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ab−c−ba−c−cb−a=0=>ab−c−ba−c−cb−a=0
=>ab−c=ba−c+cb−a=b2−ab+ac−c2(c−a)(a−b)=>ab−c=ba−c+cb−a=b2−ab+ac−c2(c−a)(a−b)
Nhân cả 2 vế với 1b−c1b−c ta được
a(b−c)2=b2−ab+ac−c2(a−b)(b−c)(c−a)(1)a(b−c)2=b2−ab+ac−c2(a−b)(b−c)(c−a)(1)
Tương tự ta có:
b(c−a)2=c2−bc+bc−a2(a−b)(b−c)(c−a)(2)b(c−a)2=c2−bc+bc−a2(a−b)(b−c)(c−a)(2)
c(a−b)2=a2−ca+cb−c2(a−b)(b−c)(c−a)(3)c(a−b)2=a2−ca+cb−c2(a−b)(b−c)(c−a)(3)
Cộng theo vế (1);(2);(3) ta có ĐPCM
Lời giải:
Ta có:
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\Rightarrow \frac{a}{b-c}=\frac{-b}{c-a}+\frac{-c}{a-b}\)
\(\Leftrightarrow \frac{a}{b-c}=\frac{-b(a-b)-c(c-a)}{(a-b)(c-a)}=\frac{b^2+ca-c^2-ab}{(a-b)(c-a)}\)
\(\Rightarrow \frac{a}{(b-c)^2}=\frac{b^2+ca-c^2-ab}{(a-b)(b-c)(c-a)}\)
Hoàn toàn tương tự:
\(\frac{b}{(c-a)^2}=\frac{c^2+ab-a^2-bc}{(a-b)(b-c)(c-a)}\)
\(\frac{c}{(a-b)^2}=\frac{a^2+bc-b^2-ac}{(a-b)(b-c)(c-a)}\)
Cộng theo vế các đẳng thức vừa thu được ta có:
\(\frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}=\frac{b^2+ac-c^2-ab+c^2+ab-a^2-bc+a^2+bc-b^2-ac}{(a-b)(b-c)(c-a)}=0\)
Ta có đpcm.
Ta có:
\(\dfrac{b-c}{1\left(a-b\right)\left(a-c\right)}+\dfrac{c-a}{\left(b-c\right)\left(b-a\right)}+\dfrac{a-b}{\left(c-a\right)\left(c-b\right)}\)
\(=\dfrac{c-b}{1\left(a-b\right)\left(c-a\right)}+\dfrac{a-c}{\left(b-c\right)\left(a-b\right)}+\dfrac{b-a}{\left(c-a\right)\left(b-c\right)}\)
Quy đồng rút gọn ta được
\(=\dfrac{2\left(ab+bc+ca-a^2-b^2-c^2\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\dfrac{2\left[\left(a-b\right)\left(b-c\right)+\left(b-c\right)\left(c-a\right)+\left(c-a\right)\left(a-b\right)\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=2\left(\dfrac{1}{a-b}+\dfrac{1}{b-c}+\dfrac{1}{c-a}\right)\)
PS: Hôm qua đi chơi nên nay mới giải nhé.
Từ \(\dfrac{a-\left(c-b\right)}{b-c}+\dfrac{b-\left(a-c\right)}{c-a}+\dfrac{c-\left(b-a\right)}{a-b}=3\)
\(=>\dfrac{a}{b-c}+1+\dfrac{b}{c-a}+1+\dfrac{c}{a-b}+1=3\)
\(=>\dfrac{a}{b-c}-\dfrac{b}{a-c}-\dfrac{c}{b-a}=0\)
\(=>\dfrac{a}{b-c}=\dfrac{b}{a-c}+\dfrac{c}{b-a}=\dfrac{b^2-ab+ac-c^2}{\left(c-a\right)\left(a-b\right)}\)
Nhân cả 2 vế với \(\dfrac{1}{b-c}\) ta được
\(\dfrac{a}{\left(b-c\right)^2}=\dfrac{b^2-ab+ac-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(1\right)\)
Tương tự ta có:
\(\dfrac{b}{\left(c-a\right)^2}=\dfrac{c^2-bc+bc-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(2\right)\)
\(\dfrac{c}{\left(a-b\right)^2}=\dfrac{a^2-ca+cb-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(3\right)\)
Cộng theo vế (1);(2);(3) ta có ĐPCM
CHÚC BẠN HỌC TỐT.........
