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a: =>3/5-x+13/20=5/6
=>11/10-x=5/6
=>x=11/10-5/6=4/15
b: =>x=4/9+5/9=1
c: =>3/2-2x=16/3:8/3=2
=>2x=-1/2
=>x=-1/4
d: =>x+1,5=-5/8*8/5=-1
=>x=-2,5
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Bài 1:
\(a,\left(\frac{1}{2}\right)^{3x-1}=\frac{1}{32}\)
\(\Rightarrow\left(\frac{1}{2}\right)^{3x-1}=\left(\frac{1}{2}\right)^5\)
\(\Rightarrow3x-1=5\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
\(b,\frac{64}{\left(-2\right)^x}=-8\)
\(\Rightarrow\left(-2\right)^x=64:\left(-8\right)\)
\(\Rightarrow\left(-2\right)^x=-8\)
\(\Rightarrow\left(-2\right)^x=\left(-2\right)^3\)
\(\Rightarrow x=3\)
\(c,\left(-2\right)^3+0,5x=1,5\)
\(\Rightarrow-8+0,5x=1,5\)
\(\Rightarrow0,5x=9,5\)
\(\Rightarrow x=19\)
Bài 2,
\(a,\frac{3}{8}.27\frac{1}{5}-51\frac{1}{5}.\frac{3}{8}+19=\frac{3}{8}\left(27\frac{1}{5}-51\frac{1}{5}\right)+19\)
\(=\frac{3}{8}\left(-24\right)+19\)
\(=-9+19=10\)
\(b,25\left(-\frac{1}{5}\right)^3+\frac{1}{5}-2.\left(-\frac{1}{2}\right)^2-\frac{1}{2}=-\frac{1}{5}+\frac{1}{5}-\left(-\frac{1}{2}\right)-\frac{1}{2}\)
\(=0\)
\(c,35\frac{1}{6}:\left(-\frac{4}{5}\right)-45\frac{1}{6}:\left(-\frac{4}{5}\right)=-\frac{4}{5}\left(35\frac{1}{6}-45\frac{1}{6}\right)\)
\(=-\frac{4}{5}.\left(-10\right)=8\)
\(d,\left(-0,25\right).8,7.2^2=\left(-0,25\right).8,7.4\)
\(=-1.8,7=-8,7\)
\(e,13\frac{1}{3}.\frac{5}{8}-23\frac{1}{3}:\frac{8}{5}=\frac{40}{3}.\frac{5}{8}-\frac{70}{3}.\frac{5}{8}\)
\(=\frac{5}{8}\left(\frac{40}{3}-\frac{70}{3}\right)\)
\(=\frac{5}{8}.-10\)
\(=-\frac{25}{4}\)
1
\(\left(x-2\right):2.3=6\)
\(\Leftrightarrow\left(x-2\right):2=2\)
\(\Leftrightarrow\left(x-2\right)=4\)
\(\Leftrightarrow x=4+2=6\)
c) ta có
\(\left[\left(2x+1\right)+1\right]m:2=625\)
\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)
\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)
\(\Leftrightarrow\left(2x+1\right)^2=1250\)
...
2
\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}:\frac{5}{3}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}.\frac{3}{5}\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)
\(a,\dfrac{-1}{8}=\dfrac{3}{x}\\ \dfrac{3}{-24}=\dfrac{3}{x}\\ x=-24\\ b,\dfrac{x}{3}=\dfrac{3}{x}\\ x.x=3.3\\ x^2=9\\ x=\pm3\\ c,\dfrac{3}{4}.x=1\dfrac{1}{2}\\ \dfrac{3}{4}.x=\dfrac{3}{2}\\ x=\dfrac{3}{2}:\dfrac{3}{4}\\ x=2\\ d,x-\dfrac{3}{10}=\dfrac{7}{15}:\dfrac{3}{5}\\ x-\dfrac{3}{10}=\dfrac{7}{9}\\ x=\dfrac{7}{9}+\dfrac{3}{10}\\ x=\dfrac{97}{90}\\ e,\dfrac{-4}{7}-x=\dfrac{-8}{3}.\dfrac{3}{7}\\ \dfrac{-4}{7}-x=\dfrac{-8}{7}\\ x=\dfrac{-4}{7}+\dfrac{8}{7}\\ x=\dfrac{4}{7}\\ \)
Bài giải:
Câu 1: a, \(\left(-2\right).4.5.38.\left(-25\right)\)
\(=\left[\left(-2\right).5\right].\left[4.\left(-25\right)\right].38\)
\(=\left(-10\right).\left(-100\right).38\)
\(=1000.38=38000\)
b,\(\frac{1}{3}+\frac{3}{8}-\frac{7}{12}\)
\(=\left(\frac{1}{3}+\frac{3}{8}\right)-\frac{7}{12}\)
\(=\frac{17}{24}-\frac{7}{12}=\frac{1}{8}\)
c, \(\frac{-5}{8}.\frac{5}{12}+\frac{-5}{8}.\frac{7}{12}+2\frac{1}{8}\)
\(=\frac{-5}{8}.\left(\frac{5}{12}+\frac{7}{12}\right)+\frac{17}{8}\)
\(=\frac{-5}{8}.1+\frac{17}{8}\)
\(=\frac{3}{2}\)
Câu 2: a, \(x-\frac{2}{5}=0,24\)
\(x-0,4=0,24\)
\(x=0,24+0,4\)
\(\Rightarrow x=0,64\left(\frac{16}{25}\right)\)
b,\(\frac{2}{3}.x+\frac{1}{12}=\frac{1}{10}\)
\(\frac{2}{3}.x=\frac{1}{10}-\frac{1}{12}\)
\(\frac{2}{3}.x=\frac{1}{60}\)
\(x=\frac{1}{60}:\frac{2}{3}\)
\(\Rightarrow x=\frac{1}{40}\)
c, \(\left(3\frac{1}{2}-2x\right).1\frac{1}{3}=7\frac{1}{3}\)
\(\frac{7}{2}-2x=\frac{22}{3}:\frac{4}{3}\)
\(\frac{7}{2}-2x=\frac{11}{2}\)
\(2x=\frac{7}{2}-\frac{11}{2}\)
\(2x=-2\)
\(\Rightarrow x=-2:2\)
\(x=-1\)
\(2,4:\left(-\frac{1}{2}-x\right)=1\frac{3}{5}\)
<=> 2,4 :(-1/2 - x) = 8/5
-1/2 - x = 2,4 : 8/5
-1/2 - x = 3/2
x = -1/2 - 3/2
x = -2