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a: \(\dfrac{x}{0.9}=\dfrac{5}{6}\)
\(\Leftrightarrow x=\dfrac{3}{4}\)
b: \(\dfrac{-6}{x}=\dfrac{9}{-15}\)
\(\Leftrightarrow x=10\)
c: \(\dfrac{\dfrac{14}{15}}{\dfrac{9}{10}}=\dfrac{x}{\dfrac{3}{7}}\)
\(\Leftrightarrow x=\dfrac{3}{7}\cdot\dfrac{14}{15}:\dfrac{9}{10}=\dfrac{2}{5}\cdot\dfrac{10}{9}=\dfrac{20}{45}=\dfrac{4}{9}\)
\(\frac{x}{7}+\frac{9}{49}=\frac{3}{14}\)
\(\Rightarrow\frac{x}{7}=\frac{3}{14}-\frac{9}{49}=\frac{3}{98}\)
\(\Rightarrow x=3.7:98=\frac{3}{14}\)
Vậy \(x=\frac{3}{14}\)
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x+3}{8}=\frac{y-9}{7}=\frac{x+3-y+9}{8-7}=\frac{\left(x-y\right)+\left(3+9\right)}{1}=\frac{-14+12}{1}=-2\)
+) \(\frac{x+3}{8}=-2\Rightarrow x=-19\)
+) \(\frac{y-9}{7}=-2\Rightarrow y=-5\)
Vậy \(x=-19,y=-5\)
\(/x-\frac{1}{2}/=\frac{1}{3}\\ =>\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=-\frac{1}{3}\end{cases}}\\ =>\orbr{\begin{cases}x=\frac{1}{3}+\frac{1}{2}\\x=-\frac{1}{3}+\frac{1}{2}\end{cases}}\\ =>\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}\)
\(a,|x-\frac{1}{2}|=\frac{1}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=-\frac{1}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}}\)
\(b,\frac{14}{15}:\frac{9}{10}=x:\frac{3}{7}\)
\(\frac{28}{27}=x:\frac{3}{7}\)
\(x=\frac{4}{9}\)
ta có:
\(x:\frac{9}{14}=\frac{7}{3}:x\)
\(\Rightarrow\frac{14x}{9}=\frac{7}{3x}\)
\(\Rightarrow42x^2=63\)
\(\Rightarrow x^2=\frac{3}{2}\)
\(\Rightarrow x=\sqrt{\frac{3}{2}}\)
chúc bn hok tốt
\(x:\frac{9}{14}=\frac{7}{3}:x\)
\(\Leftrightarrow\)\(\frac{14x}{9}=\frac{7}{3x}\)
\(\Leftrightarrow\)\(42x^2=63\)
\(\Leftrightarrow\) x2 = \(\frac{63}{42}\)
\(\Leftrightarrow\)\(x=\sqrt{\frac{63}{42}}\)