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b) \(\frac{26+x}{39-x}=\frac{6}{7}\)
=> 7( 26+ x) = 6(39-x)
=>182 +7x = 234 - 6x
=> 7x+6x = 234-182
=> 13x= 52
=> x=4
a) \(\frac{26+x}{39+x}=\frac{6}{7}\)
=> 7(26+x) = 6(39+x)
=> 182 + 7 x = 234 + 6x
=> 7x - 6x = 234 - 182
=> x = 52
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Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{5}=\dfrac{x-y+z}{3-2+5}=-\dfrac{36}{6}=-6\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-6\right).3=-18\\y=\left(-6\right).2=-12\\z=\left(-6\right).5=-30\end{matrix}\right.\)
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\(\frac{21}{13}\cdot x=-\frac{7}{26}\)
\(\Leftrightarrow x=-\frac{1}{6}\)
vậy..........
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a, \(\left(x-2,5\right)\text{ : }1\frac{1}{2}=x\text{ : }2\)
\(\left(x-2,5\right)\text{ : }\frac{3}{2}=x\text{ : }2\)
\(\left(x-2,5\right)\text{ : }\frac{3}{2}\cdot2=x\)
\(\left(x-2,5\right)\text{ : }3=x\)
\(\frac{x}{3}-\frac{2,5}{3}-x=0\)
\(\frac{-2x}{3}-\frac{2,5}{3}=0\)
\(\frac{-2x-2,5}{3}=0\)
\(\Rightarrow\text{ }-2x-2,5=0\)
\(-2x=2,5\)
\(x=\frac{-2,5}{2}\)
\(=\dfrac{2^{14}}{2^6}\cdot\dfrac{3^6}{3^9}=\dfrac{2^8}{3^3}\)
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