\(\left(-\dfrac{x}{5}+2y^2\right)^2=\dfrac{x^2}{25}-\dfrac{4xy^2}{5}+4y^4\)

\(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

Có: \(\sin B=\dfrac{AC}{BC}=0,6\)

           \(\dfrac{AC}{9}=0,6\)

    => AC=5,4 cm

Xét \(\Delta ABC\) vuông tại A có:

\(AB^2+AC^2=BC^2\)

\(AB=\sqrt{9^2-5,4^2}=7,2cm\)

\(ĐKXĐ:x\ge0;x\ne1\)

\(A=\left(\sqrt{x}-\dfrac{x+2}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-4}{1-x}\right)\)

\(A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\dfrac{x+2}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

\(A=\dfrac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\dfrac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(A=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

Ta có: \(aaa=100a+10a+a\)

                  \(=a\left(100+10+1\right)\)

                   \(=111a\)

mà \(111\) chi hết cho 37 nên 111a chia hết cho 37 hay aaa chia hết cho 37

2

\(\left(x-1\right)^3+15=55\)

\(\left(x-1\right)^3=40\)

\(x-1=\sqrt[3]{40}\)

\(x=\sqrt[3]{40}+1\)

a)ĐKXĐ: \(\begin{matrix}x+3\ne0\\x-1\ne0\end{matrix}\)\(\left\{{}\begin{matrix}x\ne-3\\x\ne1\end{matrix}\right.\)

b)\(P=\dfrac{3x^2+5x-4}{\left(x+3\right)\left(x-1\right)}-\dfrac{x+1}{x+3}-\dfrac{x+3}{x-1}\)

\(P=\dfrac{3x^2+5x-4}{\left(x+3\right)\left(x-1\right)}-\dfrac{x^2-1}{\left(x+3\right)\left(x-1\right)}-\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-1\right)}\)

\(P=\dfrac{3x^2+5x-4-x^2+1-x^2-6x-9}{\left(x+3\right)\left(x-1\right)}\)

\(P=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-1\right)}\)

\(P=\dfrac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-1\right)}=\dfrac{x-4}{x-1}\)

Tại \(P=-\dfrac{1}{2}\) giá trị của x là:

\(\dfrac{x-4}{x-1}=-\dfrac{1}{2}\)

\(2x-8=1-x\)

\(3x=9\)

x=3

Xét hình thang ABCD có :

AB//CD 

=> \(\widehat{B}+\widehat{C}=180^o\)

mà \(\widehat{B}=\widehat{C}\)

=> \(2\widehat{B}=180^o\)

\(\widehat{B}=90^o\)

=> \(\widehat{C}=90^o\)

=> ABCD là hình thang vuông

\(\sqrt{\left(3-\sqrt{7}\right)^2}=\left|3-\sqrt{7}\right|=3-\sqrt{7}\)