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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
\(\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\)
\(=\dfrac{a^2+a^2-2ac+c^2}{b^2+b^2-2bc+c^2}\)
\(=\dfrac{2a^2-2ac+c^2}{2b^2-2bc+c^2}\)
cho \(c^2+2ab-2ac-2bc\)
rút gọn biểu thức \(P=\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\)
Cái đầu ko rút gọn được
Cái sau:
\(=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a+c-b\right)}=\dfrac{a+b-c}{a-b+c}\)
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
\(c^2-2ac+a^2+2ab-2bc=a^2\)
\(\Rightarrow\left(a-c\right)^2+2b\left(a-c\right)=a^2\)
\(c^2-2bc+b^2+2a\left(b-c\right)=b^2\Rightarrow\left(b-c\right)^2+2a\left(b-c\right)=b^2\)
\(\Rightarrow B=\frac{\left(a-c\right)^2+2b\left(a-c\right)+\left(a-c\right)^2}{\left(b-c\right)^2+2a\left(b-c\right)+\left(b-c\right)^2}=\frac{2\left(a-c\right)\left(a-c+b\right)}{2\left(b-c\right)\left(b-c+a\right)}=\frac{a-c}{b-c}\)
Lời giải:
\(\frac{a^2(b-c)+b^2(c-a)+c^2(a-b)}{ab^2-ac^2-b^3+bc^2}=\frac{a^2(b-c)-b^2[(b-c)+(a-b)]+c^2(a-b)}{a(b^2-c^2)-b(b^2-c^2)}\)
\(=\frac{(a^2-b^2)(b-c)-(b^2-c^2)(a-b)}{(a-b)(b^2-c^2)}=\frac{(a-b)(b-c)(a+b-b+c)}{(a-b)(b-c)(b+c)}=\frac{(a-b)(b-c)(a-c)}{(a-b)(b-c)(b+c)}\)
\(=\frac{a-c}{b+c}\)
Ta có: \(\dfrac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
\(=\dfrac{a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)}{a\left(b^2-c^2\right)-b\left(b^2-c^2\right)}\)
\(=\dfrac{a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)}{\left(b^2-c^2\right)\left(a-b\right)}\)
\(=\dfrac{\left(b-c\right)\left(a^2-b^2\right)-\left(a-b\right)\left(b^2-c^2\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\dfrac{\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(b+c\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\dfrac{\left(a-b\right)\left(b-c\right)\left(a+b-b-c\right)}{\left(b-c\right)\left(a-b\right)\left(b+c\right)}\)
\(=\dfrac{a-c}{b+c}\)