Nếu \(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{a+b}\) (với a,b\(\ne\)0; a\(\ne\)-b) thì giá trị của biểu thức \(\dfrac{b}{a}+\dfrac{a}{b}\) là:........
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LV
0
19 tháng 12 2017
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}:\frac{1}{2}=\frac{1}{a}+\frac{1}{b}\)
\(\frac{2}{c}=\frac{a+b}{ab}\)
\(\Rightarrow2ab=ac+bc\)
\(\Rightarrow ac-ab=ab-bc\)
\(\Rightarrow a.\left(c-b\right)=b.\left(a-c\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)( đpcm )
NL
Nguyễn Lê Phước Thịnh
CTVHS
24 tháng 9 2025
a: \(\frac{y}{4}-2+\frac{15}{4y}\)
\(=\frac{y-8}{4}+\frac{15}{4y}=\frac{y^2-8y+15}{4y}=\frac{\left(y-5\right)\left(y-3\right)}{4y}\)
\(\frac{y}{2}+\frac{6}{y}-\frac72=\frac{y-7}{2}+\frac{6}{y}=\frac{y\left(y-7\right)+12}{2y}\)
\(=\frac{y^2-7y+12}{2y}=\frac{\left(y-3\right)\left(y-4\right)}{2y}\)
Ta có: \(M=\frac{\frac{y}{4}-2+\frac{15}{4y}}{\frac{y}{2}+\frac{6}{y}-\frac72}\)
\(=\frac{\left(y-5\right)\left(y-3\right)}{4y}:\frac{\left(y-3\right)\left(y-4\right)}{2y}\)
\(=\frac{\left(y-5\right)\left(y-3\right)}{4y}\cdot\frac{2y}{\left(y-3\right)\left(y-4\right)}=\frac{y-5}{2\cdot\left(y-4\right)}\)
b: \(N=\frac{3b-\frac{1}{9b^2}}{1+\frac{1}{3b}+\frac{1}{9b^2}}\)
\(=\frac{27b^3-1}{9b^2}:\frac{9b^2+3b+1}{9b^2}\)
\(=\frac{\left(3b-1\right)\left(9b^2+3b+1\right)}{9b^2}\cdot\frac{9b^2}{9b^2+3b+1}\)
=3b-1
NL
Nguyễn Lê Phước Thịnh
CTVHS
11 tháng 7 2021
Ta có: \(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}\)
\(=\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-2\left(\dfrac{c}{abc}+\dfrac{b}{abc}+\dfrac{a}{abc}\right)}\)
\(=\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-2\cdot\dfrac{a+b+c}{abc}}\)
\(=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
17 tháng 5 2018
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\) hả Lặng Thầm
8 tháng 9 2017
Ta có: \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)
\(=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}\)
Ta cần chứng minh: \(\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=0\) thật vậy:
\(\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=\dfrac{2\left(a+b+c\right)}{abc}=\dfrac{2.0}{abc}=0\)Tức là:\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\left(đpcm\right)\)
\(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{a+b}\Leftrightarrow\dfrac{a+b}{ab}=\dfrac{1}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2=ab\)
\(\Rightarrow\dfrac{b}{a}+\dfrac{a}{b}=\dfrac{a^2+b^2}{ab}=\dfrac{\left(a+b\right)^2-2ab}{ab}=\dfrac{ab-2ab}{ab}=\dfrac{-ab}{ab}=-1\)
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{a+b}{ab}\Rightarrow\dfrac{a+b}{ab}=\dfrac{1}{a+b}\)
\(\Leftrightarrow ab=\left(a+b\right)^2\)
Vậy, \(\dfrac{b}{a}+\dfrac{a}{b}=\dfrac{a^2+b^2}{ab}=\dfrac{a^2+2ab+b^2-2ab}{ab}=\dfrac{\left(a+b\right)^2-2ab}{ab}=\dfrac{ab-2ab}{ab}=\dfrac{-ab}{ab}=-1\)