Bài 1:
Ta có: \(\frac{a}{5}=\frac{b}{3}\Rightarrow\frac{a}{25}=\frac{b}{15}.\)
\(\frac{b}{5}=\frac{c}{4}\Rightarrow\frac{b}{15}=\frac{c}{12}.\)
=> \(\frac{a}{25}=\frac{b}{15}=\frac{c}{12}\) và \(a-b+c=147.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{a}{25}=\frac{b}{15}=\frac{c}{12}=\frac{a-b+c}{25-15+12}=\frac{147}{22}.\)
\(\left\{{}\begin{matrix}\frac{a}{25}=\frac{147}{22}\Rightarrow a=\frac{147}{22}.25=\frac{3675}{22}\\\frac{b}{15}=\frac{147}{22}\Rightarrow b=\frac{147}{22}.15=\frac{2205}{22}\\\frac{c}{12}=\frac{147}{22}\Rightarrow c=\frac{147}{22}.12=\frac{882}{11}\end{matrix}\right.\)
Vậy \(\left(a;b;c\right)=\left(\frac{3675}{22};\frac{2205}{22};\frac{882}{11}\right).\)
Mình chỉ làm bài 1 thôi nhé.
Chúc bạn học tốt!
Bài 2:
Đặt \(A=\left|x-2001\right|+\left|x-2\right|\)
\(\Rightarrow A=\left|2001-x\right|+\left|x-2\right|\left(\text{vì }\left|x-2001\right|=\left|2001-x\right|\text{với mọi x}\in Q\right)\)
Có: \(\left|2001-x\right|\ge2001-x\); \(\left|x-2\right|\ge x-2\)
\(\Rightarrow\left|2001-x\right|+\left|x-2\right|\ge2001-x+x-2\\ \Rightarrow A\ge2001-2=1999\)
Vậy GTNN của | x - 2001 | + | x - 2 | = 1999
\("="\Leftrightarrow\left\{{}\begin{matrix}\left|2001-x\right|=2001-x\\\left|x-2\right|=x-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2001-x\ge0\\x-2\ge0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\le2001\\x\ge2\end{matrix}\right.\\ \Leftrightarrow2\le x\le2001\)