Câu a thiếu đề nhé, -3a+b= bao nhiêu thế bạn?

b/ Theo đề ta có:

\(\frac{a}{7}=\frac{b}{4}\Rightarrow\frac{a}{14}=\frac{b}{8}\); \(\frac{b}{8}=\frac{c}{5}\)

=> \(\frac{a}{14}=\frac{b}{8}=\frac{c}{5}\)

a/d tính chất của dãy tỉ số = nhau ta có:

\(\frac{a}{14}=\frac{b}{8}=\frac{c}{5}=\frac{10a}{140}=\frac{5b}{40}=\frac{c}{5}=\frac{10a-5b+c}{140-40+5}=\frac{100}{105}=\frac{20}{21}\)

=> \(\left\{{}\begin{matrix}a=\frac{20}{21}\cdot14=\frac{40}{3}\\b=\frac{20}{21}\cdot8=\frac{160}{21}\\c=\frac{20}{21}\cdot5=\frac{100}{21}\end{matrix}\right.\)

vậy...

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

Ta có : 

\(2a=\frac{a}{\frac{1}{2}};3b=\frac{b}{\frac{1}{3}};5b=\frac{b}{\frac{1}{5}};7c=\frac{c}{\frac{1}{7}}\)

Lại có \(\hept{\begin{cases}\frac{a}{\frac{1}{2}}=\frac{b}{\frac{1}{3}}\\\frac{b}{\frac{1}{5}}=\frac{c}{\frac{1}{7}}\end{cases}}\Rightarrow\frac{a}{\frac{3}{2}}=b=\frac{c}{\frac{5}{7}}\Leftrightarrow\frac{3a}{\frac{9}{2}}=\frac{7b}{1}=\frac{5c}{\frac{25}{7}}\)

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

\(\frac{3a}{\frac{9}{2}}=\frac{7b}{1}=\frac{5c}{\frac{25}{7}}=\frac{3a-7b+5c}{\frac{9}{2}-1+\frac{25}{7}}=\frac{-30}{\frac{99}{14}}=\frac{-140}{33}\)

\(\Rightarrow\hept{\begin{cases}3a=\frac{-140}{33}\cdot\frac{9}{2}=\frac{-210}{11}\Rightarrow a=\frac{-70}{11}\\7b=\frac{-140}{33}\Rightarrow b=\frac{-20}{33}\\5c=\frac{-140}{33}\cdot\frac{25}{7}=\frac{-500}{33}\Rightarrow c=\frac{-100}{33}\end{cases}}\)

Vậy....

Chắc sai =))

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 \(a=42;b=28;c=20\)

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

câu hỏi tương tự có đó bn ơi

tick nhé !!!

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

\(5b=2c=>\frac{b}{2}=\frac{c}{5}\)

=>\(\frac{a}{3}=\frac{b}{2}=\frac{c}{5}\)

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

\(\frac{a}{3}=\frac{b}{2}=\frac{c}{5}=\frac{3a}{9}=\frac{7b}{14}=\frac{5c}{25}=\frac{3a+5c-7b}{9+25-14}=\frac{30}{20}=\frac{3}{2}\)

=>\(a=\frac{3}{2}.3=\frac{9}{2},b=\frac{3}{2}.2=3,c=\frac{3}{2}.5=\frac{15}{2}\)

\(2a=2b\Rightarrow\frac{a}{2}=\frac{b}{2}\Rightarrow\frac{a}{2}.\frac{1}{7}=\frac{b}{2}.\frac{1}{7}\Rightarrow\frac{a}{14}=\frac{b}{14}\)
\(5b=7c\Rightarrow\frac{b}{7}=\frac{c}{5}\Rightarrow\frac{b}{7}.\frac{1}{2}=\frac{c}{5}.\frac{1}{2}\Rightarrow\frac{b}{14}=\frac{c}{10}\)
(Ngoặc '}' 2 điều trên lại)
\(\Rightarrow\frac{a}{14}=\frac{b}{14}=\frac{c}{10}\)(1)
Từ (1) \(\Rightarrow\frac{3a}{3.14}=\frac{7b}{7.14}=\frac{5c}{5.10}=\frac{3a}{42}=\frac{7b}{98}=\frac{5c}{50}\)
Áp dụng tính chất DTSBN:
\(\frac{a}{14}=\frac{b}{14}=\frac{c}{10}=\frac{3a}{42}=\frac{7b}{98}=\frac{5c}{50}=\frac{3a-7b+5c}{42-98+50}=\frac{-30}{-6}=5\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{14}=5\Rightarrow a=5.14=70\\\frac{b}{14}=5\Rightarrow a=5.14=70\\\frac{c}{10}=5\Rightarrow c=5.10=50\end{cases}}\)
Vậy a = 70, b = 70, c = 50

Cảm ơn nha!