1. Đặt \(\frac{a}{b}=\frac{c}{d}=k=>a=bk,c=dk\)

Thay vào 2 vế là sẽ CM được

1. Đặt \(\frac{a}{b}=\frac{c}{d}=k>a=bk.c=dk\)

Thay vào 2 vế để chứng minh

1 ) 

Ta có : 

\(\frac{a}{b}=\frac{c}{d}\)

\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)

\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)(  Áp dụng t/c DTSBN ) 

\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\left(1\right)\)

Lại có : \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{3a^2}{3c^2}=\frac{2b^2}{2d^2}=\frac{3a^2+2b^2}{3c^2+2d^2}\) (  Áp dụng t/c DTSBN )  \(\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\)

\(\Rightarrow\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{3a^2+2b^2}{3c^2+2d^2}\left(đpcm\right)\)

2 ) 

Ta có : 

\(x+y+2xy=83\)

\(\Rightarrow2\left(x+y+2xy\right)=166\)
\(\Rightarrow2x+2y+4xy+1=167\)

\(\Rightarrow2x\left(2y+1\right)+\left(2y+1\right)=167\)

\(\Rightarrow\left(2x+1\right)\left(2y+1\right)=167\)

Do \(x;y\in Z\)
\(\Leftrightarrow2x+1;2y+1\in Z\)
\(\Leftrightarrow2x+1;2y+1\in\left\{\pm1;\pm167\right\}\)

Ta có bảng sau : 

Vậy \(\left(x;y\right)\in\left\{\left(0;83\right),\left(83;0\right),\left(-1;-84\right),\left(-84;-1\right)\right\}\)
 

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\). Ta có:

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

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

Đến đây nhìn có vẻ đề sai

\(\frac{a}{b}=\frac{c}{d}=k\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)ta có:

\(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{\left(bk-b\right)^3}{\left(dk-d\right)^3}=\frac{\left[b\left(k-1\right)\right]^3}{\left[d\left(k-1\right)\right]^3}=\frac{b^3}{d^3}\)

\(\frac{2b^2+3a^2}{2d^2+3c^2}=\frac{4.b^2+9.k^2.b^2}{4.d^2+9.d^2.k^2}=\frac{b^2\left(4+k^2.9\right)}{d^2\left(4+9.k^2\right)}=\frac{b^2}{d^2}\)

\(Taco:\frac{b^3}{d^3}=\frac{b^2}{d^2}\Leftrightarrow b=d\)

khó quá ak

ừ, bạn bik làm thì giúp mình nha ^^

mk làm câu a thôi, b dài nhưng tương tự

Gọi a/b=c/d=k =>a=bk ; c=dk

=>\(\frac{\left(2a+3b\right)^2}{\left(3a-4b\right)^2}=\frac{\left(2bk+3b\right)^2}{\left(3bk-4b\right)^2}=\frac{\left[b\left(2k+3\right)\right]^2}{\left[b\left(3k-4\right)\right]^2}=\frac{b^2\left(2k+3\right)^2}{b^2\left(3k-4\right)^2}=\frac{\left(2k+3\right)^2}{\left(3k-4\right)^2}\)(1)

=>\(\frac{\left(2c+3d\right)^2}{\left(3c-4d\right)^2}=\frac{\left(2dk+3d\right)^2}{\left(3dk-4d\right)^2}=\frac{\left[d\left(2k+3\right)\right]^2}{\left[d\left(3k-4\right)\right]^2}=\frac{\left(2k+3\right)^2}{\left(3k-4\right)^2}\)(2)

Từ (1);(2)=> đpcm

b)\(\frac{ac}{bd}=\frac{bkdk}{bd}=k.k=k^2\)

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

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

Đặt k ( với k khác 0 , thuộc Z ) sao cho \(\frac{a}{b}=\frac{c}{d}=k\) => \(a=kb\)  /  \(c=dk\) .

a) Thế vào \(\frac{5a-b}{3a+2b}\) , ta có \(\frac{5kb-3b}{3kb+2b}\)\(=\frac{b\left(5k-3\right)}{b\left(3k+2\right)}\)\(=\frac{5k-3}{3k+2}\)  /  \(\frac{5c-3d}{3c+2d}=\frac{5dk-3d}{3dk-2d}=\frac{d\left(5k-3\right)}{d\left(3k+2\right)}=\frac{\left(5k+3\right)}{\left(3k+2\right)}\)

=> VT = VP

a/ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{3a}{3c}=\frac{5b}{5d}=\frac{3a+5b}{3c+5d}=\frac{3a-5b}{3c-5d}\Rightarrow\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)

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

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

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