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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) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)
\(=\left(4x^2-3x-18+4x^2+3x\right)\left(4x^2-3x-18-4x^2-3x\right)\)
\(=\left(8x^2-18\right)\left(-6x-18\right)\)
\(=\left[2\left(4x^2-9\right)\right]\left[-6\left(x+3\right)\right]\)
\(=12\left(2x+3\right)\left(2x-3\right)\left(x+3\right)\)
\(\left(ax+by\right)^2-\left(ay+bx\right)^2\)
\(=\left(ax+by+ay+bx\right)\left(ax+by-ay-bx\right)\)
\(=\left[a\left(x+y\right)+b\left(x+y\right)\right]\left[a\left(x-y\right)-b\left(x-y\right)\right]\)
\(=\left(a+b\right)\left(a-b\right)\left(x+y\right)\left(x-y\right)\)
\(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left[\left(a^2+b^2-5\right)+2\left(ab+2\right)\right]\left[\left(a^2+b^2-5\right)-2\left(ab+2\right)\right]\)
\(=\left[a^2+b^2-5+2ab+4\right]\left[a^2+b^2-5-2ab-4\right]\)
\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)
\(=\left(a+b-1\right)\left(a+b+1\right)\left(a-b-3\right)\left(a-b+3\right)\)
a)
(ax+by)2 - (ay+bx)2
=(ax+by-ay-bx)(ax+by+ay+bx)
=[ a(x-y) -b(x-y)][ a(x+y) + b(x+y)]
=(a-b)(x-y)(a+b)(x+y)
b)(a2+b2-5)2 - 4(ab+2)2
=(a2+b2-5-2ab-4)(a2+b2-5+2ab+4)
=[ (a-b)2 -9][ (a+b)2 -1]
=(a-b-3)(a-b+3)(a+b-1)(a+b+1)
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 )
Ta có ; x2 - 11x + 24
= x2 - 3x - 8x + 24
= x(x - 3) - (8x - 24)
= x(x - 3) - 8(x - 3)
= (x - 3)(x - 8)
a) \(4x^3\left(x^2+x\right)-\left(x^2+x\right)=\left(x^2+x\right)\left(4x^3-1\right)\)
b)\(\left(1-2a+a^2\right)-\left(b^2-2bc+c^2\right)=\left(1-a\right)^2-\left(b-c\right)^2=\)\(\left(1-a+b-c\right)\left(1-a-b+c\right)\)
lm tiếp câu c
c) \(C=\left(x-7\right)\left(x-5\right)\left(x-4\right)\left(x-2\right)-72\)
\(=\left[\left(x-7\right)\left(x-2\right)\right]\left[\left(x-5\right)\left(x-4\right)\right]-72\)
\(=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\)
Đặt \(x^2-9x+17=a\) ta có:
\(C=\left(a-3\right)\left(a+3\right)-72\)
\(=a^2-9-72\)
\(=a^2-81=\left(a-9\right)\left(a+9\right)\)
Thay trở lại ta được: \(C=\left(x^2-9x++8\right)\left(x^2-9x+26\right)\)
\(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)
\(=\left[a^2-\left(b^2-2bc+c^2\right)\right].\left[\left(b^2+2bc+c^2\right)-a^2\right]\)
\(=\left[a^2-\left(b-c\right)^2\right].\left[\left(b+c\right)^2-a^2\right]\)
\(=\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)\)
\(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)
\(=\left[\left(a-b\right)^2-3^2\right].\left[\left(a+b\right)^2-1\right]\)
\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)
Tham khảo nhé~