phân tích đa thức thành nhân tử (a+b+c)^2+(a-b+c)^2-4b^2
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NM
1
17 tháng 6 2016
\(\left(a+b+c\right)^2+\left(a-b+c\right)^2-4b^2=\left[\left(a+b+c\right)^2-4b^2\right]+\left(a-b+c\right)^2=\left(a-b+c\right)\left(a+3b+c\right)+\left(a-b+c\right)^2=\left(a-b+c\right)\left(2a+2b+2c\right)=2\left(a+b+c\right)\left(a-b+c\right)\)
HN
0
NL
Nguyễn Lê Phước Thịnh
CTVHS
4 tháng 6
a: \(\left(a-2b\right)^2-4b^2\)
\(=\left(a-2b\right)^2-\left(2b\right)^2\)
=(a-2b-2b)(a-2b+2b)
=a(a-4b)
b: \(\left(a-b\right)^2-c^2=\left(a-b-c\right)\left(a-b+c\right)\)
c: \(\left(a+b\right)^2-4=\left(a+b\right)^2-2^2=\left(a+b+2\right)\left(a+b-2\right)\)
d: \(\left(a+3b\right)^2-9b^2\)
\(=\left(a+3b\right)^2-\left(3b\right)^2\)
=(a+3b-3b)(a+3b+3b)=a(a+6b)
e: \(\left(x-3\right)^3-27\)
\(=\left(x-3-3\right)\left\lbrack\left(x-3\right)^2+3\left(x-3\right)+9\right\rbrack\)
\(=\left(x-6\right)\left(x^2-6x+9+3x-9+9\right)=\left(x-6\right)\left(x^2-3x+9\right)\)
f: \(\left(x+1\right)^3-125\)
\(=\left(x+1-5\right)\left\lbrack\left(x+1\right)^2+5\left(x+1\right)+25\right\rbrack\)
\(=\left(x-4\right)\left(x^2+2x+1+5x+5+25\right)=\left(x-4\right)\left(x^2+7x+31\right)\)
HD
0
MA
0
\(\left(a+b+c\right)^2+\left(a-b+c\right)^2-4b^2\)
\(=2a^2+2b^2+2c^2+2ab+2ac+2bc-2ab-2bc+2ac-4b^2\)
\(=2a^2-2b^2+2c^2+4ac\)
\(=2\left[\left(a^2+2ac+c^2\right)-b^2\right]=2\left[\left(a+c\right)^2-b^2\right]\)
\(=2\left(a+c-b\right)\left(a+b+c\right)\)
\(\left(a+b+c\right)^2-\left(a-b+c\right)^2-4b^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2-2ab-2bc+2ca-4b^2\)
\(=2a^2-2b^2+2c^2+4ca\)
\(=2\left(a^2-b^2+c^2+2ac\right)\)
\(=2\left[\left(a+c\right)^2-b^2\right]\)
\(=2\left(a-b+c\right)\left(a+b+c\right)\)