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d) ax2 + ay - bx2 - by
= ( ax2 + ay ) - ( bx2 + by )
= a ( x2 + y ) - b ( x2 + y )
= ( x2 + y )( a - b )
c) x2y + xy2 - x - y
= ( x2y + xy2 ) - ( x + y )
= xy ( x + y ) - ( x+ y )
= ( x + y ) ( xy - 1 )
`ab(x-3) -a^2(x-3)`
`=(x-3)(a^2-ab)`
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`ax+ay+bx+by`
`=a(x+y)+b(x+y)`
`=(x+y)(a+b)`
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`ax+ay -2x-2y`
`=(ax+ay)-(2x+2y)`
`=a(x+y)-2(x+y)`
`=(x+y)(a-2_`
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`2x-2y +ax-ay`
`=2(x-y)+a(x-y)`
`=(x-y)(2+a)`
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`10ax -5ay -2x+y`
`= 5a(2x-y) -(2x-y)`
`=(2x-y)(5a-1)`
1: =(x-3)(ab-a^2)
=a(b-a)(x-3)
2: =a(x+y)+b(x+y)
=(x+y)(a+b)
3: =a(x+y)-2(x+y)
=(x+y)(a-2)
4: =2(x-y)+a(x-y)
=(x-y)(a+2)
5: =5a(2x-y)-(2x-y)
=(2x-y)(5a-1)
bn post nhiều nên mình ghi đáp án thôi nhé phần nào sai đề mình cho qua
b)\(\left(x+1\right)\left(xy+1\right)\)
c)\(\left(a+b\right)\left(x+y\right)\)
d)\(\left(x-a\right)\left(x-b\right)\)
e)\(\left(x+y\right)\left(xy-1\right)\)
f)\(\left(a-b\right)\left(x^2+y\right)\)
\(a,\left(3x-1\right)^2-16=\left(3x-5\right)\left(3x+3\right)=3\left(x+1\right)\left(3x-5\right)\)
\(b,\left(5x-4\right)^2-49x^2=\left(12x-4\right)\left(-2x-4\right)=4\left(3x-1\right)\left(-2\right)\left(x+2\right)=-8\left(3x-1\right)\left(x+2\right)\)\(c,\left(2x+5\right)^2-\left(x-9\right)^2=\left(3x-4\right)\left(x+14\right)\)
a,\(\left(3x+1\right)^2-16=\left(3x-1-16\right)\left(3x-1+16\right)\\ =\left(3x-17\right)\left(3x+15\right)\)
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: \(x^2-2xy+y^2-4m^2+4mn-n^2\)
\(=\left(x-y\right)^2-\left(4m^2-4mn+n^2\right)\)
\(=\left(x-y\right)^2-\left(2m-n\right)^2\)
\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)
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)
a) \(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc+b^2+c^2-a^2\right)\left(2bc-b^2-c^2+a^2\right)\)
\(=\left[\left(b+c\right)^2-a^2\right]\left[a^2-\left(b-c\right)^2\right]\)
\(=\left(b+c+a\right)\left(b+c-a\right)\left(a+b-c\right)\left(a-b+c\right)\)
b) \(\left(ax+by\right)^2-\left(ay+bx\right)^2\)
\(=\left(ax+by+ay+bx\right)\left(ax+by-ay-bx\right)\)
\(=\left(a+b\right)\left(x+y\right)\left(a-b\right)\left(x-y\right)\)
c) \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5+2ab+4\right)\left(a^2+b^2-5-2ab-4\right)\)
\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)
\(=\left(a+b+1\right)\left(a+b-1\right)\left(a-b+3\right)\left(a-b-3\right)\)
d) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)
\(=\left(4x^2-3x-18+4x^2+3x\right)\left(4x^2-3x-18-4x^2-3x\right)\)
\(=\left(8x^2-18\right)\left(-6x-18\right)\)
\(=\left[2\left(4x^2-9\right)\right]\left[-6\left(x+3\right)\right]\)
\(=12\left(2x+3\right)\left(2x-3\right)\left(x+3\right)\)