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Đặt \(\frac{a}{b}=\frac{c}{d}=k\) => a=bk,c=dk
Ta có: \(\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left(bk+b\right)^2}{\left(bk\right)^2+b^2}=\frac{\left[b\left(k+1\right)\right]^2}{b^2k^2+b^2}=\frac{b^2\left(k+1\right)^2}{b^2\left(k^2+1\right)}=\frac{\left(k+1\right)^2}{k^2+1}\left(1\right)\)
\(\frac{\left(c+d\right)^2}{c^2+d^2}=\frac{\left(dk+d\right)^2}{\left(dk\right)^2+d^2}=\frac{\left[d\left(k+1\right)\right]^2}{d^2k^2+d^2}=\frac{d^2\left(k+1\right)^2}{d^2\left(k^2+1\right)}=\frac{\left(k+1\right)^2}{k^2+1}\left(2\right)\)
Từ (1) và (2) => đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
Nguyễn Thị Linh Chi: Em có cách khác ạ. (cách này em làm trên lớp thường ngày.Và cũng khác đơn giản ạ)
ĐK: b,d ≠ 0 ; b≠d
Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\).Đặt \(\frac{a}{c}=\frac{b}{d}=k\Rightarrow\hept{\begin{cases}a=kc\\b=kd\end{cases}}\).Thay vào:
\(\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left(kc+kd\right)^2}{k^2c^2+k^2d^2}=\frac{\left[k\left(c+d\right)\right]^2}{k^2\left(c^2+d^2\right)}=\frac{\left(c+d\right)^2}{c^2+d^2}^{\left(đpcm\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Còn nha. Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có: \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\frac{b^2}{d^2}^{\left(1\right)}\)
Lại có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{b^2}{d^2}^{\left(2\right)}\)
Từ (1) và (2) => đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\left(1\right)\)
mà \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)
Từ (1) \(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\Rightarrow\frac{a^2-b^2}{ab}=\frac{c^2-d^2}{cd}\)
ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)
Lại có: \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
\(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\Rightarrow\frac{\left(a+b^2\right)}{a^2+b^2}=\frac{\left(c+d\right)^2}{c^2+d^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(1\right)\)
\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b^2\right)}{\left(c+d\right)^2}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\Rightarrow\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left(c+d\right)^2}{c^2+d^2}\left(dpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\)(1)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)(2)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2k^2-b^2}{d^2k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\)(3)
Từ (1),(2),(3) =>\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{a^2-b^2}{c^2-d^2}\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt a/b=c/d=k
=>a=bk; c=dk
Sửa đề: \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{a^2+b^2}{c^2+d^2}\)
\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{b^2}{d^2}\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{a^2+b^2}{c^2+d^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giả sử tất cả các tỷ lệ thức đều có nghĩa.
Từ: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)
Và suy ra: \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\)
Và Từ: \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}\)
Dễ nhất là bạn hãy đặt k đi, thay vào là nó sẽ ra thôi.
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{b^2k+2b^2k+b^2}{d^2k+2d^2k+d^2}=\frac{3b^2k}{3d^2k}=\frac{b^2}{d^2}\)
Tương tự vs mấy cái còn lại là ra ngay thôi