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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)
Sửa lại ạ!
a) \(\left(3x-1\right)^2-16\)
\(=\left(3x-1\right)^2-4^2\)
\(=\left(3x-1-4\right)\left(3x-1+4\right)\)
\(=\left(3x-5\right)\left(3x+3\right)\)
b) \(\left(5x-4\right)^2-49x^2\)
\(=\left(5x-4\right)^2-\left(7x\right)^2\)
\(=\left(5x-4-7x\right)\left(5x-4+7x\right)\)
\(=\left(-4-2x\right)\left(-4+12x\right)\)
c) \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)
\(=\left(x+14\right)\left(3x-4\right)\)
d) \(\left(3x+1\right)^2-4\left(x-2\right)^2\)
\(=\left(3x+1\right)^2-\left[2\left(x-2\right)\right]^2\)
\(=\left(3x+1\right)^2-\left(2x-4\right)^2\)
\(=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\)
\(=\left(x+5\right)\left(5x-3\right)\)
e) \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)
\(=\left[3\left(2x+3\right)\right]^2-\left[2\left(x+1\right)\right]^2\)
\(=\left(6x+9\right)^2-\left(2x+2\right)^2\)
\(=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\)
\(=\left(4x+7\right)\left(8x+11\right)\)
P/s: Ko chắc!
a) \(\left(2x+5\right)^2\)\(-\left(x-9\right)^2\)
=\(\left(2x+5+x-9\right).\left(2x+5-x+9\right)\)
=\(\left(3x-4\right).\left(x+14\right)\)
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) ( 3x -1 )2 - 16
= (3x -1 ) 2 - 42
= ( 3x -1 -4 ).( 3x -1 +4 )
b) ( 5x-4 ) 2 - 49x2
= ( 5x-4 ) 2 - (7x)2
=( 5x -4 -7x).( 5x -4 + 7x )
=( -2x -4 ) .( 12x -4 )
còn lại giống tương tự nha pạn
~ hok tốt ~
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
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) \(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc\right)^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc+b^2+c^2-a^2\right)\left(2bc-b^2-c^2+a^2\right)\)
\(=-\left[\left(b+c\right)^2-a^2\right]\left(b^2-2bc+c^2-a^2\right)\)
\(=-\left(b+c-a\right)\left(b+c+a\right)\left[\left(b-c\right)^2-a^2\right]\)
\(=-\left(b+c-a\right)\left(b+c+a\right)\left(b-c-a\right)\left(b-c+a\right)\)
d) \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5\right)^2-\left(2ab+4\right)^2\)
\(=\left(a^2+b^2-5+2ab+4\right)\left(a^2+b^2-5-2ab-4\right)\)
\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-3^2\right]\)
\(=\left(a+b+1\right)\left(a+b-1\right)\left(a-b-3\right)\left(a-b+3\right)\)
\(a,\left(3x-1\right)^2-16=\left(3x-5\right)\left(3x+3\right)=3\left(x+1\right)\left(3x-5\right)\)
\(b,\left(5x-4\right)^2-49x^2=\left(12x-4\right)\left(-2x-4\right)=4\left(3x-1\right)\left(-2\right)\left(x+2\right)=-8\left(3x-1\right)\left(x+2\right)\)\(c,\left(2x+5\right)^2-\left(x-9\right)^2=\left(3x-4\right)\left(x+14\right)\)
a,\(\left(3x+1\right)^2-16=\left(3x-1-16\right)\left(3x-1+16\right)\\ =\left(3x-17\right)\left(3x+15\right)\)