Phân tích thành nhân tử
3x2-7x-10
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\(=x\left(3x-7y\right)-\left(3x-7y\right)=\left(x-1\right)\left(3x-7y\right)\)
3x2 - 7xy - 3x + 7y
= (3x2-3x) - (7xy-7y)
= 3x(x-1) - 7y(x-1)
= (3x-7y)(x-1)
Ta có x 2 – 7x + 10
= x 2 – 2x – 5x + 10
= x(x – 2) – 5(x – 2) = (x – 5)(x – 2)
Đáp án cần chọn là: B
\(x^3+4x^2-7x-10\)
\(=x^3-2x^2+6x^2-12x+5x-10\)
\(=x^2\left(x-2\right)+6x\left(x-2\right)+5\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+6x+5\right)\)
\(=\left(x-2\right)\left(x^2+x+5x+5\right)\)
\(=\left(x-2\right)\left[x\left(x+1\right)+5\left(x+1\right)\right]\)
\(=\left(x-2\right)\left(x+1\right)\left(x+5\right)\)
x3+4x2-7x-10
= x3-2x2+6x2-12x+5x-10
= x2(x-2)+6x(x-2)+5(x-2)
= (x2+6x+5)(x-2)
= (x2+x+5x+5)(x-2)
= [x(x+1)+5(x+1)](x-2)
= (x+5)(x+1)(x-2)
\(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-10\right)\)
3x2 - 7x - 10 = 3x2 + 3x - 10x - 10 = 3x(x + 1) - 10(x + 1) = (3x - 10)(x + 1)
\(12x^2+7x-10\)
\(=12x^2-8x+15x-10\)
\(=4x\left(3x-2\right)+5\left(3x-2\right)\)
\(=\left(4x+5\right)\left(3x-2\right)\)
a, \(x^2-5x+6=x^2+x-6x+6=x\left(x-1\right)-6\left(x-1\right)=\left(x-1\right)\left(x-6\right)\)
b, \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)
\(=3\left[x\left(x-2\right)+5\left(x-2\right)\right]=3\left(x-2\right)\left(x+5\right)\)
c, \(x^2+7x+10=x^2+2x+5x+10=x\left(x+2\right)+5\left(x+2\right)=\left(x+2\right)\left(x+5\right)\)
3x2 -7x -10 = 3x2 +3x - 10x -10
= 3x(x+1) - 10(x +1)
=(x+1)(3x-10)