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Bài 1:
a) \(\left(a-b^2\right)\left(a+b^2\right)=a^2-b^4\)
b) \(\left(a^2+2a-3\right)\left(a^2+2a+3\right)=\left(a^2+2a\right)^2-9\)
c) \(\left(a^2+2a+3\right)\left(a^2-2a-3\right)=a^2-\left(2a+3\right)^2\)
d) \(\left(a^2-2a+3\right)\left(a^2+2a+3\right)=9-\left(a^2-2a\right)^2\)
e) \(\left(-a^2-2a+3\right)\left(-a^2-2a+3\right)=\left(-a^2-2a+3\right)^2\)
g) \(\left(a^2+2a+3\right)\left(a^2-2a+3\right)=\left(a^2+3\right)^2-4a^2\)
f) \(\left(a^2+2a\right)\left(2a-a^2\right)=4a^2-a^4\)
Bài 2 :
a) \(\left(x+1\right)\left(x^2-x+1\right)=x^3+1\)
b) \(\left(x+y+z\right)^2=\left(x+y+z\right)\left(x+y+z\right)=x^2+xy+xz+yx+y^2+yz+zx+zy+z^2=x^2+2xy+2yz+2xz+y^2+z^2\)
c) \(\left(x-y+z\right)^2=\left(x-y+z\right)\left(x-y+z\right)=x^2-xy+xz-xy+y^2-yz+xz-yz+z^2=x^2+y^2+z^2-2xy+2xz-2yz\)d) \(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=\left(x-2y\right)^3\)
e) \(\left(x-y-z\right)^2=\left(x-y-z\right)\left(x-y-z\right)=x^2-xy-xz-xy+y^2+yz-xz+yz+z^2=x^2-2xy-2xz+2yz+y^2+z^2\)
Bài 2:
a) =a2b - a2c + b2c - ab2 + ac2 - bc2
=(a2b - bc2) - (a2c - ac2) + (b2c - ab2)
=b(a-c)(a+c) - ac(a-c) - b2(a-c)
=(a - c)(ab -bc - ac - b2)
b)=(1 - 2a + a2) - (b2 - 2bc + c2)
=(1 - a)2 - (b - c)2
=(c - b - a + 1)(b - c - a + 1)
1) a) A= (x+2)(\(x^2-2x+4\)) -(\(x^3-2\))
=(x+2)(\(x^2-2x+2^2\))-(\(x^3-2\))
= \(x^3+2^3\)-\(x^3+2\)
=(\(x^3-x^3\))+(\(2^3+2\))
=10
b) B= (a+2)(a-2)(\(a^2+2a+4\))(\(a^2-2a+4\))
= \(a^2-2^2\)+\(a^2+\left(2a\right)^2+4^2\)
=\(a^2-4+a^2+4a^2+16\)
=(\(a^2+a^2+4a^2\))+(-4+16)
=\(6a^2\)+12
a) ( a - 3)x = a2 + 1
x = \(\dfrac{a^2+1}{a-3}\)( với a # 3 )
Vậy , ....
b) a2x + x = 2a2 - 3
x( a2 + 1) = 2a2 - 3
x =\(\dfrac{2a^2-3}{a^2+1}\)
Vậy ,....