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a) \(12\dfrac{1}{3}-\left(3\dfrac{3}{4}+4\dfrac{3}{4}\right)=\dfrac{37}{3}-\left(\dfrac{15}{4}+\dfrac{19}{4}\right)\)
\(=\dfrac{37}{3}-\dfrac{34}{4}=\dfrac{37}{3}-\dfrac{17}{2}=\dfrac{74}{6}-\dfrac{51}{6}=\dfrac{23}{6}\)
b) \(3\dfrac{5}{6}+2\dfrac{1}{6}.6=\dfrac{23}{6}+\dfrac{13}{6}.6=\dfrac{23}{6}+\dfrac{78}{6}=\dfrac{101}{6}\)
c) \(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}=\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}=\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}=-\dfrac{92}{14}=-\dfrac{46}{7}\)
d) \(4\dfrac{1}{2}+\dfrac{1}{2}:5\dfrac{1}{2}=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}=\dfrac{9}{2}+\dfrac{1}{2}.\dfrac{2}{11}=\dfrac{9}{2}+\dfrac{1}{11}=\dfrac{99}{22}+\dfrac{2}{22}=\dfrac{101}{22}\)
a. \(12\dfrac{1}{3}-\left(3\dfrac{3}{4}+4\dfrac{3}{4}\right)=\dfrac{37}{3}-\left(\dfrac{15}{4}+\dfrac{19}{4}\right)\)
\(=\dfrac{37}{3}-\dfrac{34}{4}=\dfrac{23}{6}\)
\(b.3\dfrac{5}{6}+2\dfrac{1}{6}.6=\dfrac{23}{6}+13=\dfrac{101}{6}\)
\(c.3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}=\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}=\dfrac{20}{7}\)
d \(4\dfrac{1}{2}+\dfrac{1}{2}:5\dfrac{1}{2}\)
\(=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}\)
\(=\dfrac{9}{2}+\dfrac{1}{11}\)
\(=\dfrac{101}{22}\)
a: 14/5-7/5=7/5
b: 7/8-1/3+5/4
=21/24-8/24+30/24
=43/24
c; =7/6+5/6+2/15+13/15
=2+1
=3
d: =4*5/3*11=20/33
e: =2/9*1/6*1/4=2/9*1/24=1/108
2:
a: \(=\dfrac{3}{9}\cdot\dfrac{4}{4}\cdot\dfrac{5}{5}\cdot\dfrac{6}{6}\cdot\dfrac{7}{7}=\dfrac{1}{3}\)
b: \(=\dfrac{1}{6}\left(\dfrac{22}{3}-\dfrac{2}{3}\right)=\dfrac{10}{3}\cdot\dfrac{1}{6}=\dfrac{10}{18}=\dfrac{5}{9}\)
c; \(=\dfrac{1}{3}\left(9-\dfrac{2}{5}-\dfrac{3}{5}\right)=\dfrac{8}{3}\)
1: =72/90+65/90=137/90
2: =24/56-77/56=-53/56
3: =-7/10+4/5=1/10
4: =15/100-4/100=11/100
5: =4/6-5/6=-1/6
6: =10/40-15/40-76/40=-81/40
7: =-9/10+7/18
=-81/90+35/90=-46/90=-23/45
8: =27/90-55/90=-28/90=-14/45
9: =36/60-50/60-35/60=-49/60
10: =-4/9+5/6-3/8
=-32/72+60/72-27/72
=1/72
a: =(-1)+(-1)+...+(-1)=-1011
b: =(-5)+(-5)+...+(-5)=-175
\(\dfrac{4}{5}\) : (\(\dfrac{4}{5}\) .- \(\dfrac{5}{4}\)) : (\(\dfrac{16}{25}\) - \(\dfrac{1}{5}\))
= \(\dfrac{4}{5}\) : (-1) : (\(\dfrac{16}{25}\) - \(\dfrac{5}{25}\))
= -\(\dfrac{4}{5}\) : \(\dfrac{11}{25}\)
= - \(\dfrac{4}{5}\) x \(\dfrac{25}{11}\)
= - \(\dfrac{20}{11}\)
\(\dfrac{4}{5}\): (\(\dfrac{4}{5}\).-\(\dfrac{5}{4}\)) : (\(\dfrac{16}{25}\) - \(\dfrac{1}{5}\))
=\(\dfrac{4}{5}\) x - 1: (\(\dfrac{16}{25}\) - \(\dfrac{5}{25}\))
= - \(\dfrac{4}{5}\) : \(\dfrac{11}{25}\)
= - \(\dfrac{4}{5}\) x \(\dfrac{25}{11}\)
= - \(\dfrac{20}{11}\)
\(\dfrac{11}{12}\): (\(\dfrac{7}{9}\) + - \(\dfrac{1}{3}\)) - (\(\dfrac{2}{3}\) - \(\dfrac{5}{15}\))
= \(\dfrac{11}{12}\) : (\(\dfrac{7}{9}\) - \(\dfrac{3}{9}\)) - (\(\dfrac{2}{3}\) - \(\dfrac{1}{3}\))
= \(\dfrac{11}{12}\) : \(\dfrac{4}{9}\) - \(\dfrac{1}{3}\)
= \(\dfrac{11}{12}\) x \(\dfrac{9}{4}\) - \(\dfrac{1}{3}\)
= \(\dfrac{99}{48}\) - \(\dfrac{16}{48}\)
= \(\dfrac{83}{48}\)
1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
1+5+3+4=13
5+3+6=14
giải
1+5+3+4=13
5+3+6=14