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<=> \(\frac{15a+364}{a}=41\Leftrightarrow15a+364=41a\Leftrightarrow26a=364\Rightarrow a=14\)
a) \(92.4-27=\frac{x+350}{x}+315\)
\(368-27=1+\frac{350}{x}+315\)
\(341=\frac{350}{x}+316\)
\(341-316=\frac{350}{x}\)
\(25=\frac{350}{x}\)
\(25x=350\)
\(x=14\)
b) \(\left(\frac{3}{2}-x\right).\frac{1}{3}-\frac{2}{3}.\left(\frac{1}{2}-x\right)=\frac{1}{3}\)
\(\frac{\frac{3}{2}}{3}-\frac{x}{3}-\frac{2}{3}.\left(\frac{1}{2}-x\right)=\frac{1}{3}\)
\(\frac{1}{2}-\frac{x}{2}-\frac{2}{3}.\left(\frac{1}{2}-x\right)=\frac{1}{3}\)
\(\frac{3}{2}-x-2.\left(\frac{1}{2}-x\right)=1\)
\(\frac{3}{2}-x-1+2x=1\)
\(\frac{1}{2}+x=1\)
\(x=1-\frac{1}{2}\)
\(x=\frac{1}{2}\)
2:
=1-1+1-1=0
3:
a: =>34*(100+1)/2:a=17
=>a=101
b: =>5/3(x-1/2)=5/4
=>x-1/2=5/4:5/3=3/4
=>x=5/4
1a, \(\dfrac{2005}{2001}\) = 1+\(\dfrac{4}{2001}\); \(\dfrac{2009}{2005}\)=1+\(\dfrac{4}{2005}\)vì\(\dfrac{4}{2001}\)>\(\dfrac{4}{2005}\)nên\(\dfrac{2005}{2001}\)>\(\dfrac{2009}{2005}\)
1b,\(\dfrac{1313}{1515}\)=\(\dfrac{1313:101}{1515:101}\)= \(\dfrac{13}{15}\); \(\dfrac{131313}{151515}\)=\(\dfrac{131313:10101}{151515:10101}\)=\(\dfrac{13}{15}\)
Vậy \(\dfrac{13}{15}\)=\(\dfrac{1313}{1515}\)=\(\dfrac{131313}{151515}\)