đặt bằng k thì ra hết bn ơi

\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\\ \Rightarrow cd\left(a^2+b^2\right)=ab\left(c^2+d^2\right)\\ \Rightarrow a^2cd+b^2cd=abc^2+abd^2\\ \Rightarrow a^2cd+b^2cd-abc^2-abd^2=0\\ \Rightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\\ \Rightarrow\left(ac-bd\right)\left(ad-bc\right)=0\\\Rightarrow \left[{}\begin{matrix}ac=bd\\ad=bc\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{a}{b}=\frac{d}{c}\\\frac{a}{b}=\frac{c}{d}\end{matrix}\right.\)

\(\Leftrightarrow\frac{a}{c}=\frac{b}{d}=\frac{ab}{cd}.\)

\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{matrix}\right.\left(đpcm\right).\)

Chúc bạn học tốt!

Đặt \(\frac{a}{b}\) = \(\frac{c}{d}\) = k ⇒ \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có:

\(\frac{ab}{cd}\) = \(\frac{bk.b}{dk.d}\) = \(\frac{b^2}{d^2}\)

\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\) = \(\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}\) = \(\frac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}\) = \(\frac{b^2}{d^2}\)

⇒ \(\frac{ab}{cd}\) = \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\) (đpcm)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\\ \dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\\ \Rightarrow\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)

Gọi a/b=c/d=k nên a=bk; c=dk

nên \(\frac{ab}{cd}=\frac{bk\cdot b}{dk\cdot d}=\frac{b^2\cdot k}{d^2\cdot k}=\frac{b^2}{d^2}\)(1)

nên \(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2\cdot k^2+b^2}{d^2\cdot k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)(2)

Từ (1);(2) => ab/cd=(a^2+b^2)/(c^2+d^2)

từ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

nên \(\frac{a}{c}\cdot\frac{b}{d}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)

Giải:

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

a, Ta có: \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\) (1)

\(\dfrac{ab}{cd}=\dfrac{bkb}{dkd}=\dfrac{b^2}{d^2}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)

b, Ta có: \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2}{d^2}\) (1)

\(\dfrac{ab}{cd}=\dfrac{bkb}{dkd}=\dfrac{b^2}{d^2}\) (2)

Từ (1), (2) \(\Rightarrowđpcm\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\left(1\right)\)

a) Thay (1) vào đề:

\(VT=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\)

\(VP=\dfrac{bkb}{dkd}=\dfrac{b^2}{d^2}\)

\(\Rightarrow VT=VP\)

\(\Leftrightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\rightarrowđpcm.\)

b) Thay (1) vào đề bài:

\(\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\dfrac{b^2}{d^2}\)

Theo câu a) \(\dfrac{ab}{cd}=\dfrac{b^2}{d^2}\)

\(\Rightarrow\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{ab}{cd}\rightarrowđpcm.\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

a) Ta có : \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2.k^2+b^2}{d^2.k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\) (1)

\(\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2}{k^2}\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)