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\(A=\dfrac{\left(a-b\right)^2}{ab}+\dfrac{\left(b-c\right)^2}{bc}+\dfrac{\left(c-a\right)^2}{ca}\)
\(B=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
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\(A=\dfrac{a^2+b^2-2ab}{ab}+\dfrac{b^2-2ab+c^2}{bc}+c^2+a^2-\dfrac{2ca}{ca}\)
\(A=\left(\dfrac{a}{b}+\dfrac{b}{a}-2\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}-2\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}-2\right)=\dfrac{\left(b+c\right)}{a}+\dfrac{a+c}{b}+\dfrac{a+b}{c}-6\)
\(A=\left[\dfrac{\left(b+c\right)}{a}+1\right]+\left[\dfrac{\left(a+c\right)}{b}+1\right]+\left[\dfrac{\left(a+b\right)}{c}+1\right]-9\)
\(A=\dfrac{\left(a+b+c\right)}{a}+\dfrac{\left(a+b+c\right)}{b}+\left[\dfrac{\left(a+b+c\right)}{c}\right]-9\)
\(A=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-9\)
Ket luan
\(A\ne B\) => đề sai--> hoặc mình công trừ sai
1)\(\dfrac{c-b}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}+\dfrac{a-c}{\left(b-a\right)\left(b-c\right)\left(a-c\right)}+\dfrac{b-a}{\left(b-a\right)\left(c-b\right)\left(c-a\right)}=\dfrac{c-b+a-c+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
==" tách a^2 +ac-b^2 - bc = (a-b)(a+b+c)
tương tụ mấy sô kia mày sẽ thấy kết quả :))
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.
\((\dfrac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\dfrac{1}{\left(c-a\right)\left(b^2+ba-c^2-ca\right)}+\dfrac{1}{\left(a-b\right)\left(c^2+cb-a^2-ab\right)}=0 \)
\(\Leftrightarrow\dfrac{1}{\left(b-c\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]}+\dfrac{1}{\left(c-a\right)\left[\left(b-c\right)\left(b+c\right)+a\left(b-c\right)\right]}+\dfrac{1}{\left(a-b\right)\left[\left(c-a\right)\left(c+a\right)+b\left(c-a\right)\right]}=0\)
\(\Leftrightarrow\dfrac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\dfrac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\dfrac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}=0\)
\(\Leftrightarrow\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
\(\Leftrightarrow\dfrac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)(t/m)
Suy ra ta được Đt cần chứng minh.
Chúc bạn học tốt với hoc24 nha
Lời giải:
Ta có:
\(\frac{1}{(b-c)(a^2+ac-b^2-bc)}+\frac{1}{(c-a)(b^2+bc-c^2-ca)}+\frac{1}{(a-b)(c^2+cb-a^2-ab)}\)
\(=\frac{1}{(b-c)[(a^2-b^2)+(ac-bc)]}+\frac{1}{(c-a)[(b^2-c^2)+(ba-ca)]}+\frac{1}{(a-b)[(c^2-a^2)+(cb-ab)]}\)
\(=\frac{1}{(b-c)[(a-b)(a+b)+c(a-b)]}+\frac{1}{(c-a)[(b-c)(b+c)+a(b-c)]}+\frac{1}{(a-b)[(c-a)(c+a)+b(c-a)]}\)
\(=\frac{1}{(b-c)(a-b)(a+b+c)}+\frac{1}{(c-a)(b-c)(b+c+a)}+\frac{1}{(a-b)(c-a)(c+a+b)}\)
\(=\frac{(c-a)+(a-b)+(b-c)}{(a-b)(b-c)(c-a)(a+b+c)}=\frac{0}{(a-b)(b-c)(c-a)(a+b+c)}=0\)
Ta có đpcm.