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Nii-chan

Phân tích đa thức thành nhân tử:

a) x2 - 5x +6

b) 3x2 + 9x - 30

c) x2 - 3x + 2

d) x2 - 9x + 18

e) x2 - 6x + 8

f) x2 - 5x - 14

g) x2 - 6x + 5

h) x2 - 7x + 12

i) x2 - 7x + 10

Nguyễn Lê Phước Thịnh
13 tháng 8 2020 lúc 21:13

a) Ta có: \(x^2-5x+6\)

\(=x^2-2x-3x+6\)

\(=x\left(x-2\right)-3\left(x-2\right)\)

\(=\left(x-2\right)\left(x-3\right)\)

b) Ta có: \(3x^2+9x-30\)

\(=3\left(x^2+3x-10\right)\)

\(=3\left(x^2+5x-2x-10\right)\)

\(=3\left[x\left(x+5\right)-2\left(x+5\right)\right]\)

\(=3\left(x+5\right)\left(x-2\right)\)

c) Ta có: \(x^2-3x+2\)

\(=x^2-x-2x+2\)

\(=x\left(x-1\right)-2\left(x-1\right)\)

\(=\left(x-1\right)\left(x-2\right)\)

d) Ta có: \(x^2-9x+18\)

\(=x^2-3x-6x+18\)

\(=x\left(x-3\right)-6\left(x-3\right)\)

\(=\left(x-3\right)\left(x-6\right)\)

e) Ta có: \(x^2-6x+8\)

\(=x^2-4x-2x+8\)

\(=x\left(x-4\right)-2\left(x-4\right)\)

\(=\left(x-4\right)\left(x-2\right)\)

f) Ta có: \(x^2-5x-14\)

\(=x^2-7x+2x-14\)

\(=x\left(x-7\right)+2\left(x-7\right)\)

\(=\left(x-7\right)\left(x+2\right)\)

g) Ta có: \(x^2-6x+5\)

\(=x^2-x-5x+5\)

\(=x\left(x-1\right)-5\left(x-1\right)\)

\(=\left(x-1\right)\left(x-5\right)\)

h) Ta có: \(x^2-7x+12\)

\(=x^2-3x-4x+12\)

\(=x\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x-3\right)\left(x-4\right)\)

i) Ta có: \(x^2-7x+10\)

\(=x^2-2x-5x+10\)

\(=x\left(x-2\right)-5\left(x-2\right)\)

\(=\left(x-2\right)\left(x-5\right)\)

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HUYNH NHAT TUONG VY
13 tháng 8 2020 lúc 21:21

a/\(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)b/

\(3x^2+9x-30=3\left(x^2+3x-10\right)\)

c/

\(x^2-3x+2=x^2-x-2x+2=x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(x-2\right)\)

d/\(x^2-9x+18=x^2-3x-6x+18=x\left(x-3\right)-6\left(x-3\right)=\left(x-3\right)\left(x-6\right)\)e/

\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-2\right)\left(x-4\right)\)f/\(x^2-5x-14=x^2+2x-7x-14=x\left(x+2\right)-7\left(x+2\right)=\left(x+2\right)\left(x-7\right)\)

g/

\(x^2-6x+5=x^2-x-5x+5=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)

h/

\(x^2-7x+12=x^2-4x-3x+12=x\left(x-4\right)-3\left(x-4\right)=\left(x-4\right)\left(x-3\right)\)i/\(x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)

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