}{2\sin\dfrac{B}{2}\cos\dfrac{B}{2}}+\dfrac{2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2}}{2\sin\dfrac{C}{2}\cos\dfrac{C}{2}}$
$=\dfrac{\sin B+\sin C}{\sin A}+\dfrac{\sin C+\sin A}{\sin B}+\dfrac{\sin A+\sin B}{\sin C}$
$=\dfrac{\sin B}{\sin A}+\dfrac{\sin A}{\sin B}+\dfrac{\sin C}{\sin B}+\dfrac{\sin B}{\sin C}+\dfrac{\sin C}{\sin A}+\dfrac{\sin A}{\sin C}$
$\ge2\sqrt{\dfrac{\sin B}{\sin A}.\dfrac{\sin A}{\sin B}}+2\sqrt{\dfrac{\sin C}{\sin
B}.\dfrac{\sin B}{\sin C}}+2\sqrt{\dfrac{\sin C}{\sin A}.\dfrac{\sin A}{\sin C}}=6$
Dấu bằng xảy ra khi: $\Delta ABC$ đều.
|
|
|
|
giải đáp
|
Bất đẳng thức trong tam giác.
|
|
|
|
c.
Ta có:
$\cos\dfrac{A}{2}+\cos\dfrac{B}{2}=2\cos(\dfrac{A}{4}+\dfrac{B}{4})\cos(\dfrac{A}{4}-\dfrac{B}{4})\le2\cos(\dfrac{A}{4}+\dfrac{B}{4})$
$\cos\dfrac{C}{2}+\cos\dfrac{\pi}{6}=2\cos(\dfrac{C}{4}+\dfrac{\pi}{12})\cos(\dfrac{C}{4}-\dfrac{\pi}{12})\le2\cos(\dfrac{C}{4}+\dfrac{\pi}{12})$
$\cos(\dfrac{A}{4}+\dfrac{B}{4})+\cos(\dfrac{C}{4}+\dfrac{\pi}{12})=2\cos\dfrac{\pi}{6}\cos(\dfrac{A}{8}+\dfrac{B}{8}-\dfrac{C}{8}-\dfrac{\pi}{24})\le2\cos\dfrac{\pi}{6}$
Suy ra: $\cos\dfrac{A}{2}+\cos\dfrac{B}{2}+\cos\dfrac{C}{2}\le3\cos\dfrac{\pi}{6}=\dfrac{3\sqrt3}{2}$
Ta có:
$\dfrac{\cos\dfrac{A}{2}}{1+\cos
A}+\dfrac{\cos\dfrac{B}{2}}{1+\cos
B}+\dfrac{\cos\dfrac{C}{2}}{1+\cos C}$
$=\dfrac{1}{2\cos\dfrac{A}{2}}+\dfrac{1}{2\cos\dfrac{B}{2}}+\dfrac{1}{2\cos\dfrac{C}{2}}$
$\ge\dfrac{9}{2\cos\dfrac{A}{2}+2\cos\dfrac{B}{2}+2\cos\dfrac{C}{2}}\ge\sqrt3$
Dấu bằng xảy ra khi: $\Delta ABC$ đều.
|
|
|
|
giải đáp
|
phương trình 38
|
|
|
|
Điều kiện: $x\ge1$
Ta có:
$\sqrt{x-1}\ge0$
$-x^3-4x+5=(1-x)(x^2+x+5)\le0$
Dấu bằng xảy ra khi: $x=1$
Vậy: $x=1$
|
|
|
|
giải đáp
|
phương trình 39
|
|
|
|
Điều kiện: $x\ge1$
Ta có:
$\sqrt{x-1}\ge0$
$1-2x+2x^2-x^3=(1-x)^3+x-x^2=(1-x)^3+x(1-x)\le0$
Dấu bằng xảy ra khi: $x=1$
Vậy: $x=1$
|
|
|
|
giải đáp
|
phương trình 37
|
|
|
|
Điều kiện: $\left\{ \begin{array}{l} 4x-1\ge0\\ 4x^2-1\ge0 \end{array} \right.\Leftrightarrow x\ge\frac{1}{2}$
Từ đó suy ra: $\sqrt{4x-1}+\sqrt{4x^2-1}\ge\sqrt{4.\frac{1}{2}-1}+\sqrt{4.\frac{1}{2^2}-1}=1$
Dấu bằng xảy ra khi: $x=\frac{1}{2}$
|
|
|
|
giải đáp
|
phương trình 40
|
|
|
|
Điều kiện: $x\ge1$.
Ta có:
$\sqrt{x-1}+\sqrt{x+2}=3$
$\Leftrightarrow 2x+1+2\sqrt{(x-1)(x+2)}=9$
$\Leftrightarrow \sqrt{x^2+x-2}=4-x$
$\Leftrightarrow \left\{\begin{array}{l}4-x\ge0\\x^2+x-2=x^2-8x+16\end{array}\right.$
$\Leftrightarrow x=2$, thỏa mãn.
|
|
|
|
|
|
|
|