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KN
2
30 tháng 7 2018
a, \(\left(x+y\right)^2-4\left(x+y\right)+4=\left(x+y\right)^2-2.\left(x+y\right)+2^2=\left(x+y-2\right)^2\)
b, \(4b^2c^2-\left(b^2+c^2-a^2\right)^2=\left(2bc\right)^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)
\(=\left[a^2-\left(b^2-2bc+c^2\right)\right]\left[b^2+2bc+c^2-a^2\right]\)
\(=\left[a^2-\left(b-c\right)^2\right]\left[\left(b+c\right)^2-a^2\right]\)
= \(\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)\)
Chusc bajn hojc toost.
NH
0
KN
9 tháng 9 2019
a) \(\left(6x-1\right)^2-\left(3x+2\right)^2\)
\(=\left(6x-1+3x+2\right)\left(6x-1-3x-2\right)\)
\(=\left(9x+1\right)\left(3x-3\right)\)
\(=3\left(9x+1\right)\left(x-1\right)\)
b) \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)
\(=\left(6x+9\right)^2-\left(2x+2\right)^2\)
\(=\left(6x+9+2x+2\right)\left(6x+9-2x-2\right)\)
\(=\left(8x+11\right)\left(4x+7\right)\)
c) \(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc\right)^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc+b^2+c^2-a^2\right)\left(2bc-b^2-c^2+a^2\right)\)
\(=-\left[\left(b+c\right)^2-a^2\right]\left(b^2-2bc+c^2-a^2\right)\)
\(=-\left(b+c-a\right)\left(b+c+a\right)\left[\left(b-c\right)^2-a^2\right]\)
\(=-\left(b+c-a\right)\left(b+c+a\right)\left(b-c-a\right)\left(b-c+a\right)\)
KN
9 tháng 9 2019
d) \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5\right)^2-\left(2ab+4\right)^2\)
\(=\left(a^2+b^2-5+2ab+4\right)\left(a^2+b^2-5-2ab-4\right)\)
\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-3^2\right]\)
\(=\left(a+b+1\right)\left(a+b-1\right)\left(a-b-3\right)\left(a-b+3\right)\)
23 tháng 9 2018
\(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)
\(=\left[a^2-\left(b^2-2bc+c^2\right)\right].\left[\left(b^2+2bc+c^2\right)-a^2\right]\)
\(=\left[a^2-\left(b-c\right)^2\right].\left[\left(b+c\right)^2-a^2\right]\)
\(=\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)\)
\(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)
\(=\left[\left(a-b\right)^2-3^2\right].\left[\left(a+b\right)^2-1\right]\)
\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)
Tham khảo nhé~
NH
1
LC
11 tháng 8 2015
=2ab.[a+2b]+c^2.[a+2b]- c.[a^2+4ab+4.b^2]
=.................................-c[a+2b]^2
=[a+2b].{2ab+c^2-ca-2bc]
=[a+2b]{ 2b.[a-c]-c.[a-c] }
=[a+2b].[a-c].[2b-c]
3 tháng 12 2016
a) \(\left(2x+5\right)^2\)\(-\left(x-9\right)^2\)
=\(\left(2x+5+x-9\right).\left(2x+5-x+9\right)\)
=\(\left(3x-4\right).\left(x+14\right)\)
BV
1
22 tháng 7 2016
1, \(a^3-a^2-a^2+a\)
\(=a^2\left(a-1\right)-a\left(a-1\right)\)
\(=\left(a^2-a\right)\left(a-1\right)\)
\(=a\left(a-1\right)\left(a-1\right)\)
\(=a\left(a-1\right)^2\)
2,\(=2\left(b^2+2b+1-c\right)\)
Trả lời:
\(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc\right)^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)
\(=\left[a^2-\left(b^2-2bc+c^2\right)\right]\left[\left(b^2+2bc+c^2\right)-a^2\right]\)
\(=\left[a^2-\left(b-c\right)^2\right]\left[\left(b+c\right)^2-a^2\right]\)
\(=\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)\)
\(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)
\(=\left[a^2-\left(b-c\right)^2\right]\left[\left(b+c\right)^2-a^2\right]\)
\(=\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(a+b+c\right)\)