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

ta cs a/b=c/d=>a/c=b/d

=>2a+3b/2c+3d=3a-4b/3c-4d

=>2a+3b/3a-4b=2c+3d/3c-4d

=>bai toan dc c/m

Cau b tuong tu nha ban

don't forget tick me

a) Ta có \(\frac{a}{b}=\frac{c}{d}.\)

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

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

\(\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{3b}{3d}=\frac{2a+3b}{2c+3d}\) (1).

\(\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{4b}{4d}=\frac{3a-4b}{3c-4d}\) (2).

Từ (1) và (2) \(\Rightarrow\frac{2a+3b}{2c+3d}=\frac{3a-4b}{3c-4d}.\)

\(\Rightarrow\frac{2a+3b}{3a-4b}=\frac{2c+3d}{3c-4d}\left(đpcm\right).\)

Chúc bạn học tốt!

Giải:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=b.k,c=d.k\)

a) Ta có:

\(\frac{a}{3a+b}=\frac{b.k}{3.b.k+b}=\frac{b.k}{b\left(3k+1\right)}=\frac{k}{3k+1}\) (1)

\(\frac{c}{3c+d}=\frac{dk}{3dk+d}=\frac{dk}{d\left(3k+1\right)}=\frac{k}{3k+1}\) (2)

Từ (1) và (2) suy ra \(\frac{a}{3a+b}=\frac{c}{3c+d}\)

b) Ta có:

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

\(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2}{d^2}\) (2)

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

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

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

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

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

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:

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

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

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

1, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{c}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)

2, a, Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{ab}{cd}\)

\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{b}{d}\cdot\frac{b}{d}\Rightarrow\frac{ab}{cd}=\frac{b^2}{d^2}\)

\(\Rightarrow\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)

b, Ta có: \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a-b}{c-d}\cdot\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

giúp mình với, mai mình kiểm tra cuối kỉ rồi