giúp em vs

tìm liên hệ giữa số a và b biết : \left| {a+b} \right| > \left| {a-b} \right|

Bất đẳng thức

giúp em vs

tìm liên hệ giữa số a và b biết : $\left| {a+b} \right| > \left| {a-b} \right|$

Bất đẳng thức

Đặt $t=\cos x \Rightarrow -1\le t\le 1$.Khi đó: $y=f(t)=t^2-2t+3$Ta có:$f'(t)=2t-2$$f'(t)=0 \Leftrightarrow t=1$Lập bảng biến thiên hàm $f(t)$ trong $[-1;1]$ ta được:$\min y=2 \Leftrightarrow t=1 \Leftrightarrow x=k2\pi, k\in\mathbb{Z}$$\max y=2 \Leftrightarrow t=-1 \Leftrightarrow x=\pi+k2\pi, k\in\mathbb{Z}$

Đặt $t=\cos x \Rightarrow -1\le t\le 1$.Khi đó: $y=f(t)=t^2-2t+3$Ta có:$f'(t)=2t-2$$f'(t)=0 \Leftrightarrow t=1$Lập bảng biến thiên hàm $f(t)$ trong $[-1;1]$ ta được:$\min y=2 \Leftrightarrow t=1 \Leftrightarrow x=k2\pi, k\in\mathbb{Z}$$\max y=6 \Leftrightarrow t=-1 \Leftrightarrow x=\pi+k2\pi, k\in\mathbb{Z}$

Áp dụng bổ đề: $\dfrac{1}{(n+1)\sqrt n+n\sqrt{n+1}}=\dfrac{1}{\sqrt n}+\dfrac{1}{\sqrt{n+1}}$, ta được:$S=\dfrac{1}{\sqrt 1}-\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt2}-\dfrac{1}{\sqrt3}+\dfrac{1}{\sqrt3}-\dfrac{1}{\sqrt4}+\ldots+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{100}}=\dfrac{9}{10}$

Áp dụng bổ đề: $\dfrac{1}{(n+1)\sqrt n+n\sqrt{n+1}}=\dfrac{1}{\sqrt n}-\dfrac{1}{\sqrt{n+1}}$, ta được:$S=\dfrac{1}{\sqrt 1}-\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt2}-\dfrac{1}{\sqrt3}+\dfrac{1}{\sqrt3}-\dfrac{1}{\sqrt4}+\ldots+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{100}}=\dfrac{9}{10}$

giúp mình với

cho x,y là các số dương .Tìm min của biểu thức $B=\frac{x^{2}+3xy+y^{2}}{\sqrt{xy(x+y)}}$

GTLN, GTNN

giúp mình với

cho x,y là các số dương .Tìm min của biểu thức $B=\frac{x^{2}+3xy+y^{2}}{\sqrt{xy}(x+y)}$

GTLN, GTNN

1. Số số tự nhiên thỏa mãn là: $\dfrac{6!}{3!}.C_7^4=4200$ (số).

1. Số số tự nhiên thỏa mãn là: $\dfrac{6!}{1!2!3!}.C_7^4=2100$ (số).

2.Ta có: $2000y^2\equiv 0\;($ mod $4)$$\Rightarrow 1999^2\equiv 1\;($ mod $4)$mà $1999\equiv 3\;($ mod $4) \Rightarrow x^2\equiv 3\;($ mod $4)$, vô lý.Vậy phương trình không có nghiệm nguyên.

2.Ta có: $2000y^2\equiv 0\;($ mod $4)$$\Rightarrow 1999x^2\equiv 1\;($mod$4)$mà$1999\equiv 3\;($mod$4) \Rightarrow x^2\equiv 3\;($mod$4)$, vô lý.Vậy phương trình không có nghiệm nguyên.

lg

\cos 2x-\sqrt{3}\sin 2x-\sqrt{3}\sin x-\cos x+4=0

Phương trình lượng giác cơ bản

lg

$\cos 2x-\sqrt{3}\sin 2x-\sqrt{3}\sin x-\cos x+4=0$

Phương trình lượng giác cơ bản

Ta có $\left\{\begin{array}{l}x^4-y^4=544\\6x^3+27x^2+10y^3-75y^2=-54x-250y\end{array}\right.$$\Leftrightarrow \left\{\begin{array}{l}x^4+81=y^4+625\\12x^3+54x^2+54x=-20y^3+150y^2-500y\end{array}\right. $$\Leftrightarrow \left\{\begin{array}{l}x^4-y^4=544\\(x+3)^4=(y-5)^4\end{array}\right.$$ \Leftrightarrow \left[ $$\begin{array}{l}\left\{\begin{array}{l}x^4-y^4=544\\x+3=y-5\end{array}$$ \right.\\\left\{ $$\begin{array}{l}x^4-y^4=544\\x+3=5-y\end{array}$$ \right.\end{array}\right.$$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=-5\\y=3\end{array}\right.\\\left\{\begin{array}{l}x=5\\y=-3\end{array}\right.\end{array}\right.$

