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a) \(2x\left(x-3\right)^2+5x\left(3-x\right)\)
\(=2x\left(x-3\right)^2-5x\left(x-3\right)\)
\(=\left(x-3\right)\left[2x\left(x-3\right)-5x\right]\)
\(=\left(x-3\right)\left(2x^2-6x-5x\right)\)
\(=\left(x-3\right)\left(2x^2-11x\right)\)
\(=x\left(x-3\right)\left(2x-11\right)\)
b) \(\left(x+3\right)^2-4\left(y^2-2y+1\right)\)
\(=\left(x+3\right)^2-2^2\left(y-1\right)^2\)
\(=\left(x+3\right)^2-\left[2\left(y-1\right)\right]^2\)
\(=\left[\left(x+3\right)-2\left(y-1\right)\right]\left[\left(x+3\right)+2\left(y-1\right)\right]\)
\(=\left(x+3-2y+2\right)\left(x+3+2y-2\right)\)
\(=\left(x-2y+5\right)\left(x+2y+1\right)\)
a) \(2x.\left(x-3\right)^2+5x.\left(-x+3\right)=2x.\left(x-3\right)^2-5x.\left(x-3\right)\)
\(=\left(x-3\right).\left(2x^2-11x\right)=\left(x-3\right).x.\left(2x-11\right)\)
b) \(\left(x+3\right)^2-4.\left(y^2-2y+1\right)=\left(x+3\right)^2-2^2.\left(y-1\right)^2\)
\(=\left(x+3\right)^2-\left[2.\left(y-1\right)\right]^2=\left(x-2y+1\right).\left(x+2y+5\right)\)
a) 2x2-6x-x+3 = 2x(x-3) - (x-3) = (x-3)(2x-1)
b) x2-x-5x+5 = x(x-1) - 5(x-1) = (x-1)(x-5)
c) 5x(x-2y) + 2( x-2y)2 = (x-2y)(5x+2x-2y) = (x-2y)(7x-2y)
chú ý : (A-B)2=(B-A)2
d) 7x(4-y)2 - (4-y)3 = ( 16-8y+y2) (7x-4+y)
a) \(2x^2-7x+3=2x^2-6x-x+3=2x\left(x-3\right)-\left(x-3\right)=\left(x-3\right)\left(2x-1\right)\)
b) \(x^2-6x+5=x^2-5x-x+5=x\left(x-5\right)-\left(x-5\right)=\left(x-5\right)\left(x-1\right)\)
c)\(5x\left(x-2y\right)+2\left(2y-x\right)^2=5x\left(x-2y\right)+2\left(x-2y\right)^2\\ =\left(x-2y\right)\left(5x+2x-4y\right)=\left(x-2y\right)\left(7x-4y\right)\)
d) \(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(y-4\right)+\left(y-4\right)^3=\left(y-4\right)\left(7x-y-4\right)\)
a) \(x^4-2x^2+1=\left(x^2-1\right)^2=\left(x-1\right)^2\left(x+1\right)^2\)
b) \(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)
c) \(2x^3-x^2-8x+4\)
\(=x^2\left(2x-1\right)-4\left(2x-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(2x-1\right)\)
d) \(x\left(x-y\right)^2+y\left(x-y\right)^2-xy+x^2\)
\(=\left(x+y\right)\left(x-y\right)^2+x\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-y^2+x\right)\)
e) \(2x^2-5x+2\)
\(=\left(2x^2-x\right)-\left(4x-2\right)\)
\(=x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(x-2\right)\left(2x-1\right)\)
\(\left(5x+1\right)^2=\left(2x-3\right)^2\)
\(\Rightarrow\left(5x+1\right)^2-\left(2x-3\right)^2=0\)
\(\Rightarrow\left(5x+1+2x-3\right)\left(5x+1-2x+3\right)=0\)
\(\Rightarrow\left(7x-2\right)\left(3x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7x-2=0\\3x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2}{7}\\x=\frac{-4}{3}\end{cases}}}\)
Vậy.......
