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