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Phần a là 0
Phần b là 5
Phần c là 6
Phần d là 1
Phần e là 0
a, x + 2/5 = 1/2
x = 1/2 + 2/5
x = 9/10 b,34/5 + x = 15/8
x = 34/5 - 15/8
x =197/40 c, x - 45/6 = 21/6 + 5/6
x - 45/6 = 26/6
x = 26/6 + 45/6
x =71/6 d,14/5 - x = 11/5 - 3/5
14/5 - x = 8/5
x = 14,5 - 8/5
x =6/5
a, x + 2/5 = 1/2
x = 1/2 + 2/5
x = 9/10
b,34/5 + x = 15/8
x = 34/5 - 15/8
x =197/40
c, x - 45/6 = 21/6 + 5/6
x - 45/6 = 26/6
x = 26/6 + 45/6
x =71/6
d,14/5 - x = 11/5 - 3/5
14/5 - x = 8/5
x = 14,5 - 8/5
x =6/5
- Làm lại.
a) \(x+2\frac{1}{6}=3\frac{1}{3}\)
\(x+\frac{13}{6}=\frac{10}{3}\)
\(x=\frac{10}{3}-\frac{13}{6}=\frac{7}{6}\)
b) \(x-2\frac{3}{7}=3\)
\(x-\frac{17}{7}=3\)
\(x=3+\frac{17}{7}=\frac{38}{7}\)
c) \(x.1\frac{1}{3}=2\frac{2}{3}\)(dấu chấm là dấu nhân)
\(x.\frac{4}{3}=\frac{8}{3}\)
\(x=\frac{8}{3}:\frac{4}{3}=2\)
(Nhớ k cho mình với nhé!)
\(\frac{70}{3}\left(\frac{39}{30}+\frac{39}{42}\right)-\frac{246}{7}\div\left(\frac{41}{56}+\frac{41}{72}\right)\)
\(=\frac{70}{3}\left(\frac{13}{10}+\frac{13}{14}\right)-\frac{246}{7}\div\left(\frac{41}{7\cdot8}+\frac{41}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(1+\frac{3}{10}+1-\frac{1}{14}\right)-\frac{246}{7}\div\left(\frac{40+1}{7\cdot8}+\frac{40+1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left[\left(1+1\right)+\left(\frac{3}{10}-\frac{1}{14}\right)\right]-\frac{246}{7}\div\left(\frac{5}{7}+\frac{1}{7\cdot8}+\frac{5}{9}+\frac{1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(2+\frac{8}{35}\right)-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\right]\)
\(=\frac{70}{3}\cdot\frac{78}{35}-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\right]\)
\(=\frac{35\cdot2\cdot26\cdot3}{3\cdot35}-\frac{246}{7}\div\left(\frac{5}{7}+\frac{5}{9}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=52-\frac{246}{7}\div\left[\left(\frac{5}{7}+\frac{1}{7}\right)+\left(\frac{5}{9}-\frac{1}{9}\right)\right]\)
\(=52-\frac{246}{7}\div\left(\frac{6}{7}+\frac{4}{9}\right)\)
\(=52-\frac{246}{7}\div\frac{82}{63}\)
\(=52-\frac{82\cdot3\cdot9\cdot7}{7\cdot82}\)
\(=52-27=25\)
\(\frac{57}{20}-\frac{26}{15}+\frac{139}{20}\div3\)
\(=\frac{57}{20}-\frac{26}{15}+\frac{139}{60}\)
\(=\frac{171}{60}-\frac{104}{60}+\frac{139}{60}=\frac{103}{30}\)
\(\frac{39}{4}+\frac{2}{3}\left(11-\frac{23}{4}\right)\)
\(=\frac{39}{4}+11\cdot\frac{2}{3}-\frac{23}{4}\cdot\frac{2}{3}\)
\(=\frac{39}{4}+\frac{22}{3}-\frac{56}{12}\)
\(=\frac{119}{12}+\frac{88}{12}-\frac{56}{12}=\frac{151}{12}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2002}\right)\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2001}{2002}\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2001\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2002\cdot2003\cdot2004}=\frac{1}{2004}\)
a)\(x.3-x:3=36,5.4\)
\(\Leftrightarrow x.\left(3-\frac{1}{3}\right)=146\)
\(\Leftrightarrow x.\left(\frac{2}{3}\right)=146\)
\(\Leftrightarrow x=219\)
b)\(\left(x+1\right)+\left(x+4\right)+\left(x+7\right)+...+\left(x+28\right)=156\)
\(\Leftrightarrow10x+145=156\)
\(\Leftrightarrow10x=11\)
\(\Leftrightarrow x=\frac{11}{10}\)
c)\(11.\left(x-6\right)=\left(4x\right)+11\)
\(\Leftrightarrow11x-66=4x+11\)
\(\Leftrightarrow7x=77\)
\(\Leftrightarrow x=11\)
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