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4 tháng 9 2023

=(a+b+c)(ab+bc+ca)

4 tháng 9 2023

=(a+b)(b+c)(c+a)

27 tháng 7 2023

a Đề sai: )

b

\(a^3-a^2x-ay+xy\\ =a^2\left(a-x\right)-y\left(a-x\right)\\ =\left(a-x\right)\left(a^2-y\right)\)

c

\(4x^2-y^2+4x+1\\ =\left(2x\right)^2+2.2x.1+1-y^2\\ =\left(2x+1\right)^2-y^2\\ =\left(2x+1-y\right)\left(2x+1+y\right)\)

d

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

e

\(5x^2-10xy+5y^2-5z^2\\ =5\left(x^2-2xy+y^2-z^2\right)\\ =5\left[\left(x-y\right)^2-z^2\right]\\ =5\left(x-y-z\right)\left(x-y+z\right)\)

c: =(2x+1)^2-y^2

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

d: =x^2(x^2+2x+1)

=x^2(x+1)^2

e: =5(x^2-2xy+y^2-z^2)

=5[(x-y)^2-z^2]

=5(x-y-z)(x-y+z)

a: =(x+y)^3+z^3-3xy(x+y)-3xyz

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)

b: \(=\left(x+y+y-z\right)^3-3\left(x+y\right)\left(y-z\right)\left(x+y+y-z\right)+\left(z-x\right)^3\)

\(=\left(x-z\right)^3+\left(z-x\right)^3-3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)

\(=-3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)

c: \(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)

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

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

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

d: =b^2c+bc^2+ac^2-a^2c-a^2b-ab^2

=b^2c-b^2a+bc^2-a^2b+ac^2-a^2c

=b^2(c-a)+b(c^2-a^2)+ac(c-a)

=(c-a)(b^2+ac)+b(c-a)(c+a)

=(c-a)(b^2+ac+bc+ba)

=(c-a)[b^2+bc+ac+ab]

=(c-a)[b(b+c)+a(b+c)]

=(c-a)(b+c)(b+a)

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

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

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

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

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

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

d: =(x-y)(x+y)-7(x+y)

=(x+y)(x-y-7)

a: Ta có: \(x^2-6x+9-y^2\)

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

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

b: Ta có: \(x^3+4x^2+4x\)

\(=x\left(x^2+4x+4\right)\)

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

c: Ta có: \(4xy-4x^2-y^2+9\)

\(=-\left(4x^2-4xy+y^2-9\right)\)

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

15 tháng 7 2015

(b-c)^2+b(a-c)^2+c(a-b)^2- a^3 -b^3 -c^3 +4abc 
=a[(b-c)^2-a^2)]+ b[(a-c)^2-b^2)]+c[(a-b)^2-c^2)]+4abc 
=a[(b-c)^2-a^2)]+ b[(a+c)^2-b^2)]+c[(a-b)^2-c^2)] 
=a(b-c-a)(b-c+a)+b(a+c-b)(a+b+c)+c(a+c... 
=[-a(b-c+a)+b(a+b+c)+c(a-b-c)](a+c-b) 
Bạn cứ tiếp tục phân tích cái vế trong ngoặc vuông đuọc (a+b-c)(b+c-a) là đc.

Đáp số : (a+c-b)(a+b-c)(b+c-a) 

15 tháng 7 2015

:)) Thớt không search google nên bạn í search hộ thôi =.=

Câu 1: A

Câu 21: A

 

1 tháng 11 2021

\(16,A\\ 17,C\\ 18,A\\ 19,C\\ 20,A\\ 21,A\)

NV
3 tháng 1

a.

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

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

b.

\(x^3-9x^2+6x+16=\left(x^3-7x^2-8x\right)-\left(2x^2-14x-16\right)\)

\(=x\left(x^2-7x-8\right)-2\left(x^2-7x-8\right)\)

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

\(=\left(x-2\right)\left[x\left(x+1\right)-8\left(x+1\right)\right]=\left(x-2\right)\left(x+1\right)\left(x-8\right)\)

c.

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

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

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

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

\(=\left(x^2+7x+10\right)^2-4\left(x^2+7x+10\right)+6\left(x^2+7x+10\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+10-4\right)+6\left(x^2+7x+10-4\right)\)

\(=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

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

Bài 1: 

a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)

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

\(=\left(2x+1\right)\left(8-2x\right)\)

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

b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)

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

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

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

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

Bài 2: 

a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)

\(=\left(a-b\right)\left(2a-4b\right)\)

\(=2\left(a-b\right)\left(a-2b\right)\)

f: Ta có: \(x^2-6xy+9y^2+4x-12y\)

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

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