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\(T=\frac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}+\frac{b^2}{\left(b-c\right)\left(b+c\right)-a^2}+\frac{c^2}{\left(c-a\right)\left(c+a\right)-b^2}\)
\(=\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}\)
\(=\frac{a^2}{a^2-\left(b+c\right)^2+2bc}+\frac{b^2}{b^2-\left(c+a\right)^2+2ca}+\frac{c^2}{c^2-\left(a+b\right)^2+2ab}\)
\(=\frac{a^2}{a^2-\left(-a\right)^2+2bc}+\frac{b^2}{b^2-\left(-b\right)^2+2ca}+\frac{c^2}{c^2-\left(-c\right)^2+2ab}\)
\(=\frac{a^2}{2bc}+\frac{b^2}{2ca}+\frac{c^2}{2ab}\)
\(=\frac{a^3+b^3+c^3}{2abc}\)
Từ \(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\) ( tự chứng minh nhé )
\(\Rightarrow T=\frac{3abc}{2abc}=\frac{3}{2}\)
Vậy T=3/2
a: =b-c-a+c+1-a-b+c
=-2a+1
b: =a-b-c-b+c+a+c-b-a
=c-3b+a
c: =2(a-b-b+c-c+a)
=2(2a-2b)
=4a-4b
a) \(\left(b-c\right)-\left(a-c-1\right)-\left(a+b-c\right)\)
\(=b-c-a+c+1-a-b+c\)
\(=c-2a+1\)
b) \(\left(a-b-c\right)-\left(b-c-a\right)+\left(c-b-a\right)\)
\(=a-b-c-b+c+a+c-b-a\)
\(=a-3b+c\)
c) \(2\cdot\left(a-b\right)-2\cdot\left(b-c\right)-2\cdot\left(c-a\right)\)
\(=2\cdot\left(a-b-b+c-c+a\right)\)
\(=2\cdot\left(2a-2b\right)\)
\(=4a-4b\)
=(b+c)(ac-a2+bc-ab)+(b+c)(ac-bc+a2-ab)+(c+a)(a+b)(b-c)
=(b+c)(ac-a2+bc-ab+ac-bc+a2-ab)+(a+c)(a+b)(b-c)
=(b+c)(2ac-2ab)-(a+c)(a+b)(c-b)
=(b+c).2a.(c-b)-(a2+ab+ac+bc)(c-b)
=(c-b)(2ab+2ac-a2-ab-ac-bc)
=(c-b)(-a2+ab+ac-bc)=(c-b)[a(b-a)-c(b-a)]
=(c-b)(b-a)(a-c)
Ta có: \(\left(b-c\right)^3+\left(c-a\right)^3-\left(a-b\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(=\left(b-c+c-a\right)\left[\left(b-c\right)^2-\left(b-c\right)\left(c-a\right)+\left(c-a\right)^2\right]-\left(a-b\right)\left[1+3\left(b-c\right)\left(c-a\right)\right]\)
\(=\left(b-a\right)\left(b^2-3bc+3c^2+ab-3ac+a^2\right)-\left(a-b\right)\left(1+3bc-3ab-3c^2+3ac\right)\)
\(=\left(b-a\right)\left(b^2-3bc+3c^2+ab-3ac+a^2+1+3bc-3ab-3c^2+3ac\right)\)
\(=\left(b-a\right)\left(b^2-2ab+a^2+1\right)\)
\(=\left(b-a\right)^3+\left(b-a\right)\)
\(=b^3-3b^2a+3ba^2-a^3+b-a\)
\(T=\left(a+b-c\right)-\left(b-a\right)-\left(a-b-c\right)\)
\(=a+b-c-b+a-a+b+c\)
\(=a+b\)
Vậy \(T=a+b\)
T = ( a + b - c ) - ( b - a )- ( a - b - c )
T = a + b - c - b + a - a + b + c
T =(a + a - a ) + ( b - b + b ) + ( -c + c )
T = a + b + 0
T = a + b