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BM
0
TT
0
K
2
23 tháng 10 2021
\(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5\right)^2-\left(2ab+4\right)^2\)
\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)
\(=\left[\left(a-b\right)^2-9\right]\left[\left(a+b\right)^2-1\right]\)
\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b+1\right)\left(a+b-1\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 10 2021
\(=\left(a^2+b^2-5\right)^2-\left(2ab+4\right)^2\)
\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)
\(=\left(a^2-2ab+b^2-9\right)\left(a^2+2ab+b^2-1\right)\)
\(=\left[\left(a-b\right)^2-9\right]\left[\left(a+b\right)^2-1\right]\)
\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)
NP
1
23 tháng 7 2018
\(a,-2x^3y^4-4x^4y^3+2x^3y^3\)
\(=-2x^3y^3\left(y+x-1\right)\)
\(b,ab\left(x-5\right)-a^2\left(5-x\right)\)
\(=\left(x-5\right)\left(ab+a^2\right)\)
M
31 tháng 10 2015
\(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-c\right)\)
\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b+b-c\right)\)
\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b\right)-ca\left(b-c\right)\)
\(=\left(a-b\right)\left(ab-ca\right)+\left(b-c\right)\left(bc-ca\right)\)
\(=\left(a-b\right)a\left(b-c\right)+\left(b-c\right)c\left(b-a\right)\)
\(=\left(a-b\right)a\left(b-c\right)-\left(b-c\right)c\left(a-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
mình làm vội, có chỗ nào sai bạn thông cảm nha
CG
2
11 tháng 3 2021
Ta có:
\(A=8abc+4\left(ab+bc+ca\right)+2\left(a+b+c\right)+1\)
\(A=\left(8abc+4ab\right)+\left(4bc+2b\right)+\left(4ca+2a\right)+\left(2c+1\right)\)
\(A=4ab\left(2c+1\right)+2b\left(2c+1\right)+2a\left(2c+1\right)+\left(2c+1\right)\)
\(A=\left(2c+1\right)\left(4ab+2a+2b+1\right)\)
\(A=\left(2c+1\right)\left[2a\left(2b+1\right)+\left(2b+1\right)\right]\)
\(A=\left(2a+1\right)\left(2b+1\right)\left(2c+1\right)\)
NC
11 tháng 3 2021
Ta có:\(A=8abc+4\left(ab+bc+ca\right)+2\left(a+b+c\right)+1\)
\(=8abc+4ab+4bc+4ca+2a+2b+2c+1\)
\(=\left(8abc+4ab\right)+\left(4bc+2b\right)+\left(4ca+2a\right)+\left(2c+1\right)\)
\(=4ab\left(2c+1\right)+2b\left(2c+1\right)+2a\left(2c+1\right)+\left(2c+1\right)\)
\(=\left(2c+1\right)\left(4ab+2b+2a+1\right)\)
\(=\left(2c+1\right)\left[2b\left(2a+1\right)+\left(2a+1\right)\right]\)
\(=\left(2c+1\right)\left(2b+1\right)\left(2a+1\right)\)
GL
0
22 tháng 7 2018
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) \(x^3-5x^2-14x\)
\(=x\left(x^2-5x-14\right)\)
\(=x\left(x^2+2x-7x-14\right)\)
\(=x.\left[x\left(x+2\right)-7.\left(x+2\right)\right]\)
\(=x.\left[\left(x-7\right)\left(x+2\right)\right]\)