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\(x^2+6x-y^2+9\)
\(=\left(x^2+6x+9\right)-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3-y\right)\left(x+3+y\right)\)
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)
a) \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)
b) sửa đề nhé!
\(6x-9-x^2=-\left(x^2-6x+9\right)\)
\(=-\left(x-3\right)^2\)
\(\left(5x+1\right)^2=\left(2x-3\right)^2\)
\(\Rightarrow\left(5x+1\right)^2-\left(2x-3\right)^2=0\)
\(\Rightarrow\left(5x+1+2x-3\right)\left(5x+1-2x+3\right)=0\)
\(\Rightarrow\left(7x-2\right)\left(3x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7x-2=0\\3x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2}{7}\\x=\frac{-4}{3}\end{cases}}}\)
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)x4-1=(x2-1)(x2+1)=(x-1)(x+1)(x2+1)
b)x2-y2-2x+2y=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)
c)x2-6x-y2+9=(x2-6x+9)-y2=(x-3)2-y2=(x-y-3)(x+y-3)
d)5x2+3(x+y)2-5y2
=5(x2-y2)+3(x+y)2
=5(x-y)(x+y)+3(x+y)2
=(x+y)(5x-5y+3x+3y)
=(x+y)(8x-2y)
a) \(x^2-xy+x-y\)
\(=\left(x^2-xy\right)+\left(x-y\right)\)
\(=x\left(x-y\right)+\left(x-y\right)\)
\(=\left(x+1\right)\left(x-1\right)\)
b) \(2xy-x^2-y^2+16\)
\(=16-\left(x^2-2xy+y^2\right)\)
\(=4^2-\left(x-y\right)^2\)
\(=\left(4-x+y\right)\left(4+x-y\right)\)
c) \(x^2-6x-16\)
\(=x^2-6x+9-25\)
\(=\left(x^2-6x+9\right)-25\)
\(=\left(x-3\right)^2-5^2\)
\(=\left(x-3-5\right)\left(x-3+5\right)\)
\(=\left(x-8\right)\left(x+2\right)\)
a) \(x^2-2x+x\)
\(=x^2-x=x\left(x-1\right)\)
b) \(x^2+y^2-2xy-9\)
\(=\left(x-y\right)^2-3^2\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
c) \(x^2-y^2+6x+9\)
\(=\left(x^2+6x+9\right)-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3-y\right)\left(x+3+y\right)\)
\(x^2-2x+x\)
\(=\left(x^2-2x.1+1\right)+\left(x-1\right)\)
\(=\left(x-1\right)^2+\left(x-1\right)\)
\(=\left(x-1\right)\left(x-1+1\right)\)
\(=x.\left(x-1\right)\)