1.Theo bài ra ta có 3a = 2b ; 5b = 7c

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

\(\Rightarrow\frac{a}{14}=\frac{b}{21};\frac{b}{21}=\frac{c}{15}\)

\(\Rightarrow\frac{a}{14}=\frac{b}{21}=\frac{c}{15}\)

Đặt \(\frac{a}{14}=\frac{b}{21}=\frac{c}{15}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=14k\\b=21k\\c=15k\end{matrix}\right.\)

Thay a = 14k ; b = 21k ; c = 15 k vào 3a+5b-7c = 60 ta có

3.14k + 5.21k - 7.15k =60

=> 42k + 105k - 105k = 60

=> k. (42 + 105 - 105) = 60

=> k . 42 = 60

=> \(k=\frac{60}{42}=\frac{10}{7}\)

\(\Rightarrow\left\{{}\begin{matrix}a=14.\frac{10}{7}=2.10=20\\b=21.\frac{10}{7}=3.10=30\\c=15.\frac{10}{7}=\frac{150}{7}\end{matrix}\right.\)

Vậy a = 20; b = 30 ; c = \(\frac{150}{7}\)

2. | 2x-3| - x = |2-x| (1)

+) Nếu x < \(\frac{3}{2}\) thì | 2x - 3| = 3 - 2x và |2 - x| = 2 - x

\(\Rightarrow\left(1\right)\Leftrightarrow\) 3 - 2x - x = 2 - x

\(\Leftrightarrow\) 3 - 3x = 2 - x

\(\Leftrightarrow\) 3 - 2 = 3x - x

\(\Leftrightarrow1=2x\)

\(\Leftrightarrow x=\frac{1}{2}\) ( thỏa mãn x < \(\frac{3}{2}\))

Nếu \(\frac{3}{2}\le x\le2\) thì | 2x - 3| = 2x - 3 ; |2-x| = 2 - x

\(\Rightarrow\left(1\right)\Leftrightarrow2x-3=2-x\)

\(\Leftrightarrow2x+x=2+3\)

\(\Leftrightarrow3x=5\)

\(\Leftrightarrow x=\frac{5}{3}\) ( không thỏa mãn \(\frac{3}{2}\le x\le2\))

Nếu x> 2 thì | 2x - 3| = 2x - 3 ; | 2 - x| = x - 2

\(\Rightarrow\left(1\right)\Leftrightarrow2x-3-x=x-2\)

\(\Leftrightarrow x-3=x-2\) ( vô lí vs mọi x)

Vậy \(x=\frac{1}{2}\) thỏa mãn đề bài

~ Học tốt

ta có:3a=2b;5b=7c và 3a+5b-7c=60

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

=>\(\frac{b}{7}=\frac{c}{5}\Rightarrow\frac{a}{21}=\frac{c}{15}\left(2\right)\)

từ (1) và (2) ta có :

a/14=b/21=1/15

áp dụng tính chất dãy tỉ số = nhau ta có:

\(\frac{a}{14}=\frac{b}{21}=\frac{c}{15}=\frac{3a+5b-7c}{3.14+5.21-15.7}=\frac{60}{42}=\frac{10}{7}\)

=>a=10/7.14=20

b=10/7.21=30

c=10/7.15=150/7

ta có:

3a=2b suy ra a/2=b/3 suy ra a/14=b/21

5b=7c suy ra b/7=c/5 suy ra b/21=c/15

suy ra: a/14=b/21=c/15=(3a+5b-7c)/(42+105-105)=60/42=10/7

ta có:

a=10/7x14=20

b=10/7x21=30

c=10/7x15=150/7

3a = 2b => a/2 = b/3 => a/14 = b/21 => 3a/42 = 5b/105

5b = 7c => b/7 = c/5 => b/21 = c/15 => 5b/105 = 7c/105

=> 3a/42 = 5b/105 = 7c/105 = 3a+3b-7c/42+105-105 = 60/42 = 

đúng k z

Theo bài ra ta cs

\(3a=2b\Rightarrow\frac{a}{2}=\frac{b}{3}\Rightarrow\frac{a}{14}=\frac{b}{21}\left(1\right)\)

\(5b=7c\Rightarrow\frac{b}{7}=\frac{c}{5}\Rightarrow\frac{b}{21}=\frac{c}{15}\left(2\right)\)

Từ (1) ; (2) => \(\frac{a}{14}=\frac{b}{21}=\frac{c}{15}\)

ADTC dãy tỉ số bằng nhau ta cs

\(\frac{a}{14}=\frac{b}{21}=\frac{c}{15}=\frac{3a+5b-7c}{3.14+5.21-7.15}=\frac{60}{42}=\frac{10}{7}\)

\(\Leftrightarrow\frac{a}{14}=\frac{10}{7}\Leftrightarrow a=20\)

\(\Leftrightarrow\frac{b}{21}=\frac{10}{7}\Leftrightarrow b=30\)

\(\Leftrightarrow\frac{c}{15}=\frac{10}{7}\Leftrightarrow c=\frac{150}{7}\)

