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30 tháng 7 2017

bài 1: a) \(x^2-3=x^2-\left(\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)\)

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

c) \(x^3-27b^3=\left(x-3b\right)\left(x^2+3xb+b^2\right)\)

28 tháng 10 2021

Bài 1:

\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)

Bài 2:

\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)

Bài 3:

\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)

13 tháng 9 2021

dấu ^ là mũ nha mn

 

2 tháng 10 2016

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

b) \(25-x^2+4xy-4y^2=25-\left(x-2y\right)^2=\left(5-x+2y\right)\left(5+x-2y\right)\)

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

d) \(y^2\left(x-1\right)-7y^3+7xy^3\)

\(=y^2\left(x-1-7y+7xy\right)\)

\(=y^2\left[\left(x-1\right)-7y\left(1-x\right)\right]=y^2\left(x-1\right)\left(1+7y\right)\)

2 tháng 10 2016

a)

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

 

 

6 tháng 11 2021

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

\(=\left(4-a-b\right)\left(4+a-b\right)\), đằng trước là dấu trừ thì khi bỏ ngoặc phải đổi dấu chứ nhỉ :0

1.a) (3x+1)2-4(x-2)2= (3x+1)2-[2(x-2)]2=[(3x+1)-2(x-2)][(3x+1)+2(x-2)]=(x+3)(5x-1)

b) (a2+b2-5)2-4(ab+2)2= (a2+b2-5)2-[2(ab+2)]2 = (a2+b2-5-2ab-4)(a2+b2-5+2ab+4)=[(a-b)2-9][(a+b)2-1]

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

b. x3-5x2-14x=x3+2x2-7x2-14x=x2(x+2)-7x(x+2)=(x2-7x)(x+2)

23 tháng 7 2018

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

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

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

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

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

b) \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5\right)^2-\left[2.\left(ab+2\right)\right]^2\)

\(=\left(a^2+b^2-5\right)^2-\left(2ab+4\right)^2\)

\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)

\(=\left[\left(a-b\right)^2-9\right].\left[\left(a+b\right)^2-1\right]\)

\(=\left[\left(a-b-3\right)\left(a-b+3\right)\right].\left[\left(a+b-1\right)\left(a+b+1\right)\right]\)

a) \(3x^2+9x-30\)

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

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

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

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

b) \(x^3-5x^2-14x\)

\(=x\left(x^2-5x-14\right)\)

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

\(=x.\left[x\left(x+2\right)-7.\left(x+2\right)\right]\)

\(=x.\left[\left(x-7\right)\left(x+2\right)\right]\)