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\(9\left(x-1\right)^2-16\left(x-2\right)^2\)
\(=\left(3x-3\right)^2-\left(4x-8\right)^2\)
\(=\left(3x-3-4x+8\right)\left(3x-3+4x-8\right)\)
\(=\left(-x+5\right)\left(7x-11\right)\)
\(=\left(x-y\right)\left(x-3\right)\left(x+3\right)\)
\(=x^2\left(x-y\right)-9\left(x-y\right)=\left(x-y\right)\left(x^2-9\right)=\left(x-y\right)\left(x-3\right)\left(x+3\right)\)
\(\left(x+2\right)^2-9\)
\(=\left(x+2\right)^2-3^2\)
\(=\left[\left(x+2\right)-3\right]\left[\left(x+2\right)+3\right]\)
\(=\left(x+2-3\right)\left(x+2+3\right)\)
\(=\left(x-1\right)\left(x+5\right)\)
phân tích đa thức sau thành nhân tử ( a + b )2 - ( a - 2b )2
\(\left(x^2+1\right)^2-6\left(x^2+1\right)+9\)
\(=\left[\left(x^2+1\right)-3\right]^2\)
\(=\left(x^2+1-3\right)^2\)
\(=\left(x^2-2\right)^2\)
\(=\left(3x-3y\right)^2-\left(2x+2y\right)^2=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\)
\(=\left(x-5y\right)\left(5x-y\right)\)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
= - (x2 - 2xy + y2 - 9) = - [ (x - y)2 - 32 ] = - (x - y - 3).(x - y + 3)
\(=\left(x-3\right)\left(x+3\right)+\left(x-3\right)\left(2x-5\right)\\ =\left(x-3\right)\left(x+3+2x-5\right)\\ =\left(x-3\right)\left(3x-2\right)\)
\(=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
\(=x^2-3^2=(x-3).(x+3)\)