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a)
\(3\frac{4}{5}:40\frac{8}{15}=0,25:x\)
\(\Rightarrow\frac{19}{5}:\frac{608}{15}=\frac{1}{4}.x\)
\(\Rightarrow\frac{1}{4}.x=\frac{3}{32}\)
\(\Rightarrow x=\frac{3}{8}\)
Vậy x = 3 / 8
b) \(\frac{5}{6}:x=20:3\)
\(\Rightarrow\frac{5}{6x}=\frac{20}{3}\)
\(\Rightarrow120x=15\)
\(\Rightarrow x=\frac{1}{5}\)
Vậy x = 1 / 5
c)
\(x:2,5=0,003:0,75\)
\(\Rightarrow x.\frac{2}{5}=\frac{3}{1000}.\frac{4}{3}\)
\(\Rightarrow x.\frac{2}{5}=\frac{1}{250}\)
\(\Rightarrow x=\frac{1}{100}\)
d)
\(\frac{2}{3}:0,4=x:\frac{4}{5}\)
\(\Rightarrow\frac{2}{3}:\frac{2}{5}=\frac{5x}{4}\)
\(\Rightarrow\frac{5x}{4}=\frac{5}{3}\)
\(\Rightarrow15x=20\)
\(\Rightarrow x=\frac{4}{3}\)
Vậy x = 4 / 3
a) \(2,5:2,7=x:\frac{3}{5}\)
\(\Rightarrow\frac{5}{2}:\frac{27}{10}=x.\frac{5}{3}\)
\(\Rightarrow\frac{5}{2}.\frac{10}{27}=\frac{5x}{3}\)
\(\Rightarrow\frac{25}{27}=\frac{5x}{3}\)
\(\Rightarrow5x=\frac{25.3}{27}=\frac{75}{27}=\frac{25}{9}\)
\(\Rightarrow x=\frac{25}{9}:5=\frac{25}{9}.\frac{1}{5}=\frac{5}{9}\)
b) \(\frac{19}{5}:\frac{608}{15}=0,25:x\)
\(\Rightarrow\frac{19}{5}.\frac{15}{608}=\frac{1}{4}:x\)
\(\Rightarrow\frac{3}{32}=\frac{1}{4}:x\)
\(\Rightarrow x=\frac{1}{4}:\frac{3}{32}=\frac{1}{4}.\frac{32}{3}=\frac{8}{3}\)
c/ \(\frac{8}{3}:x=\frac{16}{9}:0,2\)
\(\Rightarrow\frac{8}{3}:x=\frac{16}{9}:\frac{1}{5}=\frac{16}{9}.5=\frac{80}{9}\)
\(\Rightarrow x=\frac{8}{3}:\frac{80}{9}=\frac{8}{3}.\frac{9}{80}=\frac{3}{10}\)
\(\begin{array}{l}a)\dfrac{x}{6} = \dfrac{{ - 3}}{4}\\x = \dfrac{{( - 3).6}}{4}\\x = \dfrac{{ - 9}}{2}\end{array}\)
Vậy \(x = \dfrac{{ - 9}}{2}\)
\(\begin{array}{l}b)\dfrac{5}{x} = \dfrac{{15}}{{ - 20}}\\x = \dfrac{{5.( - 20)}}{{15}}\\x = \dfrac{{ - 20}}{3}\end{array}\)
Vậy \(x = \dfrac{{ - 20}}{3}\)
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. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(7.9=\left(x-1\right).\left(x+1\right)\)
\(63=x^2-1\)
\(x^2=63+1\)
\(x^2=64\)
\(x^2=8^2\)
\(x=8\)
2:
a: Áp dụng tính chất của DTSBN, ta được:
a/5=b/-2=(a+b)/(5-2)=12/3=4
=>a=20; b=-8
b: Áp dụng tính chất của DTSBN, ta được:
a/4=b/5=(3a-2b)/(3*4-2*5)=42/2=21
=>a=84; b=105
