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dat \(x^2-2x+2=y\)
ta co pt
\(y^4+20x^2y^2+64x^4\)
\(=\left(8x^2\right)^2+2.8x^2.\frac{10}{8}y^2+\left(\frac{10^{ }}{8^{ }}y^2\right)^2-\frac{36}{64}y^4\)
\(=\left(8x^2+\frac{10}{8}y^2\right)^2-\left(\frac{6}{8}y^2\right)^2\)
\(=\left(8x^2+\frac{y^2}{2}\right)\left(8x^2+2y^2\right)\)
bạn thay y nữa là xong
\(\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)^2+64x^4\)
\(=\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)^2+100x^4-36x^4\)
\(=\left[\left(x^2-2x+2\right)^2+10x^2\right]^2-36x^4\)
\(=\left(x^4-4x^3+18x^2-8x+4\right)^2-\left(6x^2\right)^2\)
\(=\left(x^4-4x^3+24x^2-8x+4\right)\left(x^4-4x^3+12x^2-8x+4\right)\)
Bài 6:
c: \(9x^2+6x+1=\left(3x+1\right)^2\)
d: \(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)
e: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
1/(x+2)2 -(3x-1)2=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x2+12x
2/(x4+x2)(-2x3-2x)=x2(x2+1)-2x(x2+1)=(x2+1)(x2-2x)
1: \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\)
\(=\left(2x+3-2x-5\right)^2\)
=4
a: Ta có: \(x^2-36y^2-x+6y\)
\(=\left(x-6y\right)\left(x+6y\right)-\left(x-6y\right)\)
\(=\left(x-6y\right)\left(x+6y-1\right)\)
b: Ta có: \(16x-8x^2+x^3\)
\(=x\left(x^2-8x+16\right)\)
\(=x\left(x-4\right)^2\)
c: Ta có: \(2x^2-4xy+2y^2-18\)
\(=2\left(x^2-2xy+y^2-9\right)\)
\(=2\cdot\left[\left(x-y\right)^2-9\right]\)
\(=2\left(x-y-3\right)\left(x-y+3\right)\)
d: Ta có: \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-10\right)\)
e: Ta có: \(x^4-x^2-30\)
\(=x^4-6x^2+5x^2-30\)
\(=x^2\left(x^2-6\right)+5\left(x^2-6\right)\)
\(=\left(x^2-6\right)\left(x^2+5\right)\)
f: Ta có: \(x^2-xy-2y^2\)
\(=x^2-2xy+xy-2y^2\)
\(=x\left(x-2y\right)+y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+y\right)\)
g: Ta có: \(x^4-13x^2y^2+4y^4\)
\(=x^4-4x^2y^2+4y^4-9x^2y^2\)
\(=\left(x^2-2y^2\right)^2-\left(3xy\right)^2\)
\(=\left(x^2-3xy-2y^2\right)\left(x^2-3xy+2y^2\right)\)
\(=\left(x^2-3xy-2y^2\right)\left(x^2-xy-2xy+2y^2\right)\)
\(=\left[x\left(x-y\right)-2y\left(x-y\right)\right]\left(x^2-3xy-2y^2\right)\)
\(=\left(x-y\right)\left(x-2y\right)\left(x^2-3xy-2y^2\right)\)
h: Ta có: \(\left(x^2-2x\right)^2-2\left(x^2-2x\right)-3\)
\(=\left(x^2-2x\right)^2-3\left(x^2-2x\right)+\left(x^2-2x\right)-3\)
\(=\left(x^2-2x\right)\left(x^2-2x-3\right)+\left(x^2-2x-3\right)\)
\(=\left(x^2-2x-3\right)\left(x^2-2x+1\right)\)
\(=\left(x-3\right)\left(x+1\right)\cdot\left(x-1\right)^2\)
a) Áp dụng HĐT 1 thu được ( 2 x + y ) 2 .
b) Áp dụng HĐT 3 với A = 2x + l; B = x - l thu được
[(2x +1) + (x -1)] [(2x +1) - (x -1)] rút gọn thành 3x(x + 2).
c) Ta có: 9 - 6x + x 2 - y 2 = ( 3 - x ) 2 - y 2 = (3 - x - y)(3 -x + y).
d) Ta có: -(x + 2) + 3( x 2 - 4) = -{x + 2) + 3(x + 2)(x - 2)
= (x + 2) [-1 + 3(x - 2)] = (x + 2)(3x - 7).
Chọn D.
x 4 + 8x = x( x 3 +8)= x( x 3 + 2 3 ) = x(x + 2)( x 2 − 2x + 4)
\(\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)+64x^4\)
=\(\left[\left(x^2-2x+2\right)^4+2.10x^2\left(x^2-2x+2\right)^2+100x^4\right]\)-100x4+64x2
=\(\left[\left(x^2-2x+2\right)^2+10x^2\right]^2-36x^2\)
=\(\left[\left(x^2-2x+2\right)^2+4x^2\right].\left[\left(x^2-2x+2\right)^2+16x^2\right]\)