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bài 2 : a) \(\dfrac{a-1}{2}=\dfrac{b+3}{4}=\dfrac{c-5}{6}\)
áp dụng dảy tỉ số bằng nhau
ta có : \(\dfrac{5\left(a-1\right)-3\left(b+3\right)-4\left(c-5\right)}{5.2-3.4-4.6}\)
\(=\dfrac{5a-5-3b-9-4c+20}{10-12-24}=\dfrac{\left(5a-3b-4c\right)-5-9+20}{-26}\)
\(=\dfrac{46+6}{-26}=\dfrac{52}{-26}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a-1}{2}=-2\\\dfrac{b+3}{4}=-2\\\dfrac{c-5}{6}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a-1=-4\\b+3=-8\\c-5=-12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=-3\\b=-11\\c=-7\end{matrix}\right.\)
vậy \(a=-3;b=-11;c=-7\)
b) ta có : \(3a=2b\Leftrightarrow6a=4b=5c\Leftrightarrow\dfrac{6a}{2}=\dfrac{4b}{2}=\dfrac{5c}{2}\)
áp dụng dảy tỉ số bằng nhau
ta có \(\dfrac{-60a-60b+60c}{-10.2-15.2+12.2}=\dfrac{60\left(-a-b+c\right)}{-20-30+24}\)
\(=\dfrac{60\left(-52\right)}{-26}=\dfrac{-3120}{-26}=120\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{6a}{2}=120\\\dfrac{4b}{2}=120\\\dfrac{5c}{2}=120\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6a=240\\4b=240\\5c=240\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=40\\b=60\\c=48\end{matrix}\right.\)
vậy \(a=40;b=60;c=48\)
Theo đầu bài ta có:
\(\frac{a-1}{2}=\frac{b+3}{4}=\frac{c-5}{6}\)
\(\Rightarrow\frac{5c-25}{30}=\frac{3a-3}{6}=\frac{4b+12}{16}\)
\(=\frac{\left(5c-25\right)-\left(3a-3\right)-\left(4b+12\right)}{30-6-16}\)
\(=\frac{\left(5c-3a-4b\right)-\left(25-3+12\right)}{8}\)
\(=\frac{50-34}{8}=\frac{16}{8}=2\)
\(\Rightarrow\hept{\begin{cases}a=2\cdot2+1=5\\b=2\cdot4-3=5\\c=2\cdot6+5=17\end{cases}}\)
Theo t,c dãy tỉ số bằng nhau ta có :
\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-5c}{2}=\dfrac{5\left(3a-2b\right)\left(2c-5a\right)}{5.5+3.3+}=\dfrac{-10b+6c}{34}=\dfrac{-5b+3c}{17}\)
\(\Leftrightarrow\dfrac{5b-3c}{2}=\dfrac{-5b+3c}{17}\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{3c}{5}\\a=\dfrac{2c}{5}\end{matrix}\right.\)
Mà \(a+b+c=-50\)
\(\Leftrightarrow\dfrac{2c}{5}+\dfrac{3c}{5}+c=-50\)
\(\Leftrightarrow2c=-50\)
\(\Leftrightarrow c=-25\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-10\\b=-15\end{matrix}\right.\)
Vậy ...
\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}\leftrightarrow\dfrac{5\left(3a-2b\right)}{25}=\dfrac{3\left(2c-5a\right)}{9}=\dfrac{2\left(5b-3c\right)}{4}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{5\left(3a-2b\right)}{25}=\dfrac{3\left(2c-5a\right)}{9}=\dfrac{2\left(5b-3c\right)}{4}=\dfrac{5\left(3a-2b\right)+3\left(2c-5a\right)+2\left(5b-3c\right)}{25+9+4}=0\)\(\Rightarrow\left\{{}\begin{matrix}3a-2b=0\\2c-5a=0\\5b-3c=0\end{matrix}\right.\)
⇔ 15a= 10b = 6c ⇔ \(\dfrac{a}{\dfrac{1}{15}}=\dfrac{b}{\dfrac{1}{10}}=\dfrac{c}{\dfrac{1}{6}}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{\dfrac{1}{15}}=\dfrac{b}{\dfrac{1}{10}}=\dfrac{c}{\dfrac{1}{6}}=\dfrac{a+b+c}{\dfrac{1}{15}+\dfrac{1}{10}+\dfrac{1}{6}}=-\dfrac{50}{\dfrac{1}{3}}=-150\)
\(\Rightarrow\left\{{}\begin{matrix}a=-10\\b=-15\\c=-25\end{matrix}\right.\)
Ta có : \(3a=-2b\Leftrightarrow a=\dfrac{-2b}{3}\)
\(5c=-2b\Leftrightarrow c=\dfrac{-2b}{5}\)
Thay vào \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\), ta có :
\(\Leftrightarrow\dfrac{1}{-\dfrac{2b}{3}}+\dfrac{1}{b}+\dfrac{1}{-\dfrac{2b}{5}}=2\)
\(\Leftrightarrow\dfrac{-3}{2b}+\dfrac{1}{b}+\dfrac{-5}{2b}=2\)
\(\Leftrightarrow\dfrac{-3+2-5}{2b}=2\)
\(\Leftrightarrow-3b=2\)
\(\Leftrightarrow b=-\dfrac{2}{3}\)
Ta có : \(a=\dfrac{-2b}{3}=\dfrac{-2.\left(-\dfrac{2}{3}\right)}{3}=\dfrac{4}{9}\)
\(c=\dfrac{-2b}{5}=\dfrac{-2.\left(-\dfrac{2}{3}\right)}{5}=\dfrac{4}{15}\)
Vậy \(a=\dfrac{4}{9},b=-\dfrac{2}{3},c=\dfrac{4}{15}\).
@Tuấn Anh Phan Nguyễn
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