\(a,x^3-4x^2+8x-8\)
\(=\left(x^3-8\right)-\left(4x^2-8x\right)\)
\(=\left(x^3-2^3\right)-4x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4x+4\right)-4x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4x+4-x+2\right)\)
\(=\left(x-2\right)\left(x^2+3x+6\right)\)
\(b,\left(2xy+1\right)^2-\left(2x+y\right)^2\)
\(=\left(2xy+1+2x+y\right)\left(2xy+1-2x-y\right)\)
\(=\left[\left(2xy+2x\right)+\left(y+1\right)\right]\cdot\left[\left(2xy-2x\right)-\left(y-1\right)\right]\)
\(=\left[2x\left(y+1\right)+1\cdot\left(y+1\right)\right]\cdot\left[2x\left(y-1\right)-1\cdot\left(y-1\right)\right]\)
\(=\left[\left(y+1\right)\left(2x+1\right)\right]\cdot\left[\left(y-1\right)\left(2x-1\right)\right]\)
\(=\left(y+1\right)\left(y-1\right)\left(2x-1\right)\left(2x+1\right)\)
\(=\left(y^2-1\right)\left(4x^2-1\right)\)
\(c,1+6x-6x^2-x^3\)
\(=\left(1-x^3\right)-\left(6x^2-6x\right)\)
\(=\left(1^3-x^3\right)-6x\left(x-1\right)\)
\(=\left(1-x\right)\left(1+2x+x^2\right)+6x\left(1-x\right)\)
\(=\left(1-x\right)\left(1+2x+x^2+6x\right)\)
\(=\left(1-x\right)\left(1+8x+x^2\right)\)
\(a,x^2-9-2\left(x+3\right)^2=0\)
\(\Rightarrow\left(x^2-3^2\right)-2\left(x+3\right)^2=0\)
\(\Rightarrow\left(x-3\right)\left(x+3\right)-2\left(x+3\right)\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left[x-3-2\cdot\left(x+3\right)\right]=0\)
\(\Rightarrow\left(x+3\right)\left[x-3-2x-6\right]=0\)
\(\Rightarrow\left(x+3\right)\left(-x-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\-x-9=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\-x=9\Rightarrow x=-9\end{cases}}\)
\(b,\left(5x+1\right)^2=\left(2x-3\right)^2\)
\(\Rightarrow\left(5x+1\right)^2\div\left(2x-3\right)^2=0\)
\(\Rightarrow\left[\left(5x+1\right)\div\left(2x-3\right)\right]^2=0\)
\(\Rightarrow\left(5x+1\right)\div\left(2x-3\right)=0\)
\(\Rightarrow5x+1=0\)
\(\Rightarrow x=-\frac{1}{5}\)
a: \(=6x^3-12x^2+x^2-2x+x-2\)
\(=\left(x-2\right)\left(6x^2+x+1\right)\)
b: \(=3x^4+3x^3-x^3-x^2-7x^2-7x+5x+5\)
\(=\left(x+1\right)\left(3x^3-x^2-7x+5\right)\)
\(=\left(x+1\right)\left(3x^3-3x^2+2x^2-2x-5x+5\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(3x^2+2x-5\right)\)
\(=\left(x-1\right)^2\cdot\left(x+1\right)\left(3x+5\right)\)
c: \(=4x^3+x^2+4x^2+x+4x+1\)
\(=\left(4x+1\right)\left(x^2+x+1\right)\)
a) \(6x^2-x-1\)
\(=6x^2-3x+2x-1\)
\(=3x\left(2x-1\right)+\left(2x-1\right)\)
\(=\left(3x+1\right)\left(2x-1\right)\)
a, (2xy+1)2-(2x+y)2=(2xy+1-2x-y)(2xy+1+x+y)=[(2xy-2x)-(y-1)][(2xy+2x)-(y+1)]=[2x(y-1)-(y-1)][2x(y+1) +(y+1)]
=(y-a)(2x-1)(2x+1)(y+1)
b, x2(x-2)2-(x-2)2-x2+1=[x2(x-2)2-(x-2)2] - (x2-1)=(x-2)2(x2-1) - (x2-1)=(x2-1)[(x-2)2-1]=(x-1)(x+1)(x+1)(x-3)
c,x4+2x3-2x-1=(x4-1)+(2x3-2x)=(x2-1)(x2 +1)+2x(x2-1)=(x2-1)(x2+2x+1)=(x-1)(x+1)(x+1)2
d,1+6x-6x2-x3=
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
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