Phân tích đa thức thành nhân tử
a) (4x^2 - 3x - 18)^2 - (4x^2 + 3x)^2
b) 9(x + y - 1)^2 - 4(2x + 3y +1)^2
c) -4x^2 + 12xy - 9y^2 + 25
d) x^2 - 2xy + y^2 - 4m^2 + 4mn - n^2
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2 tháng 10 2020
a) ( 4x2 - 3x - 18 )2 - ( 4x2 + 3x )2
= [ ( 4x2 - 3x - 18 ) - ( 4x2 + 3x ) ][ ( 4x2 - 3x - 18 ) + ( 4x2 + 3x ) ]
= ( 4x2 - 3x - 18 - 4x2 - 3x )( 4x2 - 3x - 18 + 4x2 + 3x )
= ( -6x - 18 )( 8x2 - 18 )
= -6( x + 3 ).2( 4x2 - 9 )
= -12( x + 3 )( 2x - 3 )( 2x + 3 )
b) 9( x + y - 1 )2 - 4( 2x + 3y + 1 )2
= 32( x + y - 1 )2 - 22( 2x + 3y + 1 )2
= [ 3( x + y - 1 ) ]2 - [ 2( 2x + 3y + 1 ) ]2
= ( 3x + 3y - 3 )2 - ( 4x + 6y + 2 )2
= [ ( 3x + 3y - 3 ) - ( 4x + 6y + 2 ) ][ ( 3x + 3y - 3 ) + ( 4x + 6y + 2 ) ]
= ( 3x + 3y - 3 - 4x - 6y - 2 )( 3x + 3y - 3 + 4x + 6y + 2 )
= ( -x - 3y - 5 )( 7x + 9y - 1 )
c) -4x2 + 12xy - 9y2 + 25
= 25 - ( 4x2 - 12xy + 9y2 )
= 52 - ( 2x - 3y )2
= [ 5 - ( 2x - 3y ) ][ 5 + ( 2x - 3y ) ]
= ( 5 - 2x + 3y )( 5 + 2x - 3y )
d) x2 - 2xy + y2 - 4m2 + 4mn - n2
= ( x2 - 2xy + y2 ) - ( 4m2 - 4mn + n2 )
= ( x - y )2 - ( 2m - n )2
= [ ( x - y ) - ( 2m - n ) ][ ( x - y ) + ( 2m - n ) ]
= ( x - y - 2m + n )( x - y + 2m - n )
TG
30 tháng 7 2020
a) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)
\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)
\(=\left(-6x-18\right)\left(8x^2-18\right)\)
b) \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left[3\left(x+y-1\right)\right]^2-\left[2\left(2x+3y+1\right)\right]^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3+4x+6y+2\right)\left(3x+3y-3-4x-6y-2\right)\)
\(=\left(7x+9y-1\right)\left(-x-3y-5\right)\)
c) \(-4x^2+12xy-9y^2+25\)
\(=-\left(2x\right)^2+2.2x.3y-\left(3y\right)^2+5^2\)
\(=-\left[\left(2x\right)^2-2.2x.3y+\left(3y\right)^2-5^2\right]\)
\(=-\left[\left(2x-3y\right)^2-5^2\right]\)
\(=-\left(2x-3y-5\right)\left(2x-3y+5\right)\)
d) \(x^2-2xy+y^2-4m^2+4mn-n^2\)
\(=\left(x^2-2xy+y^2\right)-4m\left(m-n\right)-n^2\)
\(=\left(x-y\right)^2-4m\left(m-n\right)-n^2\)
\(=\left(x-y-n\right)\left(x-y+n\right)-4m\left(m-n\right)\)
21 tháng 8 2017
a)\(\left(3x-1\right)^2-16=\left(3x-1-16\right)\left(3x-1+16\right)\)
\(=\left(3x-17\right)\left(3x+15\right)\)
c)\(\left(2x+5\right)^2-\left(x-9\right)^2=\left(2x+5+x-9\right)\left(2x+5-x+9\right)\)
\(=\left(x-4\right)\left(x+14\right)\)
Aps dungj t/c a2 - b2 = ( a-b)(a+b)
TL
18 tháng 6 2019
App giải toán không cần nhập đề chỉ cần chụp ảnh cho cả nhà đây: https://www.facebook.com/watch/?v=485078328966618
21 tháng 7 2018
\(a,\left(3x-1\right)^2-16=\left(3x-5\right)\left(3x+3\right)=3\left(x+1\right)\left(3x-5\right)\)
\(b,\left(5x-4\right)^2-49x^2=\left(12x-4\right)\left(-2x-4\right)=4\left(3x-1\right)\left(-2\right)\left(x+2\right)=-8\left(3x-1\right)\left(x+2\right)\)\(c,\left(2x+5\right)^2-\left(x-9\right)^2=\left(3x-4\right)\left(x+14\right)\)
21 tháng 7 2018
a,\(\left(3x+1\right)^2-16=\left(3x-1-16\right)\left(3x-1+16\right)\\ =\left(3x-17\right)\left(3x+15\right)\)
16 tháng 8 2020
a) Ta có: \(\left(3x-1\right)^2-16\)
\(=\left(3x-1-4\right)\left(3x-1+4\right)\)
\(=\left(3x-5\right)\left(3x+3\right)\)
\(=3\left(x+1\right)\left(3x-5\right)\)
b) Ta có: \(\left(5x-4\right)^2-49x^2\)
\(=\left(5x-4-7x\right)\left(5x-4+7x\right)\)
\(=\left(-2x-4\right)\left(12x-4\right)\)
\(=-2\left(x+2\right)\cdot4\cdot\left(3x-1\right)\)
\(=-8\left(x+2\right)\left(3x-1\right)\)
c) Ta có: \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)
\(=\left(x+14\right)\left(3x-4\right)\)
