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\(3x^4+2x^3-8x^2-2x+5\)
\(=3x^4-3x^3+5x^3-5x^2-3x^2+3x-5x+5\)
\(=3x^3\left(x-1\right)+5x^2\left(x-1\right)-3x\left(x-1\right)-5\left(x-1\right)\)
\(=\left(3x^3+5x^2-3x-5\right)\left(x-1\right)\)
\(=\left[3x\left(x^2-1\right)+5\left(x^2-1\right)\right]\left(x-1\right)\)
\(=\left(3x+5\right)\left(x^2-1\right)\left(x-1\right)\)
\(=\left(3x+5\right)\left(x+1\right)\left(x-1\right)\left(x-1\right)=\left(3x+5\right)\left(x+1\right)\left(x-1\right)^2\)
b, \(2x^4-9x^3+4x^2+21x-18\)
\(=2x^4-2x^3-7x^3+7x^2-3x^2+3x+18x-18\)
\(=2x^3\left(x-1\right)-7x^2\left(x-1\right)-3x\left(x-1\right)+18\left(x-1\right)\)
\(=\left(2x^3-7x^2-3x+18\right)\left(x-1\right)\)
phân tích đa thức ->nhân tử:
a)2x2+4x-70
b)x3-5x2+8x-4
c)x2-10+16
rút gọn:
(8x-8x3-10x2+3x4-5):(3x2-2x+1)
Bài 1:
a)2x2+4x-70
=2(x2+2x-35)
=2(x2+7x-5x-35)
=2[x(x+7)-5(x+7)]
=2(x-5)(x+7)
b)x3-5x2+8x-4
=x3-4x2+4x-x2+4x-4
=x(x2-4x+4)-(x2-4x+4)
=(x2-4x+4)(x-1)
=(x-2)2(x-1)
c)x2-10x+16
=x2-2x-8x+16
=x(x-2)-8(x-2)
=(x-8)(x-2)
Bài 2:
\(\frac{8x-8x^3-10x^2+3x^4-5}{3x^2-2x+1}=\frac{\left(x^2-2x-5\right)\left(3x^2-2x+1\right)}{3x^2-2x+1}=x^2-2x-5\)
Hãy tích cho tui đi
khi bạn tích tui
tui không tích lại bạn đâu
THANKS
a, 3x3-8x2+8x-5
= x2(3x-5)-x(3x-5)+3x-5
=(3x-5)(x2-x+1)
b, 4x3-3x2+5x-21
= x2(4x-7) +x(4x-7)+3(4x-7)
=(4x-7)(x2+x+3)
\(3x^4+2x^3-8x^2-2x+5\)
\(=3x^4-3x^3+5x^3-5x^2-3x^2+3x-5x+5\)
\(=3x^3\left(x-1\right)+5x^2\left(x-1\right)-3x\left(x-1\right)-5\left(x-1\right)\)
\(=\left(x-1\right)\left(3x^3+5x^2-3x-5\right)\)
\(=\left(x-1\right)\left[3x\left(x^2-1\right)+5\left(x^2-1\right)\right]\)
\(=\left(x-1\right)\left(3x+5\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(3x+5\right)\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)^2\left(3x+5\right)\left(x-1\right)\)