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Bài 1:
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a^2+2017b^2}{c^2+2017d^2}=\dfrac{b^2k^2+2017b^2}{d^2k^2+2017d^2}=\dfrac{b^2}{d^2}\)
\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{a^2+2017b^2}{c^2+2017d^2}=\dfrac{ab}{cd}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=kb;c=kd\)
Ta có:\(\frac{a^2+ac}{c^2-ac}=\frac{b^2k^2+bk.dk}{d^2k^2-bk.dk}=\frac{bk^2\left(b+d\right)}{dk^2\left(d-b\right)}=\frac{b\left(b+d\right)}{d\left(d-b\right)}\)(1)
\(\frac{b^2+bd}{d^2-bd}=\frac{b\left(b+d\right)}{d\left(d-b\right)}\)(2)
Từ 1 và 2 =>\(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=kb\\c=kd\end{matrix}\right.\)
Ta có: \(\frac{a^2+ac}{c^2-ac}=\frac{b^2.k^2+bk.dk}{d^2.k^2-bk.dk}=\frac{bk^2.\left(b+d\right)}{dk^2.\left(d-b\right)}=\frac{b.\left(b+d\right)}{d.\left(d-b\right)}\) (1)
\(\frac{b^2+bd}{d^2-bd}=\frac{b.\left(b+d\right)}{d.\left(d-b\right)}\) (2)
Từ (1) và (2) => \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\left(đpcm\right).\)
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có: \(\frac{a}{b}=\frac{c}{d}.\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=kb\\c=kd\end{matrix}\right.\)
Ta có:
\(\frac{a^2+ac}{c^2-ac}=\frac{b^2k^2+bk.dk}{d^2k^2-bk.dk}=\frac{bk^2.\left(b+d\right)}{dk^2.\left(d-b\right)}=\frac{b.\left(b+d\right)}{d.\left(d-b\right)}\left(1\right)\)
\(\frac{b^2+bd}{d^2-bd}=\frac{b.\left(b+d\right)}{d.\left(d-b\right)}\left(2\right)\)
Từ \(\left(1\right)và\left(2\right)\Rightarrow\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\left(đpcm\right).\)
Chúc em học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{c}{d}\right)^2=\frac{ac}{bd}=\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+ac}{b^2+bd}=\frac{c^2-ac}{d^2-bd}\)
\(\Rightarrow\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}=\frac{c-a}{d-b}\)
Điều cần CM là \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\Rightarrow\frac{a^2+ac}{b^2+bd}=\frac{c^2-ac}{d^2-bd}\)
\(=\frac{a\left(a+c\right)}{b\left(b+d\right)}=\frac{c\left(c-a\right)}{d\left(d-b\right)}\)
Mà theo chứng minh trên ta có: \(\frac{a}{b}=\frac{c}{d};\frac{a+c}{b+d}=\frac{c-a}{d-b}\)
Từ đó ta\(\Rightarrow\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Sửa lại đề là \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=kb\\c=kd\end{matrix}\right.\)
Ta có: \(\frac{a^2+ac}{c^2-ac}=\frac{b^2.k^2+bk.dk}{d^2.k^2-bk.dk}=\frac{bk^2.\left(b+d\right)}{dk^2.\left(d-b\right)}=\frac{b.\left(b+d\right)}{d.\left(d-b\right)}\left(1\right)\)
\(\frac{b^2+bd}{d^2-bd}=\frac{b.\left(b+d\right)}{d.\left(d-b\right)}\left(2\right).\)
Từ \(\left(1\right)và\left(2\right)\Rightarrow\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\left(đpcm\right).\)
Mình chỉ làm câu a) thôi nhé.
Chúc bạn học tốt!