CM \(\frac{b^2-c^2}{\left(a+b\right)\left(a+c\right)}+\frac{c^2-a^2}{\left(b+c\right)\left(b+a\right)}+\frac{a^2-b^2}{\left(c+a\right)\left(c+b\right)}=\frac{b-c}{b+c}+\frac{c-a}{c+a}+\frac{a-b}{a+b}\)\(\frac{a-b}{a+b}\)
K
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2 tháng 11 2019
a) A = \(\frac{a}{\left(a-b\right)\left(a-c\right)}+\frac{b}{\left(b-a\right)\left(b-c\right)}+\frac{c}{\left(c-a\right)\left(c-b\right)}\)
=> A = \(\frac{a}{\left(a-b\right)\left(a-c\right)}-\frac{b}{\left(a-b\right)\left(b-c\right)}+\frac{c}{\left(a-c\right)\left(b-c\right)}\)
=> A = \(\frac{a\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\frac{b\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{c\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
=> A + \(\frac{ab-ac-ab+bc+ac-bc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=0\)
KN
2 tháng 11 2019
\(B=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{a^2\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{b^2\left(c-a\right)}{\left(b-a\right)\left(b-c\right)\left(c-a\right)}\)
\(+\frac{c^2\left(a-b\right)}{\left(c-a\right)\left(c-b\right)\left(a-b\right)}\)
\(=\frac{a^2\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{b^2\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(+\frac{c^2\left(a-b\right)}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)
\(=\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
19 tháng 2 2017
1) \(M=a^2b^2c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
Em chú ý bài toán sau nhé: Nếu a+b+c=0 <=> \(a^3+b^3+c^3=3abc\)
CM: có:a+b=-c <=> \(\left(a+b\right)^3=-c^3\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
Chú ý: a+b=-c nên \(a^3+b^3+c^3=3abc\)
Do \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Thay vào biểu thwusc M ta được M=3abc (ĐPCM)
2, em có thể tham khảo trong sách Nâng cao phát triển toán 8 nhé, anh nhớ không nhầm thì bài này trong đó
Nếu không thấy thì em có thể quy đồng lên mà rút gọn
ta có:
\(\frac{b^2-c^2}{\left(a+b\right).\left(a+c\right)}=\frac{b^2-a^2+a^2-c^2}{\left(a+b\right).\left(a+c\right)}=\frac{\left(b-a\right).\left(b+a\right)+\left(a-c\right).\left(a+c\right)}{\left(a+b\right).\left(a+c\right)}=\frac{b-a}{a+c}+\frac{a-c}{a+b}\left(1\right)\)
\(\frac{c^2-a^2}{\left(b+c\right).\left(b+a\right)}=\frac{c^2-b^2+b^2-a^2}{\left(b+c\right).\left(b+a\right)}=\frac{\left(c-b\right).\left(b+c\right)+\left(b-a\right).\left(a+b\right)}{\left(b+c\right).\left(b+a\right)}=\frac{c-b}{b+a}+\frac{b-a}{b+c}\left(2\right)\)
\(\frac{a^2-b^2}{\left(c+a\right).\left(c+b\right)}=\frac{a^2-c^2+c^2-b^2}{\left(c+a\right).\left(c+b\right)}=\frac{\left(a-c\right).\left(a+c\right)+\left(c-b\right).\left(c+b\right)}{\left(c+a\right).\left(c+b\right)}=\frac{a-c}{c+b}+\frac{c-b}{c+a}\left(3\right)\)
từ (1),(2),(3)
\(\Rightarrow\frac{b^2-c^2}{\left(a+b\right).\left(a+c\right)}+\frac{c^2-a^2}{\left(b+c\right).\left(b+a\right)}+\frac{a^2-b^2}{\left(c+a\right).\left(c+b\right)}\)
\(=\frac{b-a}{a+c}+\frac{a-c}{a+b}+\frac{c-b}{a+b}+\frac{b-a}{b+c}+\frac{a-c}{c+b}+\frac{c-b}{c+a}=\frac{c-a}{a+c}+\frac{b-c}{b+c}+\frac{a-b}{a+b}\Rightarrowđpcm\)