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19 tháng 8 2019

Bài 1: a) x2 - 5x + 6 = x2 - 2x - 3x + 6 = x(x - 2) - 3(x - 2) = (x - 3)(x - 2)

b) 3x2 + 9x - 30 = 3x2 + 15x - 6x - 30 = 3x(x + 5) - 6(x + 5) = (3x - 6)(x + 5) = 2(x - 2)(x + 5)

c) x2 - 3x + 2 = x2 - 2x - x + 2  = x(x - 2) - (x - 2) = (x - 1)(x - 2)

d) x2 - 9x + 18 = x2 - 3x - 6x + 18 = x(x - 3) - 6(x - 3) = (x - 6)(x - 3)

e) x2 - 6x + 8 = x2 - 2x - 4x + 8 = x(x - 2) - 4(x - 2) = (x - 4)(x - 2)

f) x2 - 5x - 14 = x2 - 7x + 2x - 14 = x(x - 7) + 2(x - 7) = (x + 2)(x - 7)

g) x2 + 6x + 5 = x2 + 5x + x + 5 = x(x + 5) + (x + 5) = (x + 1)(x + 5)

h) x2 - 7x + 12 = x2 - 3x - 4x + 12 = x(x - 3) -  4(x - 3) = (x - 4)(x - 3)

i) x2 - 7x + 10 = x2 - 2x - 5x + 10 = x(x - 2) - 5(x - 2) = (x - 5)(x - 2)

19 tháng 8 2019

Oh Sehun:Ít một thôi bạn ơi!Làm xong đống này chắc đến tuổi già mất:(

a

\(x^2-5x+6\)

\(=\left(x^2-3x\right)-\left(2x-6\right)\)

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

b

\(3x^2+9x-30\)

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

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

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

c

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

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

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

d

\(x^2-9x+18\)

\(=\left(x^2-3x\right)-\left(6x-18\right)\)

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

e

\(x^2-6x+8\)

\(=\left(x^2-2x\right)-\left(4x-8\right)\)

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

f

\(x^2-5x-14\)

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

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

g

\(x^2+6x+5\)

\(=\left(x^2+x\right)+\left(5x+5\right)\)

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

h

\(x^2-7x+12\)

\(=\left(x^2-3x\right)-\left(4x-12\right)\)

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

i

\(x^2-7x+10\)

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

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

P/S:Cho e xin phép tách câu ạ.

19 tháng 8 2019
https://i.imgur.com/pcpV5iA.jpg
4 tháng 8 2020

a/ \(3x^2-5x-2\)

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

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

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

b/ \(2x^2+x-6\)

\(=2x^2+4x-3x-6\)

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

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

c/ \(7x^2+50x+7\)

\(=7x^2+49x+x+7\)

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

\(=\left(x+7\right)\left(7x+1\right)\)

d/ \(12x^2+7x-12\)

\(=12x^2-9x+16x-12\)

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

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

e/ \(15x^2+7x-2\)

\(=15x^2+10x-3x-2\)

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

\(=\left(3x+2\right)\left(5x-1\right)\)

f/ \(a^2-5a-14\)

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

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

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

g/ \(2m^2+10m+8\)

\(=2m^2+2m+8m+8\)

\(=2m\left(m+1\right)+8\left(m+1\right)\)

\(=\left(m+1\right)\left(2m+8\right)\)

h/ \(4p^2-36p+56\)

\(=4p^2-28p-8p+56\)

\(=4p\left(p-7\right)-8\left(p-7\right)\)

\(=\left(p-7\right)\left(4p-8\right)\)

29 tháng 7 2022

câu i) tách 5x sao vậy ạ ?

 

29 tháng 9 2016

a, x^2 + 5x +4

= x^2 + 1x + 4x + 4

= (x^2 + 1x) + (4x + 4)

= x ( x + 1 ) + 4 ( x + 1 )

= (x + 1) (x + 4)

b, x^2 - 6x + 5

= x^2 - 1x - 5x + 5

= (x^2 - 1x) - (5x - 5)

= x (x - 1) - 5 (x - 1)

= (x - 1) (x - 5)

c, x^2 + 7x + 12

= x^2 + 3x + 4x + 12 

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

= x (x + 3) + 4 (x + 3)

= (x + 3) (x + 4)

d, 2x^2 - 5x + 3

= 2^x2 - 2x - 3x + 3

= 2x (x - 1) - 3 (x - 1)

= (x-1) (2x - 3)

e, 7x  - 3x^2 - 4

= 3x + 4x - 3x^2 - 4

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

= 3x (1 - x) + 4 (x - 1)

= 3x (1-x) - 4 (1 - x)

= (1 - x) (3x - 4)

f, x^2 - 10x + 16

= x^2 - 2x - 8x + 16

= (x^2 - 2x) - (8x - 16)

= x (x - 2) - 8 (x - 2)

= (x - 2) (x - 8)

29 tháng 9 2016

a, (x+1)(x+4)

b,(x-5)(x-1)

c,(x+3)(x+4)

d,(2x-3)(x-1)

e,(-3x+4)(x-1)

f, (x-8)(x-2)

15 tháng 7 2016

a)x^2-(a+b)x+ab

= x^2 - ax - bx + ab

= (x^2 - ax) - (bx - ab)

= x(x-a) - b(x-a)

= (x-b)(x-a) 

b)7x^3-3xyz-21x^2+9z

c)4x+4y-x^2(x+y)

= 4(x + y) - x^2(x+y)

= (4-x^2) (x+y)

= (2-x)(2+x)(x+y)

d) y^2+y-x^2+x

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

= (y-x)(y+x)+ (x+y)

= (y-x+1) (x+y)

e)4x^2-2x-y^2-y

= [(2x)^2 - y^2] - (2x +y)

= (2x-y)(2x+y) - (2x+y)

= (2x -y -1)(2x+y)

f)9x^2-25y^2-6x+10y

31 tháng 8 2021

ko biết làm

 

13 tháng 8 2020

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/\(x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)

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: \(x^2-7x+10\)

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

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

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