Có: \(2a-10=5b\)

=>\(2a=5b+10\)

=>\(4a=10b+20\)

Thay 4a=10b+20 ta đc

\(10b+20+3b=150\)

=>b=10

Thay vào tính a

a=30

b=10xin hay h
 

Ta có : 4a = 3b => 28a = 21b (1)

            7b = 5c => 21b = 15c (2)

Từ (1) và (2) => 28a = 21b = 15c 

Ta có : 28a = 21b = 15c \(=\frac{a}{\frac{1}{28}}=\frac{b}{\frac{1}{21}}=\frac{c}{\frac{1}{15}}=\frac{2a}{\frac{1}{14}}=\frac{3b}{\frac{1}{7}}=\frac{2a+3b-c}{\frac{1}{14}+\frac{1}{7}-\frac{1}{15}}=\frac{186}{\frac{31}{210}}=1260\)

Nên : 28a = 1260 => a = 45

         21b = 1260 => b = 60

         15c = 1260 => c = 84

Vậy ........................

Ta có:

 \(4a=3b\)=> \(\frac{a}{3}=\frac{b}{4}\)=> \(\frac{a}{15}=\frac{b}{20}\left(1\right)\)

\(7b=5c\)=>\(\frac{b}{5}=\frac{c}{7}\) => \(\frac{b}{20}=\frac{c}{28}\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\)

=>\(\frac{a}{15}=\frac{b}{20}=\frac{c}{28}\)=>\(\frac{2a}{30}=\frac{3b}{60}=\frac{c}{28}\)

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

\(\frac{2a}{30}=\frac{3b}{60}=\frac{c}{28}=\frac{2a+3b-c}{30+60-28}=\frac{186}{62}=3\)

=>\(\frac{a}{15}=3\)=>\(a=45\)

    \(\frac{b}{20}=3\)=>\(b=60\)

    \(\frac{c}{28}=3\)=>\(c=84\)

Vậy \(a=40;b=60;c=84\)

Ta có: \(2a=3b\)=> \(\frac{a}{3}=\frac{b}{2}\)=>\(\frac{a}{21}=\frac{b}{14}\left(1\right)\)

          \(5b=7c\)=>\(\frac{b}{7}=\frac{c}{5}\) =>\(\frac{b}{14}=\frac{c}{10}\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\)

=>\(\frac{a}{21}=\frac{b}{14}=\frac{c}{10}\)=> \(\frac{3a}{63}=\frac{7b}{98}=\frac{5c}{50}\)

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

\(\frac{3a}{63}=\frac{7b}{98}=\frac{5c}{50}=\frac{3a-7b+5c}{63-98+50}=\frac{30}{15}=2\)

=>\(\frac{a}{21}=2\)=>\(a=42\)

    \(\frac{b}{14}=2\)=>\(b=28\)

    \(\frac{c}{10}=2\)=>\(c=20\)

Vậy \(a=42;b=28;c=20\)

A=\(\dfrac{5}{9}\)

\(\dfrac{a}{b}=\dfrac{3}{4}\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{2a-5b}{-14}=\dfrac{a-3b}{-9}=\dfrac{4a+b}{16}=\dfrac{8a-2b}{16}\\ \Leftrightarrow A=\dfrac{-14}{-9}-\dfrac{16}{16}=\dfrac{14}{9}-1=\dfrac{5}{9}\)

A=5/9

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=bk;c=dk\)

1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)

\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)

Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)

2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)

\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)

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

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

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

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

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

\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)

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

Ta có : \(\dfrac{a}{b}=\dfrac{3}{4}\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{4}\\ Đặt\dfrac{a}{3}=\dfrac{b}{4}=k\Rightarrow\left\{{}\begin{matrix}a=3k\\b=4k\end{matrix}\right.\\ ThayvàoA,tacó:\)

\(A=\dfrac{2a-5b}{a-3b}-\dfrac{4a+b}{8a-2b}\\ \Leftrightarrow=\dfrac{2\cdot3k-5\cdot4k}{3k-3\cdot4k}-\dfrac{4\cdot3k+4k}{8\cdot3k-2\cdot4k}\\ =\dfrac{6k-20k}{3k-12k}-\dfrac{12k+4k}{24k-8k}\\ =\dfrac{14k}{9k}-\dfrac{16k}{16k}\\ =\dfrac{14}{9}-1\\ =\dfrac{5}{9}\)

giúp mk nha !!!(mk cần gấp)

phá ngoặc lun nà

+4a-5c+3b-2b+a-7c-7b+3c-5a=(4a+a-5a)+(3b-2b-7b)+(-5c-7c+3c)=0-6b-9c=-9c-6b

-2a+3c-b-5b-4c+12a+9b+4c-4a-6a-3b-3c+d=(-2a+12a-4a-6a)+(-b-5b+9b-3b)+(3c-4c+4c-3c)+d=0+0+0+0+d=d

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

\(\left(2a+3b\right)\left(4c-5d\right)=\left(4a-5b\right)\left(2c+3d\right)\)

\(\Leftrightarrow8ac-10ad+12bc-15bd=8ac+12ad-10bc-15bd\)

\(\Leftrightarrow-10ad+12bc=12ad-10bc\)

\(\Leftrightarrow\left(-10ad+12bc\right)+\left(-12bc-12ad\right)=\left(12ad-10bc\right)+\left(-12bc-12ad\right)\)

\(\Leftrightarrow22bc=22ad\)