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\(2x.\left(9-x\right)+\left(2x+5\right).\left(x+1\right)\)
\(=2x.9+2x.\left(-x\right)+2x.x+2x.1+5x+5.1\)
\(=18x-2x^2+2x^2+2x+5x+5\)
\(=25x+5\)
\(\left(x-3\right)^2+\left(x+3\right)^2+2.\left(3-x\right).\left(3+x\right)\)
\(=\left(x-3\right)^2-2.\left(x-3\right).\left(x+3\right)+\left(x+3\right)^2\)
\(=[\left(x-3\right)-\left(x+3\right)]^2\)
\(=\left(x-3-x-3\right)^2\)
\(=36\)
\(\left(2-n\right)\left(n^2-3n+1\right)+n\left(n^2+12\right)+8\)
\(=2n^2-6n+2-n^3+3n^2-n+n^3+12n+8\)
\(=5\left(n^2+n+2\right)⋮5\)
Gọi đa thức đó là x2 + cx + d
Ta có : P = (x2 + cx + d)2 = x4 + c2x2 + d2 + 2cx3 + 2dx2 + 2cdx
=> x4 + ax3 + bx2 - 8x + 1 = x4 + 2cx3 + (c2 + 2d)x2 + 2cdx + d2
=> a = 2c ; b = c2 + 2d ; 2cd = -8 ; d2 = 1
Vì d2 = 1 => \(\orbr{\begin{cases}d=1\\d=-1\end{cases}}\)
Khi d = 1 => c = -4 ; b = 18 ; a = -8
Khi d = -1 => c = 4 ; b = 14 ; a = 8
Vậy các cặp (a;b) thỏa mãn là (-8 ; 18) ; (8 ; 14)
a) \(\left(x^2+2xy+y^2\right):\left(x+y\right)\)
\(=\left(x+y\right)^2:\left(x+y\right)\)
\(=x+y\)
b) \(\left(125x^3+1\right):\left(5x+1\right)\)
\(=\left(5x+1\right)\left(25x^2-5x+1\right):\left(5x+1\right)\)
\(=25x^2-5x+1\)
c) \(\left(x^2-2xy+y^2\right):\left(y-x\right)\)
\(=\left(x-y\right)^2:\left(y-x\right)\)
\(=\left(y-x\right)^2:\left(y-x\right)\)
\(=y-x\)
a, \(\frac{x+1}{2x+6}=\frac{x+1}{2\left(x+3\right)}\)
b, \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)
\(=\frac{3x}{2x\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{2x+6}{2x\left(x+3\right)}=\frac{2\left(x+3\right)}{2x\left(x+3\right)}=\frac{1}{x}\)
c, \(\frac{x-x-2xy+x}{x+2y}+\frac{4xy}{4y^2-x^2}=\frac{x-2xy}{x+2y}+\frac{4xy}{\left(2y-x\right)\left(x+2y\right)}\)
\(=\frac{\left(x-2xy\right)\left(2y-x\right)}{\left(x+2y\right)\left(2y-x\right)}+\frac{4xy}{\left(2y-x\right)\left(x+2y\right)}=\frac{2xy-x^2+4xy^2+2x^2y}{\left(2y-x\right)\left(x+2y\right)}\)