a) \(b^2=ac.CMR:\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)
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Ta có:
b^2=ac \(\Rightarrow\) \(\frac{a}{b}\)=\(\frac{b}{c}\)\(\Rightarrow\) \(\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{ab}{bc}=\frac{a}{c}\)(1)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a}{b}=\frac{a}{c}=\frac{a+b}{b+c}\)
\(\Rightarrow\)\(\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{a^2+b^2}{b^2+c^2}\)(2)
Từ (1) và (2)\(\Rightarrow\)đpcm
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\(b^2=ac\Leftrightarrow\frac{a}{b}=\frac{b}{c}=k\Leftrightarrow\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{a^2+b^2}{b^2+c^2}=k^2\)
mà a =bk ; b = ck => a =c k2 => k2 =a/c
=>\(\frac{a^2+b^2}{b^2+c^2}=k^2=\frac{a}{c}\)
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a) \(A=\left(1+\frac{b^2+c^2-a^2}{2bc}\right).\frac{1+\frac{a}{b+c}}{1-\frac{a}{b+c}}.\frac{b^2+c^2-\left(b-c\right)^2}{a+b+c}\)
\(=\frac{2bc+b^2+c^2-a^2}{2bc}.\frac{\frac{a+b+c}{b+c}}{\frac{b+c-a}{b+c}}.\frac{b^2+c^2-b^2+2bc-c^2}{a+b+c}\)
\(=\frac{\left(b+c+a\right)\left(b+c-a\right)}{2bc}.\frac{a+b+c}{b+c-a}.\frac{2bc}{a+b+c}\)
\(=a+b+c\)
b) \(B=\frac{\frac{3a}{a+b}}{\frac{2a}{a^2-2ab+b^2}}\)\(=\frac{3a}{a+b}.\frac{\left(a-b\right)^2}{2a}=\frac{3\left(a-b\right)^2}{2\left(a+b\right)}\)
c) \(C=\frac{\frac{a}{b}+\frac{b}{a}}{\frac{a}{b}-\frac{b}{a}}:\frac{\frac{a^2}{b^2}-\frac{b^2}{a^2}}{\left(\frac{1}{a}+\frac{1}{b}\right)^2}\)
\(=\frac{\frac{a^2+b^2}{ab}}{\frac{a^2-b^2}{ab}}:\frac{\frac{a^4-b^4}{a^2b^2}}{\frac{\left(a+b\right)^2}{a^2b^2}}\)
\(=\frac{a^2+b^2}{a^2-b^2}.\frac{\left(a+b\right)^2}{a^4-b^4}\)
\(=\frac{\left(a^2+b^2\right)\left(a+b\right)^2}{\left(a+b\right)\left(a-b\right)\left(a^2+b^2\right)\left(a+b\right)\left(a-b\right)}\)
\(=\frac{1}{\left(a-b\right)^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có \(b^2=ac\)
Có \(\frac{a^2+b^2}{b^2+c^2}=\frac{a^2+ac}{ac+c^2}=\frac{a\left(a+c\right)}{c\left(a+c\right)}=\frac{a}{c}\left(đpcm\right)\)
Ta có:\(\frac{a^2+b^2}{b^2+c^2}=\frac{a^2+ac}{ac+c^2}=\frac{a\left(a+c\right)}{c\left(a+c\right)}=\frac{a}{c}\)
=>ĐPCM