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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) (x-a)^4-(x+a)^4
=[(x-a)^2]^2-[(x+a)^2]^2
=[(x-a)^2-(x+a)^2][(x-a)^2+(x+a)^2]
=[(x-a-x-a)(x-a+x+a)][(x-a)^2+(x+a)^2]
=(-2a.2x)(x^2-2xa+a^2+x^2+2xa+a^2)
=(-2a.2x)(2x^2+2a^2)
=-4ax(2x^2+2a^2)
=-4ax.2(x^2+a^2)
a, \(\left(x+y\right)^2-4\left(x+y\right)+4=\left(x+y\right)^2-2.\left(x+y\right)+2^2=\left(x+y-2\right)^2\)
b, \(4b^2c^2-\left(b^2+c^2-a^2\right)^2=\left(2bc\right)^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)
\(=\left[a^2-\left(b^2-2bc+c^2\right)\right]\left[b^2+2bc+c^2-a^2\right]\)
\(=\left[a^2-\left(b-c\right)^2\right]\left[\left(b+c\right)^2-a^2\right]\)
= \(\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)\)
Chusc bajn hojc toost.
a/ (x-a)^4 - (x+a)^4
=((x-a)^2)^2 - ((x+a)^2)^2
=(x^2 - 2xa + a^2)^2 - (x^2 +2xa+a^2)^2
=(x^2-2xa+a^2-x^2-2xa-a^2)(x^2-2xa+a^2+x^2+2xa+a^2)
=-4xa(2x^2+2a^2)
b/ x^4 –y^2(2x-y)^2
=(x^2)^2-(y(2x-y)^2
=(x^2)^2-(2xy-y^2)^2
=(x^2-2xy+y^2)(x^2+2xy+y^2)
=(x-y)^2 (x+y)^2
c/(xy+4)^2- 4(x+y)^2
=(xy+4)^2- (2x+2y)^2
=(xy+y-2x-2y)(xy+y+2x+2y)
=(xy-y+2x)(xy+3y+2x)
sử dụng tam giác Pascal