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\(M=\left(-4x^2+2xy-\frac{y^2}{4}\right)-x^2+2x-1-\frac{3}{4}y^2+2y-2\)
\(M=-\left(4x^2-2\cdot2x\cdot\frac{y}{2}+\frac{y^2}{4}\right)-\left(x-1\right)^2-3\left(\frac{y^2}{4}-\frac{2y}{3}\right)-2\)
\(M=-\left(2x-\frac{y}{2}\right)^2-\left(x-1\right)^2-3\left(\frac{y^2}{4}-2\cdot\frac{y}{2}\cdot\frac{2}{3}+\frac{4}{9}-\frac{4}{9}\right)-2\)
\(M=-\left(2x-\frac{y}{2}\right)^2-\left(x-1\right)^2-3\left(\frac{y}{2}-\frac{2}{3}\right)^2+\frac{4}{3}-2\)
\(M\subseteq\frac{4}{3}-2=-\frac{2}{3}\)
Dấu = xr khi/.......
\(5x\left(x-2\right)-5x^2=30\)
\(\Leftrightarrow5x^2-10-5x^2=30\)
\(\Leftrightarrow-10x=30\)
\(\Leftrightarrow x=-3\)
a)\(x\left(x-3\right)-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
b)\(\left(3x-5\right)\left(5x-7\right)+\left(5x+1\right)\left(2-3x\right)=4\)
\(\Leftrightarrow15x^2-46x+35-15x^2+7x+2-4=0\)
\(\Leftrightarrow33-39x=0\Leftrightarrow33=39x\Leftrightarrow x=\frac{33}{39}\)
a) \(x\left(x-3\right)-2x+6=0\)
\(x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)
b) \((3x-5)(5x-7)+(5x+1)(2-3x)=4\)
\(15x^2-46x+35+10x-15x^2+2-3x-4=0\)
\(33-39x=0\)
\(3\left(11-13x\right)=0\)
\(11-13x=0\)
\(13x=11\)
\(x=\frac{11}{13}\)
\(x^2+4y^2-5x+10y-4xy+20\)
\(=x^2-4xy+4y^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}-\frac{25}{4}+20\)
\(=\left(x-2y\right)^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}+\frac{55}{4}\)
\(=\left(x-2y-\frac{5}{2}\right)^2+\frac{55}{4}\)Thay x - 2y = 5 ta được :
\(=\left(5-\frac{5}{2}\right)^2+\frac{55}{4}=20\)
\(B=x^2-2xy-2x+2y+y^2\)
\(=x^2-2xy+y^2-2\left(x-y\right)\)
\(=\left(x-y\right)^2-2\left(x-1\right)\)Thay x = y + 1 => x - y = 1 ta được :
\(=1-2=-1\)