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\(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy.\left(x^2-y^2-2y-1\right)\)
\(=2xy.[x^2-\left(y^2+2y+1\right)]\)
\(=2xy.[x^2-\left(y+1\right)^2]\)
\(=2xy.\left(x+y+1\right).\left(x-y-1\right)\)
Vậy chọn đáp án A
a) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)
\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)
\(=\left(-6x-18\right)\left(8x^2-18\right)\)
\(=-6\left(x+3\right)\cdot2\left(4x^2-9\right)\)
\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)
b) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=-\left(x+3y+5\right)\left(7x+9y-1\right)\)
c) Ta có: \(-4x^2+12xy-9y^2+25\)
\(=-\left(4x^2-12xy+9y^2-25\right)\)
\(=-\left[\left(2x-3y\right)^2-25\right]\)
\(=-\left(2x-3y-5\right)\left(2x-3y+5\right)\)
d) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)
\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)
\(=\left(x-y\right)^2-\left(2m-n\right)^2\)
\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)
1 \(=\left(4x^2+4x+1\right)-\left(3y\right)^2\)
\(=\left(2x+1\right)^2-\left(3y\right)^2\)
\(=\left(2x+1-3y\right)\left(2x+1+3y\right)\)
2,\(=\left(x^2+2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)
\(=\left(x+y\right)^2-\left(z-t\right)^2\)
\(=\left(x+y+z-t\right)\left(x+y-z+t\right)\)
3,\(=9x\left(x-y\right)-7\left(x-y\right)\)
\(=\left(x-y\right)\left(9x-7\right)\)
4\(=3\left(x-y\right)+a\left(x-y\right)\)
\(=\left(x-y\right)\left(3+a\right)\)
Ta có: \(4x^2-9y^2\\ =\left(2x\right)^2-\left(3y\right)^2\\ =\left(2x-3y\right)\left(2x+3y\right)\)
Vậy: Chọn D
1. \(4x^2-2x-3y-9y^2\)
\(=\left(2x\right)^2-\left(3y\right)^2-\left(2x+3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y\right)-\left(2x+3y\right)\)
\(=\left(2x+3y\right)\left(2x-3y-1\right)\)
2. \(x^2-25=6x-9\)
\(\Rightarrow x^2-6x+9=25\)
\(\Rightarrow\left(x-3\right)^2=25\)
\(\Rightarrow\orbr{\begin{cases}x-3=5\\x-3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-2\end{cases}}\)
\(4x^2-9y^2+2x-3y=\left[\left(2x\right)^2-\left(3y\right)^2\right]+\left(2x-3y\right)=\left(2x-3y\right)\left(2x+3y\right)+\left(2x-3y\right)=\left(2x-3y\right)\left(2x+3y+1\right)\)