\(\hept{\begin{cases}cosa+sina=\frac{1}{5}\\sin^2a+cos^2a=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}cosa=\frac{1}{5}-sina\\sin^2a+\left(\frac{1}{5}-sina\right)^2=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}cosa=\frac{1}{5}-sina\\50sin^2a-10sina-24=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}sina=-\frac{3}{5}\left(l\right)\\sina=\frac{4}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}cosa=\frac{4}{5}\\cosa=-\frac{3}{5}\end{cases}}\)

Vì \(a< 90^o\) nên \(\hept{\begin{cases}sina\ge0\\cosa>0\end{cases}}\)vậy không tìm được góc thỏa bài toán.

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Ccd

a

a, Ta có : \(x=81\Rightarrow\sqrt{x}=9\)

Thay \(\sqrt{x}=9\)vào biểu thức A ta được : 

\(A=\frac{2}{9+1}=\frac{2}{10}=\frac{1}{5}\)

b, Ta có : \(P=\frac{B}{A}\)hay\(P=\frac{\frac{1}{x+\sqrt{x}}+\frac{1}{\sqrt{x}+1}}{\frac{2}{\sqrt{x}+1}}\)

\(=\frac{1+\sqrt{x}}{x+\sqrt{x}}.\frac{\sqrt{x}+1}{2}=\frac{\sqrt{x}+1}{2\sqrt{x}}\)

c, Ta có \(\frac{1}{2}=\frac{\sqrt{x}}{2\sqrt{x}}\)mà \(\sqrt{x}< \sqrt{x}+1\)

nên \(P>\frac{1}{2}\)

a) \(A=\frac{2}{\sqrt{x}+1}=\frac{2}{\sqrt{81}+1}=\frac{2}{9+1}=\frac{1}{5}\)

b) \(B=\frac{1}{x+\sqrt{x}}+\frac{1}{\sqrt{x}+1}\)

\(=\frac{1+\sqrt{x}}{\left(1+\sqrt{x}\right)\sqrt{x}}=\frac{1}{\sqrt{x}}\)

\(\Rightarrow P=\frac{B}{A}=\frac{1}{\sqrt{x}}\div\frac{2}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{2\sqrt{x}}\)

c) Ta có: \(P=\frac{\sqrt{x}+1}{2\sqrt{x}}=\frac{1}{2}+\frac{1}{\sqrt{x}}+\frac{1}{2}+0=\frac{1}{2}\)

=> P>1/2

a) \(A=4\sqrt{x^2+1}-2\sqrt{16\left(x^2+1\right)}+5\sqrt{25\left(x^2+1\right).}\)

\(=4\sqrt{x^2+1}-2.4\sqrt{x^2+1}+5.5\sqrt{x^2+1}\)

\(=4\sqrt{x^2+1}-8\sqrt{x^2+1}+25\sqrt{x^2+1}\)

\(=\left(4-8+25\right)\sqrt{x^2+1}\)

\(=21\sqrt{x^2+1}\)

b) \(B=\frac{2}{x+y}\sqrt{\frac{3\left(x+y\right)^2}{4}}\)

\(B=\frac{2}{x+y}.\frac{\sqrt{3}\left(x+y\right)}{2}\)

\(B=\frac{\sqrt{3}\left(x+y\right)}{x+y}\)

\(B=\sqrt{3}\)

Trả lời :

Chữ đầu là x, chữ cuối là W

~HT~

X & W

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Bài 1:

\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)

\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)

\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)

\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)

Bài 2:

\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)

\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)

\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)

a.

Đặt \(\sqrt{1-x^2}=u\Rightarrow x^2=1-u^2\Rightarrow xdx=-udu\)

\(\left\{{}\begin{matrix}x=0\Rightarrow u=1\\x=1\Rightarrow u=0\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^0_1\left(1-u^2\right).u.\left(-udu\right)=\int\limits^1_0\left(u^2-u^4\right)du=\left(\dfrac{1}{3}u^3-\dfrac{1}{5}u^5\right)|^1_0\)

\(=\dfrac{2}{15}\)

b.

\(\int\limits^2_1\dfrac{dx}{x^2-2x+2}=\int\limits^2_1\dfrac{dx}{\left(x-1\right)^2+1}\)

Đặt \(x-1=tanu\Rightarrow dx=\dfrac{1}{cos^2u}du\)

\(\left\{{}\begin{matrix}x=1\Rightarrow u=0\\x=2\Rightarrow u=\dfrac{\pi}{4}\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{1}{tan^2u+1}.\dfrac{1}{cos^2u}du=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{cos^2u}{cos^2u}du=\int\limits^{\dfrac{\pi}{4}}_0du\)

\(=u|^{\dfrac{\pi}{4}}_0=\dfrac{\pi}{4}\)