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VP = \(\dfrac{\left(a-b\right)^2}{\left(a+c\right)\left(b+c\right)}+\dfrac{\left(b-c\right)^2}{\left(b+a\right)\left(c+a\right)}+\dfrac{\left(c-a\right)^2}{\left(c+b\right)\left(a+b\right)}\)
\(=\left(a-b\right).\dfrac{\left(a+c\right)-\left(b+c\right)}{\left(a+c\right)\left(b+c\right)}+\left(b-c\right).\dfrac{\left(b+a\right)-\left(c+a\right)}{\left(b+a\right)\left(c+a\right)}+\left(c-b\right).\dfrac{\left(c+b\right)-\left(a+b\right)}{\left(c+b\right)\left(a+b\right)}\)
\(=\left(a-b\right).\left(\dfrac{1}{b+c}-\dfrac{1}{a+c}\right)+\left(b-c\right)\left(\dfrac{1}{c+a}-\dfrac{1}{b+a}\right)+\left(c-a\right).\left(\dfrac{1}{a+b}-\dfrac{1}{c+b}\right)\)
\(=\left(a-b\right).\dfrac{1}{b+c}-\left(a-b\right).\dfrac{1}{a+c}+\left(b-c\right).\dfrac{1}{c+a}-\left(b-c\right).\dfrac{1}{b+a}+\left(c-a\right).\dfrac{1}{a+b}-\left(c-a\right).\dfrac{1}{c+b}\)
\(=\left(2a-b-c\right).\dfrac{1}{b+c}+\left(2b-c-a\right).\dfrac{1}{c+a}+\left(2c-a-b\right).\dfrac{1}{a+b}\)
\(=\dfrac{2a}{b+c}-\left(b+c\right).\dfrac{1}{b+c}+\dfrac{2b}{c+a}-\left(c+a\right).\dfrac{1}{c+a}+\dfrac{2c}{a+b}-\left(a+b\right).\dfrac{1}{a+b}\)
\(=2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)-3\left(đpcm\right)\)
\(VT=\dfrac{2a^3-a^2b-a^2c-ab^2-ac^2+2b^3-b^2c-bc^2+2c^3}{(a+b)(b+c)(c+a)} \)
\(\\=\dfrac{a^3+a^2b-2a^2b-2ab^2+ab^2+b^3+b^3+b^2c-2b^2c-2bc^2+bc^2+c^3+c^3+c^2a-2c^a+2ca^2-ca^2+a^3}{(a+b)(b+c)(c+a)}\)
\(\\=\dfrac{(a-b)^2(a+b)+(b-c)^2(b+c)+(c-a)^2(c+a)}{(a+b)(b+c)(c+a)}\)
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.
1)\(\dfrac{c-b}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}+\dfrac{a-c}{\left(b-a\right)\left(b-c\right)\left(a-c\right)}+\dfrac{b-a}{\left(b-a\right)\left(c-b\right)\left(c-a\right)}=\dfrac{c-b+a-c+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
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.
Ta có \(\dfrac{2}{a-b}\)+\(\dfrac{2}{b-c}\)+\(\dfrac{2}{c-a}\)
= (\(\dfrac{1}{a-b}\)+\(\dfrac{1}{c-a}\))+(\(\dfrac{1}{b-c}\)+\(\dfrac{1}{a-b}\))+(\(\dfrac{1}{c-a}\)+\(\dfrac{1}{b-c}\))
=(\(\dfrac{1}{a-b}\)- \(\dfrac{1}{a-c}\))+(\(\dfrac{1}{b-c}\)- \(\dfrac{1}{b-a}\))+(\(\dfrac{1}{c-a}\) - \(\dfrac{1}{c-b}\))
=\(\dfrac{\left(a-c\right)-\left(a-b\right)}{\left(a-b\right).\left(a-c\right)}\)+\(\dfrac{\left(b-a\right)-\left(b-c\right)}{\left(b-a\right).\left(b-c\right)}\)+\(\dfrac{\left(c-b\right)-\left(c-a\right)}{\left(c-b\right).\left(c-a\right)}\)
= \(\dfrac{a-c-a+b}{\left(a-b\right).\left(a-c\right)}\)+\(\dfrac{b-a-b+c}{\left(b-a\right).\left(b-c\right)}\)+\(\dfrac{c-b-c+a}{\left(c-b\right).\left(c-a\right)}\)
= \(\dfrac{-c+b}{\left(a-b\right).\left(a-c\right)}\)+ \(\dfrac{-a+c}{\left(b-a\right).\left(b-c\right)}\)+\(\dfrac{-b+a}{\left(c-b\right).\left(c-a\right)}\)
= \(\dfrac{b-c}{\left(a-b\right).\left(a-c\right)}\)+\(\dfrac{c-a}{\left(b-a\right).\left(b-c\right)}\)+\(\dfrac{a-b}{\left(c-b\right).\left(c-a\right)}\)
Chúc bạn học tốt.
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
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