đặt a/b =c/d = k => a =bk; c = dk(*)

ta có: ac/bd =bk.dk/bd = k^2  (1)

thay (*) vào (a+c)^2/(b+d)^2 = k^2  (2)

từ 1 và 2 suy ra điều cần CM

mình xin lỗi. mình không giúp bạn đc.hỏi sony tiểu bàng í.ai giỏi toán giúp bạn ấy đi .có vẻ gấp rồi đó

Cút ngay

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

Vì \(\frac{a}{b}=k\)\(\Rightarrow a=bk\)

Vì\(\frac{c}{d}=k\)\(\Rightarrow c=dk\)

Có \(\frac{ac}{bd}=\frac{bk.dk}{bd}=\frac{bd.k^2}{bd}=k^2\)\(\left(1\right)\)

Vì \(a=bk,c=dk\Rightarrow\)\(\frac{\left(a+b\right)^2}{\left(b+d\right)^2}\)\(=\frac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\frac{[k\left(b+d\right)]^2}{\left(b+d\right)^2}=\frac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\left(2\right)\)

Từ (1) và (2)\(\Rightarrow\)đpcm

mình sửa đề thì ms lm đc

 không pk sửa đề

Giải:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

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

Ta có:

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

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

Từ (1) và (2) suy ra \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)

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

theo đề bài ta có
\(ab\left(c^2+d^2\right)=ab.c^2+ab.d^2=\left(a.c\right).\left(b.c\right)+\left(a.d\right).\left(b.d\right)\\ cd\left(a^2+b^2\right)=cd.a^2+cd.b^2=\left(c.a\right).\left(d.a\right)+\left(c.b\right).\left(d.b\right)\)
\(\left(a.c\right)\left(b.c\right)+\left(a.d\right)\left(b.d\right)=\left(c.a\right)\left(d.a\right)+\left(c.b\right)\left(d.b\right)\) vì mỗi vế đều bằng nhau
- Cnứng minh \(\frac{\left(a^2+b^2\right)}{c^2+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
ta có vì \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{\left(a+b\right)}{\left(c+d\right)}=\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2}{c^2}=\frac{b^2}{d^2}\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(a^2+b^2\right)}{\left(c^2+d^2\right)}\)

Gọi a/b=c/d=k(k khác 0)

Ta có:

a=bk

c=dk

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

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

Từ (1) và (2) suy ra bằng nhau

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

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

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

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

Từ (1) và (2) => \(\frac{a^2+c^2}{b^2+d^2}=\frac{ac}{bd}\)(đpcm)

Có \(\dfrac{a}{b}=\dfrac{c}{d}< =>\dfrac{a}{c}=\dfrac{b}{d}\)

Áp dụng dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)

<=> \(\left(\dfrac{a}{c}\right)^2=\left(\dfrac{b}{d}\right)^2=\left(\dfrac{a+b}{c+d}\right)^2\)

<=> \(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\left(\dfrac{a+b}{c+d}\right)^2\)(1)

Có \(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\)

Áp dụng DTSBN ta có: 

\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}\)(2)

Từ (1) (2) => đpcm.