đặt \(\frac{a}{b}=\frac{c}{d}=k\)=> a=bk c=dk 

ta có : \(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2k^2+b^2}{d^2k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)(1)

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

từ (1:2) => \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)

Cái này dựa trên mạng dác dặt bút làm lắm nha

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=b.k;c=d.k\)

Ta có \(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{kb^2}{kd^2}=\frac{b^2}{d^2}\left(1\right)\)

Ta lại có \(\frac{a^2+b^2}{c^2+d^2}=\frac{k^2.b^2+b^2}{k^2.d^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\left(2\right)\)

Từ \(\left(1\right)và\left(2\right)\)ta được

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

\(\frac{a}{b}=\frac{c}{d}=>\frac{a^2}{b^2}=\frac{c^2}{d^2}\)\(=>\frac{a^2}{c^2}=\frac{b^2}{d^2}.\) .  .Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\).Xin lỗi mình chưa nghĩ ra tiếp

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

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

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

Từ (1) và (2), ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)

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

\(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\frac{\left[b.\left(k-1\right)\right]^2}{\left[d.\left(k-1\right)\right]^2}=\frac{b^2}{d^2}\)(1)

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

Từ (1) và (2), ta có: \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\)

a) Từ \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)

mà \(\left(\frac{a}{c}\right)^2=\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)

\(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)

b) Từ \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{a-b}{c-d}\right)^2=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

mà \(\left(\frac{a}{c}\right)^2=\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)

\(\Rightarrow\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=b.k

c=d.k

ta có Vế Phải : \(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(b.k\right)^2+b^2}{\left(d.k\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{b^2}{d^2}\)

Vế Trái :\(\frac{ab}{cd}=\frac{b.k.b}{d.k.d}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}\)

vì \(\frac{b^2}{d^2}=\frac{b^2}{d^2}\)

=>VP=VT

=>\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=> a=b.k; c=d.k

Suy ra:

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

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

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

Chúc bạn học tốt!

"." là nhân nha

bài 1 sai đề ko bạn

đề nào và mình ghi sai thứ tự bài

bài 1 thiếu cho ở đàu

1, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{c}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)

2, a, Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{ab}{cd}\)

\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{b}{d}\cdot\frac{b}{d}\Rightarrow\frac{ab}{cd}=\frac{b^2}{d^2}\)

\(\Rightarrow\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)

b, Ta có: \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a-b}{c-d}\cdot\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

có nhiều cách giải,cách đặt k:

a/b=c/d=k  thì a=bk;c=dk

thay vào:

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

ab/cd=.....   (2)

từ (1) và (2) =>đpcm

a/b=c/d

=>a/c=b/d

=>a/c.a/c=b/d.a/c=b/d.b/d

=>ab/cd=a²/c²=b²/d²=(a²+b²)/(c²+d²)

vậy ...