\(\frac{a}{c}=\frac{a-b}{b-c}\left(a;c\ne0;a\ne b;b\ne c\right)\)
\(Cmr:\frac{1}{a}+\frac{1}{a-b}=\frac{1}{b-c}-\frac{1}{c}\)
K
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MN
1 tháng 2 2020
Ta có :
\(VT=\frac{1}{2}\left[\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}\right]\)
\(=\frac{1}{2}\left[\frac{\left(b-c\right)^2}{\left(a-b\right)\left(a-c\right)}+\frac{\left(a-c\right)^2}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}+\frac{\left(a-b\right)^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)
\(=\frac{1}{2}\left[\frac{\left(b-c\right)^2+\left(a-c\right)^2+\left(a-b\right)^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)
\(=\frac{1}{2}\left[\frac{b^2-2bc+c^2+a^2-2ac+c^2+a^2-2ab+b^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)
\(=\frac{1}{2}\left[\frac{2a^2+2b^2+2c^2-2ab-2bc-2ac}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)
\(=\frac{a^2+b^2+c^2-ab-bc-ac}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)(1)
Lại có :
\(VP=\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\)
\(=\frac{\left(b-c\right)\left(a-c\right)+\left(a-b\right)\left(a-c\right)-\left(a-b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{ab-bc-ac+c^2+a^2-ac-ab+bc-ab+ac+b^2-bc}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{a^2+b^2+c^2-ab-ac-bc}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)(2)
Từ (1) và (2) \(\RightarrowĐPCM\)
20 tháng 2 2019
Ta có:\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{\left(b-a\right)+\left(a-c\right)}{\left(a-b\right)\left(a-c\right)}=\frac{b-a}{\left(a-b\right)\left(a-c\right)}+\frac{a-c}{\left(a-b\right)\left(a-c\right)}=\frac{1}{a-b}+\frac{1}{c-a}\)
Chứng minh tương tự,ta được:
\(\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{a-b}+\frac{1}{b-c}\)
\(\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{b-c}+\frac{1}{c-a}\)
\(\Rightarrow\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)\left(đpcm\right)\)
N
5 tháng 12 2018
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)
\(\Rightarrow\frac{b^2-c^2}{\left(a+b\right).\left(a+c\right)}+\frac{c^2-a^2}{\left(b+c\right).\left(b+a\right)}+\frac{a^2-b^2}{\left(c+a\right).\left(c+b\right)}\)
\(=\frac{b-a}{a+c}+\frac{a-c}{a+b}+\frac{c-b}{a+b}+\frac{b-a}{b+c}+\frac{a-c}{c+b}+\frac{c-b}{c+a}=\frac{c-a}{a+c}+\frac{b-c}{b+c}+\frac{a-b}{a+b}\Rightarrowđpcm\)
31 tháng 8 2017
Câu hỏi của Bùi Minh Quân - Toán lớp 9 - Học toán với OnlineMath
19 tháng 8 2017
\(VT=\frac{b-a+a-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-b+b-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-c+c-b}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{-1}{a-c}+\frac{1}{a-b}+\frac{-1}{b-a}+\frac{1}{b-c}+\frac{-1}{c-b}+\frac{1}{c-a}\)
\(=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}=VP\)
BT
19 tháng 8 2017
Ta có:
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{b-a+a-c}{\left(a-b\right)\left(a-c\right)}=\frac{b-a}{\left(a-b\right)\left(a-c\right)}+\frac{a-c}{\left(a-b\right)\left(a-c\right)}=\frac{1}{c-a}+\frac{1}{a-b}\)
Tương tự:
\(\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{c-b+b-a}{\left(b-c\right)\left(b-a\right)}=\frac{c-b}{\left(b-c\right)\left(b-a\right)}+\frac{b-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{a-b}+\frac{1}{b-c}\)
Và: \(\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{a-c+c-b}{\left(c-a\right)\left(c-b\right)}=\frac{a-c}{\left(c-a\right)\left(c-b\right)}+\frac{c-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{b-c}+\frac{1}{c-a}\)
=> \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{c-a}+\frac{1}{a-b}+\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{b-c}+\frac{1}{c-a}\)
=> \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}\)
=> đpcm
BA
14 tháng 11 2019
Ta có
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{b-a}{\left(a-b\right)\left(a-c\right)}+\frac{a-c}{\left(a-b\right)\left(a-c\right)}=\frac{1}{a-b}+\frac{1}{c-a}\left(1\right)\)
Tương tự ta có
\(\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{b-c}+\frac{1}{a-b}\left(2\right)\)
\(\frac{a-b}{\left(c-b\right)\left(c-a\right)}=\frac{1}{b-c}+\frac{1}{c-a}\left(3\right)\)
Từ (1) (2) và (3) ta có
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-b\right)\left(c-a\right)}\)
\(=\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}+\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\)
\(=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}\left(đpcm\right)\)
14 tháng 11 2019
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{c-b}{\left(a-b\right)\left(c-a\right)}=\frac{\left(c-a\right)+\left(a-b\right)}{\left(a-b\right)\left(c-a\right)}=\frac{1}{a-b}+\frac{1}{c-a}\)
Làm tương tự ta được:\(\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{b-c}+\frac{1}{a-b}\)
\(\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{c-a}+\frac{1}{b-c}\)
\(\Rightarrow\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}+\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}\)
\(=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}\)
\(\Rightarrow\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}\left(ĐPCM\right)\)
