A
💋Amanda💋
19 tháng 8 2019
https://i.imgur.com/pcpV5iA.jpg
Đúng(0)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Dưới đây là một vài câu hỏi có thể liên quan tới câu hỏi mà bạn gửi lên. Có thể trong đó có câu trả lời mà bạn cần!
TG
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/ \(4p^2-36p+56\)
\(=4p^2-28p-8p+56\)
\(=4p\left(p-7\right)-8\left(p-7\right)\)
\(=\left(p-7\right)\left(4p-8\right)\)
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.
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/\(x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)
13 tháng 8 2020
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: \(x^2-7x+10\)
\(=x^2-2x-5x+10\)
\(=x\left(x-2\right)-5\left(x-2\right)\)
\(=\left(x-2\right)\left(x-5\right)\)
4 tháng 7 2016
a) \(A=\left(x^3+x^2\right)-\left(x+1\right)=x\left(x+1\right)-\left(x+1\right)=\left(x-1\right)\left(x+1\right)\)
4 tháng 7 2016
b) \(B=\left(x^3-3x^2\right)-\left(4x-12\right)\)
\(=x^2\left(x-3\right)-4\left(x-3\right)=\left(x^2-4\right)\left(x-3\right)=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
HT
22 tháng 8 2016
a) \(x^2\)\(-5x+6\)
=\(x^2\)\(-3x-2x+6\)
=\(x\left(x-3\right)-2\left(x-3\right)\)
=\(\left(x-2\right)\left(x-3\right)\)
b) \(3x^2\)\(+9x-30\)
=\(3x^2\)\(-6x+15x-30\)
=\(3x\left(x-2\right)+15\left(x-2\right)\)
=\(\left(x-2\right)\left(3x+15\right)\)
c)\(x^2\)\(-3x+2\)
=\(x^2\)\(-2x-x+2\)
=\(x\left(x-2\right)-\left(x-2\right)\)
=\(\left(x-2\right)\left(x-1\right)\)
d) \(12x^2\)\(+7x-12\)
=\(12x^2\)\(-9x+16x-12\)
=\(3x\left(4x-3\right)+4\left(4x-3\right)\)
=\(\left(3x+4\right)\left(4x-3\right)\)
e) \(15x^2\)\(+7x-2\)
=\(15x^2\)\(-3x+10x-2\)
=\(3x\left(5x-1\right)+2\left(5x-1\right)\)
=\(\left(3x+2\right)\left(5x-1\right)\)
f) \(a^2\)\(-5a-14\)
=\(a^2\)\(-7a+2a-14\)
=\(a\left(a-7\right)+2\left(a-7\right)\)
=\(\left(a+2\right)\left(a-7\right)\)
g) \(x^2\)\(-\left(a+b\right)x+ab\)
=\(x^2\)\(-ax-bx+ab\)
=\(x\left(x-a\right)-b\left(x-a\right)\)
=\(\left(x-a\right)\left(x-b\right)\)