Ta có $\left\{\begin{array}{l}x^4-y^4=544\\6x^3+27x^2+10y^3-75y^2=-54x-250y\end{array}\right.$$\Leftrightarrow \left\{\begin{array}{l}x^4+81=y^4+625\\12x^3+54x^2+108x=-20y^3+150y^2-500y\end{array}\right. $$\Leftrightarrow \left\{\begin{array}{l}x^4-y^4=544\\(x+3)^4=(y-5)^4\end{array}\right.$$ \Leftrightarrow \left[ $$\begin{array}{l}\left\{\begin{array}{l}x^4-y^4=544\\x+3=y-5\end{array}$$ \right.\\\left\{ $$\begin{array}{l}x^4-y^4=544\\x+3=5-y\end{array}$$ \right.\end{array}\right.$$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=-5\\y=3\end{array}\right.\\\left\{\begin{array}{l}x=5\\y=-3\end{array}\right.\end{array}\right.$

giới hạn

lim (\sqrt{2n+3} -\sqrt{n-1}lim (\sqrt{n+1} +\sqrt{n})lim (\sqrt{n^{2}+2n+2} -n)

Giới hạn của dãy số

giới hạn

$\lim (\sqrt{2n+3} -\sqrt{n-1})$$\lim (\sqrt{n+1} +\sqrt{n})$$\lim (\sqrt{n^{2}+2n+2} -n)$

Giới hạn của dãy số

Đặt: $t=\sqrt{1+3\cos x} \Rightarrow t^2=1+3\cos x\Rightarrow 2tdt=-3\sin xdx$Ta có: $\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin2x+\sin x}{\sqrt{1+3\cos x}}dx$$=\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin x(2\cos x+1)}{\sqrt{1+3\cos x}}dx$$=-\dfrac{2}{3}\int\limits_2^1\dfrac{2\dfrac{t^2-1}{3}+1}{t}tdt$$=\dfrac{2}{9}\int\limits_1^2(2t^2+1)dt$$=\dfrac{2}{9}\left(\dfrac{2t^3}{3}+t\right)\left|\begin{array}{l}2\\1\end{array}\right.$$=\dfrac{44}{45}$

Đặt: $t=\sqrt{1+3\cos x} \Rightarrow t^2=1+3\cos x\Rightarrow 2tdt=-3\sin xdx$Ta có: $\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin2x+\sin x}{\sqrt{1+3\cos x}}dx$$=\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin x(2\cos x+1)}{\sqrt{1+3\cos x}}dx$$=-\dfrac{2}{3}\int\limits_2^1\dfrac{2\dfrac{t^2-1}{3}+1}{t}tdt$$=\dfrac{2}{9}\int\limits_1^2(2t^2+1)dt$$=\dfrac{2}{9}\left(\dfrac{2t^3}{3}+t\right)\left|\begin{array}{l}2\\1\end{array}\right.$$=\dfrac{34}{27}$

Đặt: $t=1+\cos x \Rightarrow dt=-\sin xdx$Ta có: $\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin2x\cos x}{1+\cos x}dx$$=2\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin x\cos^2x}{1+\cos x}dx$$=-2\int\limits_2^1\dfrac{(t-1)^2}{t}dt$$=2\int\limits_1^2\dfrac{(t-1)^2}{t}dt$$=2\int\limits_1^2\left(t-2+\dfrac{1}{t}\right)dt$$=2\left(\dfrac{t^2}{2}-2t+\ln t\right)\left|\begin{array}{l}2\\1\end{array}\right.$$=2\ln2-3$

Đặt: $t=1+\cos x \Rightarrow dt=-\sin xdx$Ta có: $\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin2x\cos x}{1+\cos x}dx$$=2\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin x\cos^2x}{1+\cos x}dx$$=-2\int\limits_2^1\dfrac{(t-1)^2}{t}dt$$=2\int\limits_1^2\dfrac{(t-1)^2}{t}dt$$=2\int\limits_1^2\left(t-2+\dfrac{1}{t}\right)dt$$=2\left(\dfrac{t^2}{2}-2t+\ln t\right)\left|\begin{array}{l}2\\1\end{array}\right.$$=2\ln2-1$

