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\(a)5871+[247-82]\times5\) \(b)1,71:[4,418+1,582]+1,94\)
\(=5871+\)\(165\) \(\times5\) \(=1,71:\) \(6\) \(+1,94\)
\(=5871\)\(+\) \(825\) \(=\) \(0,285\) \(+1,94\)
\(=\) \(6696\) \(=\) \(2,225\)
\(c)\frac{7}{2}-\frac{3}{4}:\frac{1}{2}+\frac{2}{5}\) \(d)6\times2\frac{1}{3}-4\frac{3}{7}\)
\(=\)\(\frac{7}{2}-\frac{3}{2}+\frac{2}{5}\) \(=6\times2-\frac{31}{7}\)
\(=2+\frac{2}{5}\) \(=12-\frac{31}{7}\)
\(=\frac{12}{5}\) \(=\frac{53}{7}\)
a) \(x+\frac{1}{5}-\frac{2}{7}=4\frac{1}{2}\)
=> \(x+\frac{1}{5}-\frac{2}{7}=\frac{9}{2}\)
=> \(x=\frac{9}{2}+\frac{2}{7}-\frac{1}{5}=\frac{321}{70}\)
b) \(1\frac{3}{4}+x-\frac{7}{8}=5\)
=> \(\frac{7}{4}+x-\frac{7}{8}=5\)
=> \(\frac{7}{4}+x=\frac{47}{8}\)
=> \(x=\frac{33}{8}\)
c) \(\frac{1}{2}-\frac{1}{3}\cdot x=\frac{3}{4}\)
=> \(\frac{1}{3}\cdot x=\frac{1}{2}-\frac{3}{4}=-\frac{1}{4}\)
=> \(x=\left(-\frac{1}{4}\right):\frac{1}{3}=\left(-\frac{1}{4}\right)\cdot3=-\frac{3}{4}\)
2/3+5/8=16/24+15/24=31/24
4 và 2/5-2 và 2/3=22/5-7/3=66/15-35/15=31/15
x + 2 và 1/5 =4 và 2/3
x + 11/5 = 14/3
x = 14/3-11/5
x = 37/15
x : 3/2 = 4/15
x = 4/15*3/2
x = 2/5
Bài 3 :
b) Ta có 1+ 2 + 3 +4 + ...+ x =15
Nên \(\frac{x\left(x+1\right)}{2}=15\)
\(x\left(x+1\right)=30\)
=> \(x\left(x+1\right)=5.6\)
=> x = 5
Bài 2:
h; \(\dfrac{2}{3}\)\(x\) + 50% + \(x\) = \(\dfrac{1}{10}\)
\(\dfrac{2}{3}\)\(x\) + \(\dfrac{1}{2}\) + \(x\) = \(\dfrac{1}{10}\)
(\(\dfrac{2}{3}\)\(x\) + \(x\)) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) (\(\dfrac{2}{3}\) + 1) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{1}{10}\) - \(\dfrac{1}{2}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{-2}{5}\)
\(x\) = \(\dfrac{-2}{5}\): \(\dfrac{5}{3}\)
\(x\) = - \(\dfrac{6}{25}\)
Lớp 5 chưa học số âm em nhé.
\(5871:\left[928-\left(247-82.5\right)\right]\)
\(=5871:\left[928-\left(247-460\right)\right]\)
\(=5871:\left[928-\left(-213\right)\right]\)
\(=5871:1141\)