Phân tích các đa thức sau thành nhân tử :
a) \(\left(a^2+b^2-5\right)^2-2\left(ab+2\right)^2\)
b) \(\left(4a^2-3a-18\right)^2-\left(4a^2+3a\right)^2\)
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QT
Quoc Tran Anh Le
Giáo viên
29 tháng 7 2018
\(\left(4a^2-3a-18\right)^2-\left(4a+3a\right)^2\)
\(=\left(4a^2-3a-18-4a^2-3a\right)\left(4a^2-3a-18+4a^2+3a\right)\)
\(=\left(-6a-18\right)\left(8a^2-18\right)\)
22 tháng 7 2023
`a, 4a^2 + 4a + 1 = (2a+1)^2`
`b, -3x^2 + 6xy - 3y^2`
` = -3(x-y)^2`
`c, (x+y)^2 - 2(x+y)z + z^2`
`= (x+y-z)^2`
12 tháng 10 2018
1. \(4x^2-17xy+13y^2=4x^2-4xy-13xy+13y^2=4x\left(x-y\right)-13y\left(x-y\right)=\left(x-y\right)\left(4x-13y\right)\)
2. \(2x\left(x-5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)
3. \(A=\left(2a-3b\right)^2+2\left(2a-3b\right)\left(3a-2b\right)+\left(2b-3a\right)^2\)
\(\Leftrightarrow\left(2a-3b\right)^2-2\left(2a-3b\right)\left(2b-3a\right)+\left(2b-3a\right)^2=\left(2a-3b-2b+3a\right)^2=\left(5a-5b\right)^2\)
\(=25\left(a-b\right)^2=25\cdot100=2500\)
ZS
2
KT
6 tháng 8 2018
\(4a^2b^2-\left(a^2+b^2-1\right)^2\)
\(=\left[2ab-\left(a^2+b^2-1\right)\right].\left[2ab+\left(a^2+b^2-1\right)\right]\)
\(=\left(2ab-a^2-b^2+1\right)\left(2ab+a^2+b^2+-1\right)\)
\(=\left[1-\left(a-b\right)^2\right]\left[\left(a+b\right)^2-1\right]\)
\(=\left(1-a+b\right)\left(1+a-b\right)\left(a+b+1\right)\left(a+b-1\right)\)
A
17 tháng 10 2018
\(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4.\)
\(=\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(x+3a\right)+a^4.\)
\(=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4.\)
\(=\left(x+5ax+4a^2+a^2\right)^2.\)
\(=\left(x+5ax+5a^2\right)^2.\)
18 tháng 10 2018
\(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\)
\(=\)\(\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(x+3a\right)+a^4\)
\(=\)\(\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\)
\(=\)\(\left[\left(x^2+5ax+5a^2\right)-a^2\right].\left[\left(x^2+5ax+5a^2\right)-a^2\right]+a^4\)
\(=\)\(\left(x^2+5ax+5a^2\right)^2-a^4+a^4\)
\(=\)\(\left(x^2+5ax+5a^2\right)^2\)
Chúc bạn học tốt ~
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~
DN
2
13 tháng 7 2016
4a2b2-(a2+b2-c2)2
= (4ab-a2-b2+c2)(4ab+a2+b2-c2)
= -[(a-b)2-c2][(a+b)2-c2]
=-(a-b+c)(a-b-c)(a+b-c)(a+b+c)
=(b-a-c)(b+c-a)(a+b-c)(a+b+c)
13 tháng 7 2016
\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)
\(=\left(2ab\right)^2-\left(a^2+b^2-c^2\right)^2\)
\(=\left(2ab-a^2-b^2+c^2\right)\left(2ab+a^2+b^2-c^2\right)\)
a) \(\left(a^2+b^2-5\right)^2-2\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5\right)^2-\left(\sqrt{2}.ab+\sqrt{2}.2\right)^2\)
\(=\left(a^2+b^2-5-\sqrt{2}.ab-\sqrt{2}.2\right).\left(a^2+b^2-5+\sqrt{2}.ab+\sqrt{2}.2\right)\)
b) \(\left(4a^2-3a-18\right)^2-\left(4a^2+3a\right)^2\)
\(\left(4a^2-3a-18-4a^2-3a\right).\left(4a^2-3a-18+4a^2+3a\right)\)
\(=\left(-6a-18\right).\left(8a^2-18\right)\)
\(=\left(-6\right).\left(a+3\right).2.\left(4a^2-9\right)\)
\(=\left(-12\right).\left(a+3\right).\left(2a-3\right).\left(2a+3\right)\)
a) Xem lại đề
b) ( 4a2 - 3a - 18 )2 - ( 4a2 + 3a )2
= [ ( 4a2 - 3a - 18 ) - ( 4a2 + 3a ) ][ ( 4a2 - 3a - 18 ) + ( 4a2 + 3a ) ]
= ( 4a2 - 3a - 18 - 4a2 - 3a )( 4a2 - 3a - 18 + 4a2 + 3a )
= ( -6a - 18 )( 8a2 - 18 )
= -6( a + 3 ).2( 4a2 - 9 )
= -12( a + 3 )( 4a2 - 9 )
= -12( a + 3 )( 2a - 3 )( 2a + 3 )