Từ giả thiết suy ra:
\(\left\{{}\begin{matrix}\dfrac{a}{b-c}=\dfrac{-b}{c-a}+\dfrac{-c}{a-b}=\dfrac{-ab+b^2-c^2+ac}{\left(c-a\right)\left(a-b\right)}\\\dfrac{b}{c-a}=\dfrac{-c}{a-b}+\dfrac{-a}{b-c}=\dfrac{-bc+c^2-a^2+ab}{\left(a-b\right)\left(b-c\right)}\\\dfrac{c}{a-b}=\dfrac{-a}{b-c}+\dfrac{-b}{c-a}=\dfrac{-ac+a^2-b^2+bc}{\left(b-c\right)\left(c-a\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{\left(b-c\right)^2}=\dfrac{-ab+b^2-c^2+ac}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\\\dfrac{b}{\left(c-a\right)^2}=\dfrac{-bc+c^2-a^2+ab}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\\\dfrac{c}{\left(a-b\right)^2}=\dfrac{-ac+a^2-b^2+bc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\end{matrix}\right.\)
Cộng theo vế suy ra đpcm
\(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0\)
\(\Rightarrow\dfrac{a}{b-c}=\dfrac{b}{a-c}+\dfrac{c}{b-a}=\dfrac{b^2-ab+ac-c^2}{\left(a-b\right)\left(c-a\right)}\)
\(\Leftrightarrow\dfrac{a^2}{\left(b-c\right)^2}=\dfrac{ab^2-a^2b+a^2c-ac^2}{\left(a-b\right)\left(c-a\right)\left(b-c\right)}\)
Tương tự ta có:
\(\dfrac{b^2}{\left(c-a\right)^2}=\dfrac{bc^2-b^2c+b^2a-a^2b}{\left(b-c\right)\left(c-a\right)\left(a-b\right)}\)
\(\dfrac{c^2}{\left(a-b\right)^2}=\dfrac{a^2c-c^2a+c^2b-cb^2}{\left(c-a\right)\left(a-b\right)\left(b-c\right)}\)
Cộng 3 đẳng thức trên có:
==" xl mk ko bt tài làm để có bình phương đc :)) mk chỉ can chứng minh
\(\dfrac{a}{\left(b-c\right)^2}+\dfrac{b}{\left(c-a\right)^2}+\dfrac{c}{\left(a-b\right)^2}=0đcthui\)
Đặt x = a - b, y = b - c, z = c - a
\(\Rightarrow\left\{{}\begin{matrix}x+y+z=0\\ay+bz+cx=ab-ac+bc-ab+ac-bc=0\end{matrix}\right.\)
+ \(ay+bz+cx=0\)
\(\Rightarrow\dfrac{1}{y}\left(\dfrac{a}{y}+\dfrac{b}{z}+\dfrac{c}{x}\right)=0\)
\(\Rightarrow\dfrac{a}{y^2}+\dfrac{bx}{xyz}+\dfrac{cz}{xyz}=0\)
\(\Rightarrow\dfrac{a}{y^2}=\dfrac{-bx-cz}{xyz}\)
+ Tương tự : \(\dfrac{b}{z^2}=\dfrac{-cy-ax}{xyz}\)
\(\dfrac{c}{x^2}=\dfrac{-az-by}{xyz}\)
Do đó : \(\dfrac{a}{y^2}+\dfrac{b}{z^2}+\dfrac{c}{x^2}=\dfrac{-a\left(x+z\right)-b\left(x+y\right)-c\left(y+z\right)}{xyz}\)
\(=\dfrac{ay+bz+cx}{xyz}\) ( do x + y + z = 0)
\(=0\) ( do ay + bz + cx = 0 )