có học mà bạn

đặt \(\frac{a}{b}\)=  \(\frac{c}{d}=k\Rightarrow\hept{\begin{cases}k=ab\\k=cd\end{cases}}\)

ta có :   \(\frac{7a-4b}{3a+5b}\)= \(\frac{7ak-4b}{3ak-5b}=\frac{a\left(7k-4\right)}{a\left(3k-5\right)}=\frac{7k-4}{3k-5}\left(1\right)\)

\(\frac{7c-4d}{3c+5d}\)=\(\frac{7ck-4d}{3ck+5d}\)= \(\frac{c\left(7k-4\right)}{c\left(3k+5\right)}\)= \(\frac{7k-4}{3k+5}\)(  2 ) 

từ (1) và ( 2) => \(\frac{7a-4b}{3a+5b}=\frac{7c-4d}{3c+5d}\)( điều phải chứng minh ) 

bài này có đúng ko vậy

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

Đặt \(\frac{a}{c}=\frac{b}{d}\) = k

=> a = ck , b = dk

Thay a = ck , b = dk vào  \(\frac{7a-11b}{4a+5b}\)ta có :

\(\frac{7a-11b}{4a+5b}=\frac{7.ck-11dk}{4ck+5dk}=\frac{k\left(7c-11d\right)}{k\left(4c+5d\right)}=\frac{7c-11d}{4c+5d}\)

Vậy \(\frac{7a-11b}{4a+5b}=\frac{7c-11d}{4c+5d}\)

đúng rồi

a, Áp dụng t/c dtsbn:

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

b, Áp dụng t/c dtsbn:

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

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

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

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

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

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

Ta có \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)

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

Do đó \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{7a-4b}{3a+5b}=\dfrac{7bk-4b}{3bk+5b}=\dfrac{7k-4}{3k+5}\)

\(\dfrac{7c-4d}{3c+5d}=\dfrac{7dk-4d}{3dk+5d}=\dfrac{7k-4}{3k+5}\)

Do đó: \(\dfrac{7a-4b}{3a+5b}=\dfrac{7c-4d}{3c+5d}\)

b: \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)

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

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

1/

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

b,\(\frac{a}{b}=\frac{c}{d}=\frac{4a}{4b}=\frac{7c}{7d}=\frac{4a+7c}{4b+7d}\)

2/

Gọi số học sinh tham gia của mỗi lớp lần lượt là a,b,c

Ta có: \(2a=3b=4c\)

\(\Rightarrow\frac{2a}{12}=\frac{3b}{12}=\frac{4c}{12}\Rightarrow\frac{a}{6}=\frac{b}{4}=\frac{c}{3}=\frac{a+b+c}{6+4+3}=\frac{130}{13}=10\)

=> a/6 = 10 => a = 60

b/4 = 10 => b = 40

c/3 = 10 => c = 30

Vậy số học sinh mỗi lớp lần lượt là 60 hs, 40 hs, 30hs

mn ơi giúp mk với