NM
2
K
Khách
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DN
1
LQ
28 tháng 2 2016
sử dụng hàng đẳng thức a^2-b^2
=(2bc+b^2+c^2-a^2)(2ab-b^2-C^2+a^20)
8 tháng 9 2019
\(x^2-y^2+4x+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(4x^2-y^2+8\left(y-2\right)\)
\(=4x^2-\left(y^2-8y+16\right)\)
\(=4x^2-\left(y-4\right)^2\)
\(=\left(2x+y-4\right)\left(2x-y+4\right)\)
13 tháng 10 2016
a) = (xyz+xy) +(z+1) +(yz+zx)+(x+y)
= xy(z+1) +(z+1)+z(x+y)+(x+y)
= (z+1)(xy+1)+(x+y)(Z+1)
=(z+1)(xy+1+x+y)
TN
16 tháng 7 2018
\(4x^2-4x-35\) \(=\left(2x\right)^2-2.2x.1+1-36\)
\(=\left(2x-1\right)^2-6^2\)
\(=\left(2x-7\right)\left(2x+5\right)\)
\(18x^2-5x-2\) \(=\left(x-\frac{1}{2}\right)\left(x+\frac{2}{9}\right)\)
\(8x^3-26x^2+13x+5=\) \(8x^3-8x^2-18x^2+18x-5x+5\)
\(=8x^2\left(x-1\right)-18x\left(x-1\right)-5\left(x-1\right)\)
\(=\) \(\left(8x^2-18x-5\right)\left(x-1\right)\)
\(=\left(x-\frac{5}{2}\right)\left(x+\frac{1}{4}\right)\)\(\left(x-1\right)\)
NN
1
7 tháng 10 2019
\(\left(a+b+c\right)^3-a^3-\left(b^3+c^3\right)=\left(b+c\right)\left[\left(a+b+c\right)^2+a\left(a+b+c\right)+a^2\right]-\left(b+c\right)\left(b^2-bc+c^2\right)\)\(=\left(b+c\right)\left(3a^2+3ab+3bc+3ca\right)=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
TV
0
LP
1
KT
2 tháng 2 2018
\(\left(a+b\right)\left(a^2-b^2\right)+\left(b+c\right)\left(b^2-c^2\right)+\left(c+a\right)\left(c^2-a^2\right)\)
\(=\left(a+b\right)\left(a^2-b^2\right)-\left(b+c\right)\left(a^2-b^2\right)-\left(b+c\right)\left(c^2-a^2\right)+\left(a+c\right)\left(c^2-a^2\right)\)
\(=\left(a^2-b^2\right)\left(a+b-b-c\right)-\left(c^2-a^2\right)\left(b+c-c-a\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(a-c\right)-\left(c-a\right)\left(c+a\right)\left(b-a\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(a+b-c-a\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
DA
1
CC
22 tháng 2 2020
\(a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2+8abc\)
\(=a\left(b^2-2bc+c^2\right)+b\left(c^2-2ac+a^2\right)+c\left(a^2-2ab+b^2\right)+8abc\)
\(=ab^2-2abc+ac^2+bc^2-2abc+ba^2+ca^2-2abc+cb^2+8abc\)
\(=ab^2+ac^2+bc^2+ba^2+ca^2+cb^2+2abc\)
\(=\left(ac^2+bc^2\right)+\left(ab^2+ba^2\right)+\left(ca^2+cb^2+2abc\right)\)
\(=c^2\left(a+b\right)+ab\left(a+b\right)+c\left(a^2+b^2+2ab\right)\)
\(=c^2\left(a+b\right)+ab\left(a+b\right)+c\left(a+b\right)^2\)
\(=\left(a+b\right)\left[c^2+ab+c\left(a+b\right)\right]=\left(a+b\right)\left(c^2+ab+ca+bc\right)\)
\(=\left(a+b\right)\left[\left(c^2+ca\right)+\left(ab+bc\right)\right]=\left(a+b\right)\left[c\left(c+a\right)+b\left(a+c\right)\right]\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
ta có: \(\left(a+b+c\right)^2+\left(a+b-c\right)^2-4c^2=\left(a+b+c\right)^2+\left(a+b-c-2c\right)\left(a+b-c+2c\right).\)
\(=\left(a+b+c\right)^2+\left(a+b-3c\right)\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a+b+c+a+b-3c\right)\)
\(=2\left(a+b+c\right)\left(a+b-c\right)\)
(a+b+c)^2+(a+b-c)^2-4c^2
=(a^2+b^2+c^2+2ab+2bc+2ac)+(a^2-2ab+b^2-2ac+c^2-abc)-4c^2
=a^2+b^2+c^2+2ab+2bc+2ac+a^2-2ab+b^2-2ac+c^2-abc-4c^2
=(a^2+a^2)+(b^2+b^2)+(c^2+c^2)+(2ab-2ab)+(2bc-2bc)+(2ac-2ac)-4c^2
=2a^2+2b^2+2c^2-4c^2
=(2a^2+2b^2)+(2c^2-4c^2)
=2*(a^2+b^2)+2c^2*(1-2)