Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

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

Ta có: \(VT=\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7bk^2+3bkb}{11bk^2-8b^2}=\dfrac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}=\dfrac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\)

\(VP=\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7dk^2+3dkd}{11dk^2-8d^2}=\dfrac{7d^2k^2+3d^2k}{11d^2k^2-8d^2}=\dfrac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\)

\(\Rightarrow VT=VP\)

Vậy \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\left(đpcm\right)\)

Nâng cao r 

mk chịu :)

* VT là vế trái // VP là vế phải *

\(#tutuuu..\)

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

a) \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2}{d^2}\)\(=\dfrac{\dfrac{a}{k}.b}{\dfrac{c}{k}.d}=\dfrac{ab}{cd}=VT\)

Vậy...

b) \(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{5k+3}{5k-3}\)

\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{5k+3}{5k-3}\)

Suy ra \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)

c) \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7\left(bk\right)^2+3\left(bk\right).b}{11\left(bk\right)^2-8b^2}\)\(=\dfrac{7k^2+3k}{11k^2-8}\)

\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7\left(dk\right)^2+3\left(dk\right).d}{11\left(dk\right)^2-8d^2}=\dfrac{7k^2+3k}{11k^2-8}\)

Suy ra \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)

a) Có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

=> \(ad=bc\)

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

(theo tính chất dãy tỉ số bằng nhau)

=> (đpcm)

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

=> \(\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)(theo tính chất dãy tỉ số bằng nhau)

=> \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\) (đpcm)

c) Có: \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

=> \(\dfrac{a^2}{c^2}=\dfrac{ab}{cd}=\dfrac{b^2}{d^2}\)          => \(\dfrac{7a^2}{7c^2}=\dfrac{3ab}{3cd}=\dfrac{11a^2}{11c^2}=\dfrac{8b^2}{8d^2}\)

=> \(\dfrac{7a^2+3ab}{7c^2+3cd}=\dfrac{11a^2-8b^2}{11c^2-8d^2}\) (theo tính chất dãy tỉ số bằng nhau)

=> \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)(đpcm)

#Ayumu

Gạevdhbdvd

Gkykdyhlculxys

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

\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}=\dfrac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\left(1\right)\)

\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7d^2k^2+3d^2k}{11d^2k^2-8d^2}=\dfrac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\left(2\right)\)

\(\left(1\right)\left(2\right)\RightarrowĐpcm\)

Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7b^2k^2+3\cdot bk\cdot b}{11\cdot b^2k^2-8b^2}=\dfrac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}=\dfrac{7k^2+3k}{11k^2-8}\)

\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7d^2k^2+3\cdot dk\cdot d}{11d^2k^2-8d^2}=\dfrac{7k^2+3k}{11k^2-8}\)

Do đó: \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)

đặt \(\frac{a}{b}=\frac{c}{d}=k\)==> a=b.k va c= d.k

ve trai : \(\frac{a.b}{c.d}=\frac{b.k.b}{d.k.d}=\frac{b^2}{b^2}\)

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

vế trái = vế phải ---> điều phải cm cau b tuong tu

sao mình chẳng hiểu gì cả

Câu này đặt k là xong

a)

b)

Ta có: \(\frac{a}{b}=\frac{c}{d}.\)

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

\(\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{c}{d}\right)^2=\frac{ac}{bd}.\)

\(\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{ac}{bd}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}.\)

\(\Rightarrow\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\left(đpcm\right).\)

Chúc bạn học tốt!

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