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

a)\(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\)(1)

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

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

b)\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)(1)

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

Từ (1) và(2)\(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

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

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

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

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

\(\Rightarrow\frac{2a}{2c}=\frac{3b}{3d}\)

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

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

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

d ở đâu ra vậy?

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

Suy ra: \(\frac{2a+3b}{2a-3b}=\frac{2.bk+3b}{2.bk-3b}=\frac{b.\left(2k+3\right)}{b.\left(2k-3\right)}=\)\(\frac{2k+3}{2k-3}\)

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

Vậy \(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)

Ta có:\(\frac{a}{b}=\frac{c}{d}\)=>\(\frac{a}{c}=\frac{b}{d}\)=>\(\frac{2a}{2c}=\frac{3b}{3d}\)

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

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

=>\(\frac{2a+3b}{2c+3d}=\frac{2a-3b}{2c-3d}\)=>\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)

Vậy\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)

Ta có:

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

a) \(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\left(1\right)\)

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

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

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

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

Từ (1) , (2) \(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

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

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

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

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

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

Từ (1) và (2) có \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\left(đpcm\right)\)

các câu còn lại bạn tự làm đi! HI.......

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

Áp dụng tính chất dãy tỉ số bằng nhau

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

Có: \(\frac{2a-3b}{2c-3d}=\frac{2a+3b}{2c+3d}\Leftrightarrow\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)

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

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

Áp dụng tính chất dãy tỉ số bằng nhau

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

Tu (1)&(2),có: \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

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

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{2a}{2c}=\frac{3b}{3d}\Rightarrow\frac{2a+3b}{2c+3d}=\frac{2a-3b}{2c-3d}\)(T/c dãy tỷ số bằng nhau)

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

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=> a = b.k ; c= d.k

\(\frac{2a+3b}{2a-3b}=\frac{2.\left(b.k\right)+3.b}{2.\left(b.k\right)-3b}=\frac{2b.k+3b}{2b.k-3b}=\frac{2b.\left(k+1,5\right)}{2b.\left(k-1,5\right)}=\frac{k+1,5}{k-1,5}\left(1\right)\)

\(\frac{2c+3d}{2c-3d}=\frac{2.\left(d.k\right)+3d}{2.\left(d.k\right)-3d}=\frac{2d.k+3d}{2d.k-3d}=\frac{2d.\left(k+1,5\right)}{2d.\left(k-1,5\right)}=\frac{k+1,5}{k-1,5}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) => đpcm

bài này đúng không bạn

a/c=c/d suy ra 2a/2c=3b/3d = (2a+3b)/(2c+3d)=(2a-3d)/(2c-3d)

                                       suy ra (2a+3b)/(2a-3b)=(2c+3d)/(2c-3d)