Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 \(\left(\frac{a}{c}\right)^2=\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)
\(\Rightarrow\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\)
Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
=> \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)(Tính chất dãy tỉ số bằng nhau)
=> \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)(Đpcm)
Có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)(Tính chất dãy tỉ số bằng nhau)
=> \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\)
=> \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\)(Đpcm)
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!
Đặt \(\frac{a}{b}\)=\(\frac{c}{d}\)= k ( k \(\in\)Z , k khác 0 )
=> a = bk ; c = dk
Ta có:
\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}\) (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)
Từ (1) và (2) suy ra: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)
Vậy nếu \(\frac{a}{b}=\frac{c}{d}\)thì \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
\(\Leftrightarrow\left(a^2+b^2\right).cd=ab.\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd+b^2cd=abc^2+abd^2\)
\(\Leftrightarrow a^2cd-abd^2=abc^2-b^2cd\)
\(\Leftrightarrow ad\left(ac-bd\right)=bc\left(ac-bd\right)\)
\(\Leftrightarrow ad=bc\)
\(\Leftrightarrow\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)
