Tìm Sht x trong tỉ lê thức sau
37-x/x+13. = 3/7
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a) \(\left(\dfrac{1}{4}x\right):3=\dfrac{5}{6}:\dfrac{1}{8}\\ \left(\dfrac{1}{4}x\right):3=\dfrac{20}{3}\\ \dfrac{1}{4}x=\dfrac{20.3}{3}\\ \dfrac{1}{4}x=20\\ x=80\)
a: Ta có: \(\dfrac{0.25x}{3}=\dfrac{5.6}{0.125}\)
\(\Leftrightarrow x\cdot\dfrac{1}{4}=134.4\)
hay x=537,6
b: Ta có: \(\dfrac{2.5}{7.5}=x:\dfrac{3}{5}\)
\(\Leftrightarrow x:\dfrac{3}{5}=\dfrac{1}{3}\)
hay \(x=\dfrac{1}{5}\)
a)\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)
\(\Rightarrow x^2+2x-3=x^2-4\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\frac{1}{2}\)
b)\(\frac{37-x}{x+13}=\frac{3}{7}\)\(\Rightarrow7\left(37-x\right)=3\left(x+13\right)\)
\(\Rightarrow259-7x=3x+39\)
\(\Rightarrow10x=220\)
\(\Rightarrow x=22\)
a, 0,4 : x = x : 0,9
<=> x2 = 0,4 . 0,9
<=> x2 = 0,36
<=> x = 0,6 hoặc -0,6
b, \(13\frac{1}{3}\div1\frac{1}{3}=26\div\left(2x-1\right)\)
\(\Leftrightarrow\frac{40}{3}\div\frac{4}{3}=26\div\left(2x-1\right)\)
\(\Leftrightarrow10=26\div\left(2x-1\right)\)
\(\Leftrightarrow2x-1=\frac{13}{5}\)
\(\Leftrightarrow2x=\frac{18}{5}\)
\(\Leftrightarrow x=\frac{9}{5}\)
c, \(0,2\div1\frac{1}{5}=\frac{2}{3}\div\left(6x+7\right)\)
\(\Leftrightarrow\frac{1}{5}\div\frac{6}{5}=\frac{2}{3}\div\left(6x+7\right)\)
\(\Leftrightarrow\frac{1}{6}=\frac{2}{3}\div\left(6x+7\right)\)
\(\Leftrightarrow6x+7=4\)
\(\Leftrightarrow6x=-3\)
\(\Leftrightarrow x=\frac{-1}{2}\)
d, \(\frac{37-x}{x+13}=\frac{3}{7}\)
\(\Leftrightarrow7\left(37-x\right)=3\left(x+13\right)\)
\(\Leftrightarrow259-7x=3x+39\)
\(\Leftrightarrow-10x=-220\)
\(\Leftrightarrow x=22\)
\(\text{a)}\frac{37-x}{x+13}=\frac{3}{7}\)
\(\Leftrightarrow3\left(x+13\right)=7\left(37-x\right)\)
\(\Leftrightarrow3x+39=259-7x\)
\(\Leftrightarrow3x+7x=259-39\)
\(\Leftrightarrow10x=220\)
\(\Leftrightarrow x=22\)
\(\text{b)}\frac{x+4}{20}=\frac{5}{x+4}\)
\(\Leftrightarrow\left(x+4\right)^2=100\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=10\\x+4=-10\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=6\\x=-14\end{cases}}\)