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\(\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)
\(=a^2-2a\left(b-c\right)+\left(b-c\right)^2-\left(b-c\right)^2+2a\left(b-c\right)\)
\(=a^2-2a\left(b-c\right)+2a\left(b-c\right)\)
\(=a^2\)
ta có : a+b+c=0=>a+b=-c ; b+c=-a ; a+c=-b
ta có: M= \(\frac{2ab}{a^2+\left(b+c\right)\left(b-c\right)}+\frac{2bc}{b^2+\left(c+a\right)\left(c-a\right)}+\frac{2ca}{c^2+\left(a+b\right)\left(a-b\right)}\)
M=\(\frac{2ab}{a^2-a\left(b-c\right)}+\frac{2bc}{b^2-b\left(c-a\right)}+\frac{2ca}{c^2-c\left(a-b\right)}\)
M=\(\frac{2ab}{a\left(a-b+c\right)}+\frac{2bc}{b\left(b-c+a\right)}+\frac{2ca}{c\left(c-a+b\right)}\)
M=\(\frac{2ab}{-ab+\left(a+c\right)}+\frac{2bc}{-bc+\left(a+b\right)}+\frac{2ac}{-ac+\left(b+c\right)}\)
M=\(\frac{2ab}{-2ab}+\frac{2bc}{-2bc}+\frac{2ca}{-2ca}\)
M=-1-1-1=-3
Vậy với a+b+c=0 thì M=-3
b: Ta có: \(\left(4x-y\right)\left(4x+y\right)-2\left(3x-2y\right)^2+\left(x-3y\right)^2\)
\(=16x^2-y^2-2\left(9x^2-12xy+4y^2\right)+x^2-6xy+9y^2\)
\(=17x^2-6xy+8y^2-18x^2+24xy-8y^2\)
\(=-x^2+18xy\)
c: Ta có: \(\left(2a-3b+4c\right)\left(2a-3b-4c\right)\)
\(=\left(2a-3b\right)^2-16c^2\)
\(=4a^2-12ab+9b^2-16c^2\)
Tham khảo:
Cho a≠b≠c, a+b≠c và c2+2ab-2ac-2bc=0 Hãy rút gọn \(B=\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\) - Hoc24
\(\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\)
\(=\dfrac{a^2+a^2-2ac+c^2}{b^2+b^2-2bc+c^2}\)
\(=\dfrac{2a^2-2ac+c^2}{2b^2-2bc+c^2}\)
\(\left(a-b+c\right)^2=\left[a+\left(-b\right)+c\right]^2\)
\(=a^2+\left(-b^2\right)+c^2+2.a.\left(-b\right)+2.\left(-b\right)\left(-c\right)+2.c.a\)
\(=a^2+b^2+c^2-2ab-2bc+2ca\)