Ta có: $\int\limits_1^2\dfrac{\ln x}{x^3}dx$$=-\dfrac{1}{2}\int\limits_1^2\ln xd\left(\dfrac{1}{x^2}\right)$$=-\dfrac{\ln x}{2x^2}\left|\begin{array}{l}2\\1\end{array}\right.-\dfrac{1}{2}\int\limits_1^2\dfrac{1}{x^2}d(\ln x)$$=-\dfrac{\ln2}{8}-\dfrac{1}{2}\int\limits_1^2\dfrac{1}{x^3}dx$$=-\dfrac{\ln2}{8}+\dfrac{1}{4x^2}\left| $$\begin{array}{l}2\\1\end{array}$$ \right.$$=-\dfrac{\ln2}{8}-\dfrac{3}{16}$

Ta có: $\int\limits_1^2\dfrac{\ln x}{x^3}dx$$=-\dfrac{1}{2}\int\limits_1^2\ln xd\left(\dfrac{1}{x^2}\right)$$=-\dfrac{\ln x}{2x^2}\left|\begin{array}{l}2\\1\end{array}\right.+\dfrac{1}{2}\int\limits_1^2\dfrac{1}{x^2}d(\ln x)$$=-\dfrac{\ln2}{8}+\dfrac{1}{2}\int\limits_1^2\dfrac{1}{x^3}dx$$=-\dfrac{\ln2}{8}-\dfrac{1}{4x^2}\left| $$\begin{array}{l}2\\1\end{array}$$ \right.$$=-\dfrac{\ln2}{8}+\dfrac{3}{16}$

Ta có: $\int\limits_0^{\frac{\pi}{4}}\dfrac{1-2\sin^2x}{1+\sin2x}dx$$=\int\limits_0^{\frac{\pi}{4}}\dfrac{\cos2x}{1+\sin2x}dx$$=\dfrac{1}{2}\int\limits_0^{\frac{\pi}{4}}\dfrac{d(1+\sin2x)}{1+\sin2x}dx$$=\dfrac{\ln(1+\sin2x)}{2}\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.=\dfrac{\ln2}{2}$

Ta có: $\int\limits_0^{\frac{\pi}{4}}\dfrac{1-2\sin^2x}{1+\sin2x}dx$$=\int\limits_0^{\frac{\pi}{4}}\dfrac{\cos2x}{1+\sin2x}dx$$=\dfrac{1}{2}\int\limits_0^{\frac{\pi}{4}}\dfrac{d(1+\sin2x)}{1+\sin2x}$$=\dfrac{\ln(1+\sin2x)}{2}\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.=\dfrac{\ln2}{2}$

Đặt: $t=1+\dfrac{1}{x} \Rightarrow dt=-\dfrac{1}{x^2}dx$Đổi cận: $x=\dfrac{1}{2} \Rightarrow t=3$ $x=1 \Rightarrow t=2$Ta có: $\int\limits_0^2|x^2-x|dx$$=\int\limits_0^1(x-x^2)dx+\int\limits_1^2(x^2-x)dx$$=\left(\dfrac{x^2}{2}-\dfrac{x^3}{3}\right)\left|\begin{array}{l}1\\0\end{array}\right.+\left(\dfrac{x^3}{3}-\dfrac{x^2}{2}\right)\left|\begin{array}{l}2\\1\end{array}\right.=1$

Ta có: $\int\limits_0^2|x^2-x|dx$$=\int\limits_0^1(x-x^2)dx+\int\limits_1^2(x^2-x)dx$$=\left(\dfrac{x^2}{2}-\dfrac{x^3}{3}\right)\left|\begin{array}{l}1\\0\end{array}\right.+\left(\dfrac{x^3}{3}-\dfrac{x^2}{2}\right)\left|\begin{array}{l}2\\1\end{array}\right.=1$

Đặt: $t=1+\dfrac{1}{x} \Rightarrow dt=-\dfrac{1}{x^2}dx$Đổi cận: $x=\dfrac{1}{2} \Rightarrow t=3$ $x=1 \Rightarrow t=2$Ta có: $\int\limits_{\frac{1}{2}}\dfrac{1}{x^2}\left(1+\dfrac{1}{x}\right)^{2007}dx$$=-\int\limits_3^2t^{2007}dt$$=\dfrac{t^{2008}}{2008}\left|\begin{array}{l}3\\2\end{array}\right.=\dfrac{3^{2008}-2^{2008}}{2008}$

Đặt: $t=1+\dfrac{1}{x} \Rightarrow dt=-\dfrac{1}{x^2}dx$Đổi cận: $x=\dfrac{1}{2} \Rightarrow t=3$ $x=1 \Rightarrow t=2$Ta có: $\int\limits_{\frac{1}{2}}^1\dfrac{1}{x^2}\left(1+\dfrac{1}{x}\right)^{2007}dx $$=-\int\limits_3^2t^{2007}dt$$ =\dfrac{t^{2008}}{2008}\left| $$\begin{array}{l}3\\2\end{array}$$ \right.=\dfrac{3^{2008}-2^{2008}}{2008}$