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
