Tìm a,b,c biết tổng của chúng bằng -40 và \(\dfrac{a}{3}=\dfrac{b}{2},\dfrac{b}{6}=\dfrac{c}{5}\)
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Ta có: \(\dfrac{a}{3}=\dfrac{b}{4}\)
\(\dfrac{b}{2}=\dfrac{c}{5}\Rightarrow\dfrac{b}{4}=\dfrac{c}{10}\)
\(\Rightarrow\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{10}=\dfrac{a-c+b}{3-10+4}=\dfrac{3}{-3}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}a=\left(-1\right).3=-3\\b=\left(-1\right).4=-4\\c=\left(-1\right).10=-10\end{matrix}\right.\)
\(\dfrac{b}{2}=\dfrac{c}{5}\Rightarrow\dfrac{b}{4}=\dfrac{c}{10}\)
\(\Rightarrow\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{10}=\dfrac{a-c+b}{3-10+4}=\dfrac{3}{-3}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}a=\left(-1\right).3=-3\\b=\left(-1\right).4=-4\\c=\left(-1\right).10=-10\end{matrix}\right.\)
b) Ta có : \(\dfrac{2a}{3}=\dfrac{3b}{4}=\dfrac{4c}{5}\)
\(\Leftrightarrow\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{4}{3}}=\dfrac{c}{\dfrac{5}{4}}=\dfrac{a+b+c}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)
Khi đó \(a=12.\dfrac{3}{2}=18;b=12.\dfrac{4}{3}=16;c=12.\dfrac{5}{4}=15\)
Vậy (a,b,c) = (18,16,15)
a: Ta có: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)
\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)
\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)
b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)
\(\Leftrightarrow x+1=\dfrac{50}{9}\)
hay \(x=\dfrac{41}{9}\)
c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
hay \(x\in\left\{8;-8\right\}\)
c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(7.9=\left(x-1\right).\left(x+1\right)\)
\(63=x^2-1\)
\(x^2=63+1\)
\(x^2=64\)
\(x^2=8^2\)
\(x=8\)
b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)
Đặt \(x=15k;y=20k;z=24k\)
Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}=\dfrac{a-b-c}{8-12-15}=\dfrac{28}{-19}=\dfrac{-28}{19}\)
Do đó: \(\left\{{}\begin{matrix}a=\dfrac{-224}{19}\\b=\dfrac{-336}{19}\\c=\dfrac{-420}{19}\end{matrix}\right.\)
Ta có \(\dfrac{a-b}{ab+1}+\dfrac{b-c}{bc+1}+\dfrac{c-a}{ca+1}=\dfrac{\left(a-b\right)\left(bc+1\right)\left(ca+1\right)+\left(b-c\right)\left(ca+1\right)\left(ab+1\right)+\left(a-b\right)\left(bc+1\right)\left(ca+1\right)}{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}=\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}\).
\(\dfrac{a}{3}=\dfrac{b}{2};\dfrac{b}{6}=\dfrac{c}{5}\)
\(\Rightarrow\dfrac{a}{9}=\dfrac{b}{6};\dfrac{b}{6}=\dfrac{c}{5}\)
\(\Rightarrow\dfrac{a}{9}=\dfrac{b}{6}=\dfrac{c}{5}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{9}=\dfrac{b}{6}=\dfrac{c}{5}=\dfrac{a+b+c}{9+6+5}=\dfrac{-40}{20}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}a=-2\cdot9=-18\\b=-2\cdot6=-12\\c=-2\cdot5=-10\end{matrix}\right.\)