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\(\left(a-b\right).\left(b-c\right).\left(c-a\right)\)
\(=\left(ab-ac-b^2+bc\right).\left(c-a\right)\)
\(=abc-a^2b-ac^2+a^2c-b^2c+ab^2+bc^2-abc\)
\(=-a^2b-ac^2+a^2c-b^2c+ab^2+bc^2.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{\left(a-b\right)\cdot\left(b-c\right)}-\frac{1}{\left(a-c\right)\cdot\left(b-c\right)}-\frac{1}{\left(a-b\right)\cdot\left(a-c\right)}\)
\(=\frac{a-c-\left(a-b\right)-\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{a-c-a+b-b+c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
\(\frac{1}{\left(a-b\right).\left(b-c\right)}-\frac{1}{\left(a-c\right).\left(b-c\right)}-\frac{1}{\left(a-b\right).\left(a-c\right)}\)
=\(\frac{a-c}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}-\frac{a-b}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}-\frac{b-c}{\left(a-b\right).\left(b-c\right).\left(c-a\right)}\)
=\(\frac{\left(a-c\right)-\left(a-b\right)-\left(b-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)
=\(\frac{a-c-a+b-b+c}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)
=\(\frac{\left(a-a\right)+\left(b-b\right)+\left(c-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)
=\(\frac{0}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)
=0
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a) (A + B)2 = A2 + 2AB + B2
b) (A - B)2 = A2 - 2AB + B2
c) (A + B)(A - B) = A2 - B2
d) (A + B + C)2 = A2 + B2 + C2 + 2AB + 2BC + 2AC
e) (A + B - C)2 = A2 + B2 + C2 + 2AB - 2BC - 2AC
f) (A - B - C)2 = A2 + B2 + C2 - 2AB + 2BC - 2AC
a) \(\left(A+B\right)^2=A^2+2AB+B^2\)
b) \(\left(A-B\right)^2=A^2-2AB+B^2\)
c) \(\left(A+B\right)\left(A-B\right)=A^2-B^2\)
d) .........đây là các hằng đẳng thức thôi mà
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a) x 3 – 3 x 2 + 3x – 1;
b) – x 4 + 7 x 3 – 11 x 2 + 6x – 5;
c) c 3 + 2 c 2 – 5c – 6.
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a) (x - 2)(x2 - 2x + 4)(x - 2)( x2 + 2x + 4)
= (x - 2)2(x - 2)2(x + 2)2
= (x - 2)4(x + 2)2
b) (a + b + c)3 - (b + c - a)3 - (a - b + c)3 - (a + b - c)3
Đặt a+b-c=x, c+a-b=y, b+c-a=z
=>x+y+z=a+b-c+c+a-b+b+c-a=a+b+c
Ta có hằng đẳng thức:
(x+y+z)^3-3x-3y-3z=3(x+y)(x+z)(y+z)
=>(a+b+c)^3-(b+c-a)^3-(a+c-b)^3-(a+b-c)^3=(x+y+z)^3-x^3-y^3-z^3
=3(x+y)(x+z)(y+z)
=3(a+b-c+c+a-b)(c+a-b+b+c-a)(b+c-a+a+b-c)
=3.2a.2b.2c
=24abc
c) (a + b)3 + (b + c)3 + (c + a)3 - 3(a + b)(b + c)(c + a)
Đặt x = a+b; y = b+c; z = c+a ta có:
x3+y3+z3−3xyz
= (x+y)3−3xy(x−y)+z3−3xyz
=[(x+y)3+z3]−3xy(x+y+z)
=(x+y+z)3−3z(x+y)(x+y+z)−3xy(x−y−z)
=(x+y+z)[(x+y+z)2−3z(x+y)−3xy]
=(x+y+z)(x2+y2+z2+2xy+2xz+2yz−3xz−3yz−3xy)
=(x+y+z)(x2+y2+z2−xy−yz−yx)
Thay vào ta có:
(a+b+b+c+c+a)[(a+b)2+(b+c)2+(c+a)2−(a+b)(b+c)−(b+c)(c+a)−(c+a)(a+b)]
=(2a+2b+2c)(a2−ab−ac+b2−bc+c2)
=2(a+b+c)(a2−ab−ac+b2−bc+c2)
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a)\(\left(x-2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x^2+2x-4\right)\)
\(=\left(x-2\right)^2\left(x^2-2x+4\right)\left(x^2+2x-4\right)\)
\(=\left(x-2\right)^2\left(x^4+4x^2+16\right)\)
\(=x^6-4x^5+8x^4-16x^3+32x^2-64x+64\)
=2*(c^2+b^2+a^2)