..... khảo thí ???

a)ABE = 180 độ - 35 độ = 145 độ

b) Vì DBC + BCy = 180 độ 

=>Cy // DE

mà DE // Ax 

=>Ax//Cy

Do BAx so le trong vs ABD

=>Bax = ABD = 35 độ 

ABD và DBC kề bù 

=> ABD + DBC = 35+55=90 độ 

=>AB vuông góc vs BC 

tick đi nhé

1.

\(cos\left(\dfrac{2\pi}{3}+2x\right)+cos\left(\dfrac{\pi}{3}+x\right)+1=0\)

\(\Leftrightarrow2cos^2\left(\dfrac{\pi}{3}+x\right)+cos\left(\dfrac{\pi}{3}+x\right)=0\)

\(\Leftrightarrow cos\left(\dfrac{\pi}{3}+x\right)\left[2cos\left(\dfrac{\pi}{3}+x\right)+1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\dfrac{\pi}{3}+x\right)=0\\cos\left(\dfrac{\pi}{3}+x\right)=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{3}+x=\dfrac{\pi}{2}+k\pi\\\dfrac{\pi}{3}+x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

40: Ta có: \(A=27x^3+8y^3-3x-2y\)

\(=\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)-\left(3x+2y\right)\)

\(=\left(3x+2y\right)\left(9x^2-6xy+4y^2-1\right)\)

\(\lim\limits_{x\rightarrow5}\left(x^3+5x^2-10x+8\right)=5^3+5.5^2-10.5+8=...\)

\(\lim\limits_{x\rightarrow-2}\dfrac{x^3-x^2-2x-8}{x^2+3x+2}=\dfrac{-16}{0}=-\infty\)

\(\lim\limits_{x\rightarrow-\infty}\dfrac{x^2-5x+2}{2\left|x\right|+1}=\lim\dfrac{\left|x\right|-5+\dfrac{2}{\left|x\right|}}{2+\dfrac{1}{\left|x\right|}}=\dfrac{+\infty}{2}=+\infty\)

\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt[3]{x^3+4x-3}-4x}{\sqrt{9x^2-5x+1}-4x}=\lim\limits_{x\rightarrow+\infty}\dfrac{x\left(\sqrt[3]{1+\dfrac{4}{x^2}-\dfrac{3}{x^3}}-4\right)}{x\left(\sqrt[]{9-\dfrac{5}{x}+\dfrac{1}{x^2}}-4\right)}=\dfrac{1-4}{3-4}=3\)

Lời giải:

a.

\(\lim\limits_{x\to 5}(x^3+5x^2-10x+8)=5^3+5.5^2-10.5+8=208\)

b. 

\(L=\lim\limits_{x\to -2}\frac{x^3-x^2-2x-8}{x^2+3x+2}\lim\limits_{x\to -2}\frac{x^3-x^2-2x-8}{x+1}.\frac{1}{x+2}=16\lim\limits_{x\to -2}\frac{1}{x+2}\)\(\lim\limits_{x\to -2-}\frac{1}{x+2}=-\infty \Rightarrow L=-\infty ; \lim\limits_{x\to -2+}\frac{1}{x+2}=+\infty \Rightarrow L=+\infty \)

c.

\(\lim\limits_{x\to -\infty}\frac{x^2-5x+2}{2|x|+1}=\lim\limits_{x\to -\infty}\frac{|x|-\frac{5x}{|x|}+\frac{2}{|x|}}{2+\frac{1}{|x|}}\)

\(=\lim\limits_{x\to -\infty}\frac{|x|-\frac{5x}{-x}}{2}=\frac{1}{2}\lim\limits_{x\to -\infty}(|x|+5)=+\infty \)

a.

Đặt \(sinx+cosx=t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Rightarrow sinx.cosx=\dfrac{t^2-1}{2}\)

Phương trình trở thành:

\(2t+t^2-1+1=0\)

\(\Rightarrow t\left(t+2\right)=0\Rightarrow\left[{}\begin{matrix}t=0\\t=-2< -\sqrt{2}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow sinx+cosx=0\)

\(\Rightarrow tanx=-1\)

\(\Rightarrow x=-\dfrac{\pi}{4}+k\pi\)

a, Đặt \(sinx+cosx=t\left(t\in\left[-\sqrt{2};\sqrt{2}\right]\right)\)

\(pt\Leftrightarrow2t+t^2-1+1=0\)

\(\Leftrightarrow t^2+2t=0\)

\(\Leftrightarrow t\left(t+2\right)=0\)

\(\Leftrightarrow t=0\)

\(\Leftrightarrow sinx+cosx=0\)

\(\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\)

b.

Đặt \(sinx-cosx=t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Rightarrow sinx.cosx=\dfrac{1-t^2}{2}\)

Phương trình trở thành:

\(2\sqrt{2}t-2\left(1-t^2\right)=1\)

\(\Leftrightarrow2t^2+2\sqrt{2}t-3=0\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{\sqrt{2}}{2}\\t=-\dfrac{3\sqrt{2}}{2}< -\sqrt{2}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{5}+k2\pi\end{matrix}\right.\)

\(\Rightarrow...\)