a/ Đặt :

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

Ta có :

\(VT=\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\)\(\left(1\right)\)

\(VP=\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

b/ Đặt :

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

\(VT=\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(1\right)\)

\(VP=\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

a) Từ \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

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

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

b) Từ \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{5b}{5d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{2a}{2c}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{5b}{5d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\)

Từ \(\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\) \(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)

Helpp 

a: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

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

Ta có: \(\dfrac{5a+2b}{5a-2b}=\dfrac{5bk+2b}{5bk-2b}=\dfrac{5k+2}{5k-2}\)

\(\dfrac{5c+2d}{5c-2d}=\dfrac{5dk+2d}{5dk-2d}=\dfrac{5k+2}{5k-2}\)

Do đó: \(\dfrac{5a+2b}{5a-2b}=\dfrac{5c+2d}{5c-2d}\)

giải nốt câu b

Đặt :

\(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\d=ck\end{matrix}\right.\)

Ta có :

\(VT=\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2}{d^2}\left(1\right)\)

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

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

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

Ta có:\(\dfrac{ab}{cd}\)=\(\dfrac{bk.b}{dk.d}\)=\(\dfrac{b^2.k}{d^2.k}\)=\(\dfrac{b^2}{d^2}\)=\(\left(\dfrac{b}{d}\right)^2\)(vì k khác 0) 1

\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)=\(\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}\)=\(\dfrac{\left[b.\left(k+1\right)\right]^2}{\left[d.\left(k+1\right)\right]^2}\)\(\left(\dfrac{b}{d}\right)^2\)=(vì k+1 khác 0) 2

Từ 1 và 2:

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

Vậy \(\dfrac{ab}{cd}\)=\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)(điều cần chứng minh)

Đặt a/b=c/d=k

=>a=bk; c=dk

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

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

Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)

b: \(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\left(\dfrac{b}{d}\right)^2\)

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

Do đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)

c: \(\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{k-1}{k+1}\)

\(\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{k-1}{k+1}\)

Do đó: \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)

Ta có:

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

a) \(\dfrac{3a+5c}{3b+5d}=\dfrac{3\cdot bk+5\cdot dk}{3b+5d}=\dfrac{k\left(3b+5d\right)}{3b+5d}=k\) (1)

\(\dfrac{a-2c}{b-2d}=\dfrac{bk-2dk}{b-2d}=\dfrac{k\left(b-2d\right)}{b-2d}=k\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{3a+5c}{3b+5d}=\dfrac{a-2c}{b-2d}\left(dpcm\right)\)

b) \(\dfrac{a^2-b^2}{ab}=\dfrac{\left(bk\right)^2-b^2}{bk\cdot b}=\dfrac{b^2k^2-b^2}{b^2k}=\dfrac{b^2\left(k-1\right)}{b^2k}=\dfrac{k-1}{k}\)(1)

\(\dfrac{c^2-d^2}{cd}=\dfrac{\left(dk\right)^2-d^2}{dk\cdot d}=\dfrac{d^2k^2-d^2}{d^2k}=\dfrac{d^2\left(k-1\right)}{d^2k}=\dfrac{k-1}{k}\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{a^2-b^2}{ab}=\dfrac{c^2-d^2}{cd}\left(dpcm\right)\)

c) \(\left(\dfrac{a+b}{c+d}\right)^3=\left(\dfrac{bk+b}{dk+d}\right)^3=\dfrac{b^3\left(k+1\right)^3}{d^3\left(k+1\right)^3}=\dfrac{b^3}{d^3}\) (1)

\(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\dfrac{b^3}{d^3}\) (2)

Từ (1) và (2) \(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^3=\dfrac{a^3+b^3}{c^3+d^3}\left(dpcm\right)\)

Cứu mình với mình đang cần gấp!~

giúp mình câu d) luôn nha phong

cảm ơn phong nha

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

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

Ta có VT:

\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}\)

\(=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\) (1)

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

Từ (1) và (2) 

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

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

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

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

Vậy...

*a/b=c/d=k=>a=bk;c=dk

Thay a=bk vào 2a+3b/2a-3b=2bk+3b/2bk-3b=2k+3/2k-3

Tương tự thay c=dk vào 2c+3d/2c-3d=2dk+3d/2dk-3d=2k+3/2k-3

=>2a+3b/2a-3b=2c+3d/2c-3d

*a/b=c/d=>a/c=b/d=k

=>k^2=a^2/c^2=c^2/d^2=a^2-b^2/c^2-d^2 (1)

k^2=a/c.b/d=ab/cd (2)

Từ (1) và (2)=>ab/cd=a^2-b^2/c^2-d^2

*a/b=c/d=>a/c=b/d=k=a+b/c+d

=>k^2=(a+b/c+d)^2

k^2=a^2/c^2=b^2/d^2=a^2+b^2/c^2+d^2

=>(a+b/c+d)^2=a^2+b^2/c^2+d^2

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

a)Ta có:\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\)(1)

\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}\dfrac{2k+3}{2k-3}\)(2)

Từ (1),(2)ta có:\(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)

b)Ta có:\(\dfrac{ab}{cd}=\dfrac{bk\times b}{dk\times d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)(1)

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

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

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

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

Từ (1), (2) ta có \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)

d: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

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

Ta có: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{3\cdot\left(dk\right)^2+5\cdot\left(bk\right)^2}{3d^2+5b^2}=k^2\)

\(\dfrac{c^2}{d^2}=\dfrac{\left(dk\right)^2}{d^2}=k^2\)

Do đó: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{c^2}{d^2}\)