a/ \(3a=2b;4b=3c\)

=> \(6a=4b=3c\)

=> \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{4b}{12}=\dfrac{5c}{20}=\dfrac{a+4b-5c}{2+12-20}=\dfrac{-30}{-6}=5\)

=> \(\left\{{}\begin{matrix}a=10\\b=15\\c=20\end{matrix}\right.\)

=> B

B

\(\Rightarrow\)B

a)  đặt \(\frac{a}{b}=\frac{c}{d}=k\Leftrightarrow a=b.k;c=d.k\)

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

    b)          bó tay

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 :

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

Ta có :

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

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

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

hỏi 24h.vn đi bn mik ko bik lm

Ta có: \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{d}{5}\Rightarrow\frac{3a}{6}=\frac{b}{3}=\frac{2c}{8}=\frac{4d}{20}=\frac{3a+b-2c+4d}{6+3-8+20}=\frac{105}{21}=5\)

=> a=10,b=15,c=20,d=25

\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{d}{5}=\frac{3a}{6}=\frac{2c}{8}=\frac{4d}{20}=\frac{3a+b-2c+4d}{6+3-8+20}=\frac{105}{21}=5\Rightarrow a=10,b=15,c=20,d=25\)

= 25 nha bạn