Ta có : 

\(3a=2b\Rightarrow\frac{a}{2}=\frac{b}{3}\Rightarrow\frac{a}{14}=\frac{b}{21}\left(1\right)\)

\(5b=7c\Rightarrow\frac{b}{7}=\frac{c}{5}\Rightarrow\frac{b}{21}=\frac{c}{15}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\)

\(\Rightarrow\frac{a}{14}=\frac{b}{21}=\frac{c}{15}\)

\(\Rightarrow\frac{3a}{42}=\frac{5b}{105}=\frac{7c}{105}\)

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

\(\frac{3a}{42}=\frac{5b}{105}=\frac{7c}{105}=\frac{3a+5b-7c}{42+105-105}=\frac{60}{42}=\frac{10}{7}\)

\(\Rightarrow\hept{\begin{cases}\frac{3a}{42}=\frac{10}{7}\\\frac{5b}{105}=\frac{10}{7}\\\frac{7c}{105}=\frac{10}{7}\end{cases}\Rightarrow\hept{\begin{cases}\frac{a}{14}=\frac{10}{7}\\\frac{b}{21}=\frac{10}{7}\\\frac{c}{15}=\frac{10}{7}\end{cases}\Rightarrow}\hept{\begin{cases}a=\frac{10}{7}.14=20\\b=\frac{10}{7}.21=30\\c=\frac{10}{7}.15=\frac{150}{7}\end{cases}}}\)

Vậy \(a=20;b=30;c=\frac{150}{7}\)

~ Ủng hộ nhé 

\(3a=2b\Rightarrow\frac{a}{2}=\frac{b}{3}\Rightarrow\frac{a}{14}=\frac{b}{21}\)

\(5b=7c\Rightarrow\frac{b}{7}=\frac{c}{5}\Rightarrow\frac{b}{21}=\frac{c}{15}\)

\(\Rightarrow\frac{a}{14}=\frac{b}{21}=\frac{c}{15}\)

\(\Rightarrow\frac{3a}{42}=\frac{5b}{105}=\frac{7c}{105}=\frac{3a+5b-7c}{42+105-105}\)

\(=\frac{60}{42}=\frac{10}{7}\)

\(\Leftrightarrow\frac{a}{14}=\frac{b}{21}=\frac{c}{15}=\frac{10}{7}\)

3a=2b => a/2 = b/3  => a/14 = b/21

5b = 7c => b/7 = c/5 => b/21 = c/15

=> a/14 = b/21= c/15

= \(\frac{3a}{42}=\frac{7b}{147}=\frac{5c}{75}=\frac{3a+5c-7b}{42+75-147}=\frac{60}{-30}=-2\)

=> a = -2.14 = -28

b = -2.21 = -42

c= -2 . 15= -30

kết quả là 1,2

ta có 3a + 5b - 7c =60 mà 3a=2b;5b=7c

suy ra 3a + 5b -5b=60=> a=20 mà 3a=2b=> b=30=> c =\(\frac{150}{7}\)

ta có: 3a=2b--> \(\frac{a}{2}=\frac{b}{3}\)--> \(\frac{a}{14}=\frac{b}{21}\)(1)

          5b=7c-->\(\frac{b}{7}=\frac{c}{5}\)--> \(\frac{b}{21}=\frac{c}{15}\)(2)

từ (1)và(2)--> \(\frac{a}{14}=\frac{b}{21}=\frac{c}{15}\)và 3a+5b-7c=60

áp dụng tính chất của dãy tỉ số bằng nhau ta có

\(\frac{a}{14}=\frac{b}{21}=\frac{c}{15}=\frac{3a+5b-7c}{42+105-105}=\frac{60}{42}=\frac{10}{7}\)

--> a=\(\frac{10}{7}.14=20\)

      b=\(\frac{10}{7}.21=30\)

      c=\(\frac{10}{7}.15=\frac{150}{7}\)

vậy a=20 , b=30 và  c=\(\frac{150}{7}\)

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

            \(5b=7c\Rightarrow\frac{b}{7}=\frac{c}{5}\Rightarrow\frac{b}{21}=\frac{c}{15}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{a}{14}=\frac{b}{21}=\frac{c}{15}\) và \(3a+5b-7c=60\) 

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

     \(\frac{a}{14}=\frac{b}{21}=\frac{c}{15}=\frac{3a}{42}=\frac{5b}{105}=\frac{7c}{105}=\frac{3a+5b-7c}{42+105-105}=\frac{60}{42}=\frac{10}{7}\)

\(\Rightarrow\frac{a}{14}=\frac{10}{7}\Rightarrow a=\frac{10.14}{7}=\frac{140}{7}=20\)

\(\Rightarrow\frac{b}{21}=\frac{10}{7}\Rightarrow b=\frac{10.21}{7}=\frac{210}{7}=30\)

\(\Rightarrow\frac{c}{15}=\frac{10}{7}\Rightarrow c=\frac{10.15}{7}=\frac{150}{7}\)

 Vậy:  \(a=20;b=30;c=\frac{150}{7}\)