d) Ta có: \(\left(3x+1\right)^2-4\left(x-2\right)^2\)
\(=\left(3x+1\right)^2-\left(2x-4\right)^2\)
\(=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\)
\(=\left(x+5\right)\left(5x-3\right)\)
e) Ta có: \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)
\(=\left(6x+9\right)^2-\left(2x+2\right)^2\)
\(=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\)
\(=\left(4x+7\right)\left(8x+11\right)\)
f) Ta có: \(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)
\(=-\left(b^2-2bc+c^2-a^2\right)\left[\left(b^2+2bc+c^2\right)-a^2\right]\)
\(=-\left[\left(b-c\right)^2-a^2\right]\cdot\left[\left(b+c\right)^2-a^2\right]\)
\(=-\left(b-c-a\right)\left(b-c+a\right)\left(b+c-a\right)\left(b+c+a\right)\)
g) Ta có: \(\left(ax+by\right)^2-\left(ay+bx\right)^2\)
\(=\left(ax+by-ay-bx\right)\left(ax+by+ay+bx\right)\)
\(=\left[a\left(x-y\right)+b\left(y-x\right)\right]\left[a\left(x+y\right)+b\left(x+y\right)\right]\)
\(=\left[a\left(x-y\right)-b\left(x-y\right)\right]\left(x+y\right)\left(a+b\right)\)
\(=\left(x-y\right)\left(a-b\right)\left(x+y\right)\left(a+b\right)\)
h) Ta có: \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5\right)^2-\left(2ab+4\right)^2\)
\(=\left(a^2+b^2-5+2ab+4\right)\left(a^2+b^2-5-2ab-4\right)\)
\(=\left[\left(a^2+2ab+b^2\right)-1\right]\left[\left(a^2-2ab+b^2\right)-9\right]\)
\(=\left(a+b-1\right)\left(a+b+1\right)\left(a-b-3\right)\left(a-b+3\right)\)
i) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)
\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)
\(=\left(-6x-18\right)\left(8x^2-18\right)\)
\(=-6\left(x+3\right)\cdot2\left(x^2-9\right)\)
\(=-12\left(x+3\right)^2\cdot\left(x-3\right)\)
k) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)
l) Ta có: \(-4x^2+12xy-9y^2+25\)
\(=-\left(4x^2-12xy+9y^2-25\right)\)
\(=-\left[\left(2x-3y\right)^2-5^2\right]\)
\(=-\left(2x-3y-5\right)\left(2x-3y+5\right)\)
m) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)
\(=\left(x-y\right)^2-\left(4m^2-4mn+n^2\right)\)
\(=\left(x-y\right)^2-\left(2m-n\right)^2\)
\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)
21 tháng 5 2022
a: \(=\left(3x+1\right)^2-\left(2x-4\right)^2\)
\(=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\)
\(=\left(x+5\right)\left(5x-3\right)\)
b: \(=\left(6x+9\right)^2-\left(2x+2\right)^2\)
\(=\left(6x+9+2x+2\right)\left(6x+9-2x-2\right)\)
\(=\left(8x+11\right)\left(4x+7\right)\)
c: \(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)
\(=\left(-6x-18\right)\left(8x^2-18\right)\)
\(=-6\cdot2\cdot\left(x+3\right)\left(4x^2-9\right)\)
\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)
25 tháng 1 2023
a: =(2x-3y)^2-4(2x-3y)
=(2x-3y)(2x-3y-4)
b: =3x^2+21x-x-7
=(x+7)(3x-1)
c: =(3x-1)^4+2(3x-1)^2+1
=[(3x-1)^2+1]^2
d: =2x^3-2x^2-x^2+x+x-1
=(x-1)(2x^2-x+1)
a) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)
\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)
\(=\left(-6x-18\right)\left(8x^2-18\right)\)
\(=-6\left(x+3\right)\cdot2\left(4x^2-9\right)\)
\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)
b) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=-\left(x+3y+5\right)\left(7x+9y-1\right)\)
c) Ta có: \(-4x^2+12xy-9y^2+25\)
\(=-\left(4x^2-12xy+9y^2-25\right)\)
\(=-\left[\left(2x-3y\right)^2-25\right]\)
\(=-\left(2x-3y-5\right)\left(2x-3y+5\right)\)
d) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)
\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)
\(=\left(x-y\right)^2-\left(2m-n\right)^2\)
\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)