Ta có: 

\(\left(x-y\right)^2\ge0\)

\(\Leftrightarrow x^2+y^2\ge2xy\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge x^2+y^2+2xy\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2=1\)

\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)

Dấu \(=\)khi \(x=y=\frac{1}{2}\)

\(x^2-7x+2\left(x-7\right)=0\)

\(x\left(x-7\right)+2\left(x-7\right)=0\)

\(\left(x+2\right)\left(x-7\right)=0\)

\(x+2=0\)hoặc \(x-7=0\)

\(x=-2\)hoặc \(x=7\)

Vậy x=-2 hoặc x=7

Bài làm

x² - 7x + 2( x - 7 ) = 0

<=> ( x² - 7x ) + 2( x - 7 ) =0

<=> x( x - 7 ) + 2( x - 7 ) = 0

<=> ( x - 7 )( x + 2 ) = 0

<=> \(\orbr{\begin{cases}x-7=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-2\end{cases}}\)

Vậy x = 7 hoặc x = -2

Tôi yêu mọi người!

I love everyone!

Bài làm

\(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\)( có sửa )

ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne\pm2\\x\ne4\end{cases}}\)

\(=\frac{x^2}{x\left(x^2-4\right)}-\frac{6}{3x-6}+\frac{1}{x+2}\)

\(=\frac{x}{x^2-4}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\)

\(=\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2}{x-2}+\frac{1}{x+2}\)

\(=\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2x+4}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\)

Nếu mình sửa đề sai thì ib mình làm lại cho