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20 tháng 8 2020

a (x+1)(x+2)(x+3)(x+4)+1

= ((x+1)(x+4)). ((x+2)(x+3))+1

= (x^2+5x+4)(x^2+5x+6)+1

Đặt x^2+5x+4= t thì ta có:

t(t+2)+1= t^2+2t+1=(t+1)^2

Thay vào ta có:

(t+1)^2= (x^2+5x+4+1)^2 =(x^2+5x+5)^2

tương tự lm phần b, c, d

24 tháng 7 2020

Bài làm:

a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+1\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)

Đặt \(x^2+5x+5=t\)\(\Rightarrow\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2\)

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

b) Tương tự như a phân tích và đặt ra được: \(t^2-1-24=t^2-25=\left(t-5\right)\left(t+5\right)\)

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

c) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(x^2+8x+11=t\)\(\Rightarrow\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1\)

\(=\left(t-1\right)\left(t+1\right)=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)

\(=\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)

d) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x+11=t\)\(\Rightarrow\left(t-1\right)\left(t+1\right)-24=t^2-1-24=t^2-25\)

\(=\left(t-5\right)\left(t+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

24 tháng 7 2020

Làm mẫu cho 1 vd:

a, (x+1)(x+2)(x+3)(x+4)+1

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)(1)

Đặt \(y=x^2+5x+5\)

Khi đó ::

(1) = \(\left(y-1\right)\left(y+1\right)+1\)

\(=y^2-1+1=y^2\)

Thay vào ta được: \(\left(x^2+5x+5\right)^2\)

25 tháng 8 2017

b)\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-24\)4

\(=\left[\left(x-1\right)\left(x-4\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]-24\)

\(=\left(x^2-4x-x+4\right)\left(x^2-3x-2x+6\right)-24\)

\(=\left(x^2-5x+4\right)\left(x^2-5x+4+2\right)-24\)

\(\)Đặt  \(x^2-5x+4\)là a,ta có

\(=a\left(a+2\right)-24\)

\(=a^2+2a-24\)

\(=a^2+6a-4a-24\)

\(=a\left(a+6\right)-4\left(a+6\right)\)

\(=\left(a+6\right)\left(a-4\right)\)

Hay  \(\left(x^2-5x+4+6\right)\left(x^2-5x+4-4\right)\)

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

Câu hỏi của Huỳnh Bảo Nguyên - Toán lớp 8 - Học toán với OnlineMath

Mk làm òi nhé !

18 tháng 2 2021

 a) 3x2 – 7x + 2

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

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

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

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

 b) a(x2 + 1) – x(a2 + 1)

\(=ax^2+a-\left(a^2x+x\right)\)

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

.......?

 

 

 

 

a) Ta có: \(3x^2-7x+2\)

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

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

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

b) Ta có: \(a\left(x^2+1\right)-x\left(a^2+1\right)\)

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

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

\(=xa\left(x-a\right)-\left(x-a\right)\)

\(=\left(x-a\right)\left(xa-1\right)\)

c) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)^2+16\left(x^2+7x\right)+6\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)\left(x^2+7x+16\right)+6\left(x^2+7x+16\right)\)

\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

d) Ta có: \(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+105+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)^2+12\left(a^2+8a\right)+10\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)\left(a^2+8a+12\right)+10\left(a^2+8a+12\right)\)

\(=\left(a^2+8a+12\right)\left(a^2+8a+10\right)\)

\(=\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)

9 tháng 7 2018

6 tháng 10 2019

\(x^8+x^7+1\)

\(=x^8+x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-xx+1\)

\(=\left(x^8-x^6+x^5-x^3+x^2\right)\)

\(+\left(x^7-x^5+x^4-x^2+x\right)\)

\(+\left(x^6-x^4+x^3-x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

6 tháng 10 2019

\(x^5+x+1\)

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

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

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

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

21 tháng 7 2016

d ) 

=(x2-3x)(x2-3x+2)-24

đặt x2-3x+1=a ta đc 

(a-1)(a+1)-24

=a2-1-24=a2-25

=(a-5)(a+5)

=(x2-3x+1+5)(x2-3x+1-5)

=(x2-3x+6)(x2-3x-4)

=(x2-3x+6)(x2-4x+x-4)

=(x2-3x+1)[x(x-4)+(x-4)]

=(x-4)(x+1)(x2-3x+1)

mấy câu kia làm tương tự nhé 

21 tháng 9 2019

a/\(\left(x^2-x\right)^2+4\left(x^2-x\right)-12.\)

cho \(\left(x^2-x\right)=a\)

\(\Rightarrow a^2+4a-12\)

\(=a^2+6a-2a-12\)

\(=\left(a^2+6a\right)-\left(2a+12\right)\)

\(=a\left(a+6\right)-2\left(a+6\right)\)

\(=\left(a+6\right)\left(a-2\right)\)

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

b/ \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24\)

\(=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

Gọi \(x^2+5x+5=a\)

\(\Rightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=\left(a-1\right)\left(a+1\right)-24\)

                                                                                 \(=a^2-1-24\)

                                                                                \(=a^2-25\)

                                                                                \(=\left(a-5\right)\left(a+5\right)\)

                                                                               \(\Rightarrow\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)

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

\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)