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(2x + y)2 - (2x - 3)2 = (2x + y - 2x + 3)(2x + y + 2x - 3) = (y + 3)(4x + y - 3)
\(\left(2x+y\right)^2-4x^2+12x-9\)
\(=\left(2x+y\right)^2-\left(2x-3\right)^2\)
\(=\left(2x+y-2x+3\right)\left(2x+y+2x-3\right)\)
\(=\left(y+3\right)\left(4x+y-3\right)\)
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a) Sửa đề: \(x^2+3x+1\rightarrow x^2+2x+1\)
\(x^2+2x+1=\left(x+1\right)^2\)
b) \(x^2+y^2+2xy=\left(x+y\right)^2\)
c) \(9x^2+12x+4=\left(3x+2\right)^2\)
d) \(-4x^2-9-12x=-\left(4x^2+12x+9\right)=-\left(2x+3\right)^2\)
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a) Ta có P = ( 4 x 2 − 1 ) ( 2 x + 1 ) − ( 2 x − 1 ) − ( 4 x 2 − 1 ) ( 2 x + 1 ) ( 2 x − 1 ) = 3 − 4 x 2
b) Ta có Q = 3 x ( x + 3 ) . ( x + 3 ) ( x − 3 ) − x = 9 − 3 x x + 3
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\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(=\left(2x-3\right)^2\)
\(\left(2x-3\right)^2\)