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3(a - b)2 - 2(a + b)2
= 3(a2 - 2ab + b2) - 2(a2 + 2ab + b2)
= 3a2 - 6ab + 3b2 - 2a2 - 4ab - 2b2
= a2 - 2ab + b2
= (a - b)2
a) ( a+ b )(a^4 - a^3.b + a^2.b^2 - ab^3 + b^4)
= a^5 - a^4b + a^3b^2 - a^2b^3 + ab^4 + a^4b - a^3b^2 + a^2b^3 - ab^4 + b^5
= a^5 + b^5
Thật ra đây là HĐT mở rộng
b) ( x - a )(x - b)( x- c)
= ( x^2 - bx - ax + ab )(x - c)
= x^3 - bx^2 - ax^2 + abc - cx^2 + bcx + acx - abc
\(\frac{1}{\left(a-b\right)\cdot\left(b-c\right)}-\frac{1}{\left(a-c\right)\cdot\left(b-c\right)}-\frac{1}{\left(a-b\right)\cdot\left(a-c\right)}\)
\(=\frac{a-c-\left(a-b\right)-\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{a-c-a+b-b+c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
\(\frac{1}{\left(a-b\right).\left(b-c\right)}-\frac{1}{\left(a-c\right).\left(b-c\right)}-\frac{1}{\left(a-b\right).\left(a-c\right)}\)
=\(\frac{a-c}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}-\frac{a-b}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}-\frac{b-c}{\left(a-b\right).\left(b-c\right).\left(c-a\right)}\)
=\(\frac{\left(a-c\right)-\left(a-b\right)-\left(b-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)
=\(\frac{a-c-a+b-b+c}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)
=\(\frac{\left(a-a\right)+\left(b-b\right)+\left(c-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)
=\(\frac{0}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)
=0
a) MTC = a3 + b3 = (a+b)(a2 – ab + b2)