Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(3x^4+2x^3-8x^2-2x+5\)
\(=3x^4-3x^3+5x^3-5x^2-3x^2+3x-5x+5\)
\(=3x^3\left(x-1\right)+5x^2\left(x-1\right)-3x\left(x-1\right)-5\left(x-1\right)\)
\(=\left(x-1\right)\left(3x^3+5x^2-3x-5\right)\)
\(=\left(x-1\right)\left[3x\left(x^2-1\right)+5\left(x^2-1\right)\right]\)
\(=\left(x-1\right)\left(3x+5\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(3x+5\right)\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)^2\left(3x+5\right)\left(x-1\right)\)
2x3 - x2 - 8x + 4 = x2(2x - 1) - 4(2x - 1) = (2x - 1)(x - 2)(x + 2)
2x3-x2 -8x+4=(2x3-x2)-(8x+4)
=x2(2x-1)-4(2x-1)
= (2x-1)(x2-4)
= (2x-1)(x-2)(x+2)
\(2x^3-2xy^2-8x^2+8xy\)
\(=2x\left(x^2-y^2-4x+4y\right)\)
\(=2x\left[\left(x^2-y^2\right)-4\left(x-y\right)\right]\)
\(=2x\left[\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\right]\)
\(=2x\left(x-y\right)\left(x+y-4\right)\)
2x^3 - 2xy^2 - 8x^2 + 8xy
= 2x^2 ( x - y ) - 8x ( x - y )
= ( x - y ) ( 2x^2 - 8x )
= ( x - y ) 2x ( x - 4 )
a) ta có 3x^2 -11x+6= 3x^2 - 9x -2x + 6 = 3x (x-3) - 2(x-3) =(x-3) (3x-2)
b) ta có 8x^2 + 10x - 3 = 2x (4x-1) - 3(4x- 1) = (2x-3)(4x-1)
c) ta có 8x^2 -2x - 1 = 8x^2 - 4x +2x-1 = 4x(2x-1) + 2x- 1 = (4x+1)(2x-1) k nha bạn
a, 3x2 - 11x+6 = 3x2 - 9x - 2 x + 6 = 3x(x-3) - 2(x-3) = (3x-2)(x-3)
3x-2 = 0 và x - 3 =0
x = 2/3 x = 3
1 (x-3)(3x-2)
2 (4x-1)(2x+3)
câu 3 bị sai thì phải bạn cứ nhân ra thì biết cách phân tích thôi mà
\(a,3x^2-11x+6=3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(3x-2\right)\left(x-3\right)\)
\(b,8x^2+10x-3=8x^2+12x-2x-3=4x\left(2x+3\right)-\left(2x+3\right)=\left(4x-1\right)\left(2x+3\right)\)
\(c,8x^2-2x-1=9x^2-x^2-2x-1=9x^2-\left(x+1\right)^2=\left(3x-x-1\right)\left(3x+x+1\right)\)
\(=\left(2x-1\right)\left(4x+1\right)\)
\(8x^3\left(y+z\right)-y^3\left(z+2x\right)-z^3\left(2x-y\right)\)
\(=8x^3\left(y+z\right)-y^3\left[\left(y+z\right)+\left(2x-y\right)\right]-z^3\left(2x-y\right)\)
\(=8x^3\left(y+z\right)-y^3\left(y+z\right)-y^3\left(2x-y\right)-z^3\left(2x-y\right)\)
\(=\left(y+z\right)\left(8x^3-y^3\right)-\left(2x-y\right)\left(y^3+z^3\right)\)
\(=\left(y+z\right)\left(2x-y\right)\left(4x^2+4xy+y^2\right)-\left(2x-y\right)\left(y+z\right)\left(y^2-xy+z^2\right)\)
\(=\left(y+z\right)\left(2x-y\right)\left(4x^2+4xy+y^2-y^2+xy-z^2\right)\)
\(=\left(y+z\right)\left(2x-y\right)\left(4x^2+5xy-z^2\right)\)
8x2 - 2x - 3 = 8x2 + 4x - 6x - 3
= 4x( 2x + 1 ) - 3( 2x + 1)
= ( 2x + 1 )( 4x - 3 )