cho a,b ,c khác 0 và a^2=bc. CMR: \(\frac{a^2+c^2}{b^2+a^2}\)=\(\frac{c}{d}\)
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b) a2=ac\(\Rightarrow\) \(\frac{a}{b}=\frac{b}{c}\)
c2=bd\(\Rightarrow\) \(\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\)\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\) = \(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\) = \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
=\(\frac{a.b.c}{b.c.d}=\frac{a}{d}\)
=> đpcm
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Ta có:a2 = bc \(\Rightarrow\frac{a}{b}=\frac{c}{a}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{a^2}=\frac{a}{b}.\frac{c}{a}=\frac{c}{b}\left(1\right)\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a^2}{b^2}=\frac{c^2}{a^2}=\frac{a^2+c^2}{b^2+a^2}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{a^2+c^2}{b^2+c^2}=\frac{c}{b}\left(đpcm\right)\)
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a) Mk sửa lại chỗ \(\frac{5a-7b}{5a-7d}\) nhé, đề đúng phải là \(\frac{5a-7b}{5c-7d}\)
Ta có: \(ad=bc\Leftrightarrow\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{7b}{7d}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{5a}{5c}=\frac{7b}{7d}=\frac{5a+7b}{5c+7d}=\frac{5a-7b}{5c-7d}\left(đpcm\right)\)
b) Ta có: \(ad=bc\Leftrightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{a^2-b^2}{c^2-d^2}\left(đpcm\right)\)
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Ta có : \(\frac{a^2-bc}{a}+\frac{b^2-ac}{b}+\frac{c^2-ab}{c}=0\)
=> \(a-\frac{bc}{a}+b-\frac{ac}{b}+c-\frac{ab}{c}=0\)
=> \(a+b+c=\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\)
=> \(a+b+c=abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
=> \(\frac{a+b+c}{abc}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
=> \(\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
=> \(\frac{2}{bc}+\frac{2}{ac}+\frac{2}{ab}=\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}\)
=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{bc}-\frac{2}{ac}-\frac{2}{ac}=0\)
=> \(\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\left(\frac{1}{a^2}-\frac{2}{ac}+\frac{1}{c^2}\right)+\left(\frac{1}{b^2}-\frac{1}{bc}+\frac{1}{c^2}\right)=0\)
=> \(\left(\frac{1}{a}-\frac{1}{b}\right)^2+\left(\frac{1}{a}-\frac{1}{c}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2=0\)
=> \(\hept{\begin{cases}\frac{1}{a}-\frac{1}{b}=0\\\frac{1}{a}-\frac{1}{c}=0\\\frac{1}{b}-\frac{1}{c}=0\end{cases}}\Rightarrow\hept{\begin{cases}\frac{1}{a}=\frac{1}{b}\\\frac{1}{a}=\frac{1}{c}\\\frac{1}{b}=\frac{1}{c}\end{cases}}\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\left(\text{đpcm}\right)\)
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Từ \(\frac{\overline{ab}}{a+b}=\frac{\overline{bc}}{b+c}\Rightarrow\left(10a+b\right).\left(b+c\right)=\left(10b+c\right).\left(a+b\right)\)
\(\Rightarrow10ab+b^2+10ac+bc=10ab+ac+10b^2+bc\)
\(\Rightarrow b^2+10ac=ac+10b^2\)
\(\Rightarrow10ac-ac=10b^2-b^2\)
\(\Rightarrow9ac=9b^2\)
\(\Rightarrow ac=b^2\Rightarrow\frac{a}{b}=\frac{b}{c}\left(đpcm\right)\)
\(\frac{\overline{ab}}{a+b}=\frac{\overline{bc}}{b+c}\)
<=> \(\frac{\overline{ab}}{\overline{bc}}=\frac{a+b}{b+c}\)
<=> \(\frac{a.10+b}{b.10+c}=\frac{a+b}{b+c}\)
Áp dụng dãy tỉ số bằng nhau ta có:
\(\frac{a.10+b}{b.10+c}=\frac{a+b}{b+c}=\frac{\left(10a+b\right)-\left(a+b\right)}{\left(10b+c\right)-\left(b+c\right)}=\frac{9a}{9b}=\frac{a}{b}\)
Áp dụng dãy tỉ số bằng nhau ta có:
\(\frac{a+b}{b+c}=\frac{a}{b}=\frac{\left(a+b\right)-a}{\left(b+c\right)-b}=\frac{b}{c}\)
=> \(\frac{a}{b}=\frac{b}{c}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\Rightarrow\frac{a^2+b^2}{ab}=\frac{c^2+d^2}{cd}\)
=> \(\frac{a^2}{ab}+\frac{b^2}{ab}=\frac{c^2}{cd}+\frac{d^2}{cd}\)
=> \(\frac{a}{b}+\frac{b}{a}=\frac{c}{d}+\frac{d}{c}\)
Mình chỉ làm được tới khúc này
Ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\frac{a+b}{c+d}\right)^2\left(1\right)\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{2ab}{2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\left(\frac{a-b}{c-d}\right)^2\left(2\right)\)
Từ (1) và (2) suy ra:
\(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{a-b}{c-d}\right)^2\)
Trường hợp 1: \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)+\left(a-b\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2a}{2c}=\frac{a}{c}\left(3\right)\)
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)-\left(a-b\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2b}{2d}=\frac{b}{d}\left(4\right)\)
Từ (3) và (4) suy ra \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\)
Trường hợp 2: \(\frac{a+b}{c+d}=\frac{-\left(a-b\right)}{c-d}=\frac{b-a}{c-d}=\frac{\left(a+b\right)+\left(b-a\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2b}{2c}=\frac{b}{c}\left(5\right)\)
\(\frac{a+b}{c+d}=\frac{b-a}{c-d}=\frac{\left(a+b\right)-\left(b-a\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2a}{2d}=\frac{a}{d}\left(6\right)\)
Từ (5) và (6) suy ra \(\frac{b}{c}=\frac{a}{d}\Rightarrow\frac{a}{b}=\frac{d}{c}\)