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\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=2^2\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^2+2\left(\frac{1}{a}+\frac{1}{b}\right)\frac{1}{c}+\left(\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\left(\frac{1}{a}\right)^2+2\frac{1}{a}.\frac{1}{b}+\left(\frac{1}{b}\right)^2+2\left(\frac{1}{ac}+\frac{1}{bc}\right)+\left(\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\left(\frac{1}{a}\right)^2+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2+2\frac{1}{ab}+2\left(\frac{1}{ac}+\frac{1}{bc}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a}{abc}+\frac{b}{abc}+\frac{c}{abc}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a+b+c}{abc}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a+b+c}{a+b+c}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
Ta có : a + b + c = abc
\(\frac{\Rightarrow\left(a+b+c\right)}{abc}=\frac{abc}{abc}\)
\(\Rightarrow\frac{1}{ac}+\frac{1}{bc}+\frac{1}{ab}=1\)
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=2^2\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
\(\text{Ta có: }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{a+b+c}{abc}=\frac{abc}{abc}=1\left(\text{vì }a+b+c=abc\right)\)
\(\text{Lại có: }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2.1=2\left(\text{ vì }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\right)\)
Vậy ...
a)Do bd>0 (do b>0, d>0) nên nếu \(\frac{a}{b}< \frac{c}{d}\) thì ad<bc
b)Ngược lại, nếu ad<bc thì \(\frac{ad}{bd}< \frac{bc}{bd}\Leftrightarrow\frac{a}{b}< \frac{c}{d}\)
THEO PHÂN SỐ : \(\frac{a+b}{c}=\frac{6}{5}\) \(\Rightarrow\) \(\hept{\begin{cases}a+b=6\\c=5\end{cases}}\)1
THEO PHÂN SỐ:\(\frac{b+c}{a}=\frac{9}{2}\Rightarrow\hept{\begin{cases}b+c=9\\a=2\end{cases}}\)2
THEO 1 VÀ 2 , TA CÓ : \(\frac{a+c}{b}=\frac{2+5}{4}=\frac{7}{4}\)
ĐÁP SỐ \(\frac{a+c}{b}=\frac{7}{4}\)
~ HOK TỐT ~
\(\frac{a+b}{c}=\frac{6}{5}\Rightarrow\frac{a+b}{6}=\frac{c}{5}=\frac{a+b+c}{6+5}=\frac{a+b+c}{11}\left(1\right)\)
\(\frac{b+c}{a}=\frac{9}{2}\Rightarrow\frac{b+c}{9}=\frac{a}{2}=\frac{a+b+c}{9+2}=\frac{a+b+c}{11}\left(2\right)\)
từ \(\left(1\right)\left(2\right)\Rightarrow\frac{a+b}{6}=\frac{c}{5}=\frac{b+c}{9}=\frac{a}{2}=\frac{a+b+c}{11}\Rightarrow\frac{c}{5}=\frac{a}{2}\Rightarrow2c=5a\Rightarrow c=\frac{5}{2}a\)
\(\frac{a+b}{6}=\frac{b+c}{9}\Rightarrow\frac{3\left(a+b\right)}{6}=\frac{3\left(b+c\right)}{9}=\frac{a+b}{2}=\frac{b+c}{3}=\frac{a}{2}+\frac{b}{2}=\frac{b}{3}+\frac{c}{3}\)
\(\Rightarrow\frac{b}{2}-\frac{b}{3}=\frac{c}{3}-\frac{a}{2}=\frac{3b-2b}{6}=\frac{2c-3a}{6}=\frac{b}{6}=\frac{2c-3a}{6}\Rightarrow b=2c-3a\)mà \(c=\frac{5}{2}a\)
\(\Rightarrow b=2c-3a=2\cdot\frac{5}{2}a-3a=5a-3a=2a\)
\(\Rightarrow\frac{a+c}{b}=\frac{a+\frac{5}{2}a}{2a}=\frac{\frac{7}{2}a}{2a}=\frac{7}{4}\)