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\(\left(6+\left(\frac{1}{2}\right)^3-\left|-\frac{1}{2}\right|\right):\frac{3}{12}\)
\(=\left(6+\frac{1}{8}+\frac{1}{2}\right):\frac{1}{4}\)
=\(\frac{53}{8}:\frac{1}{4}\)
\(=\frac{53}{2}\)
a)30/60,-40/60,45/60,48/60
45/60>30/60>-40/60>-48/60
=3/4>1/2>-2/3>-4/5
a) \(\frac{7}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.\frac{-1}{2^3}=\frac{7.\left(-31\right).1.10.\left(-1\right)}{5.2.125.17.2^3}=\frac{31.7}{17.125.2^3}=\frac{217}{17000}\)
b) \(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).0=0\)
c) \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{3.4.5...100}{2.3.4...99}=\frac{100}{2}=50\)
d) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}=\frac{-\left(1.2.3..99\right)}{2.3.4...100}=-\frac{1}{100}\)
e) \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}=\frac{\left(1.2.3..29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}\)
\(=\frac{1.31}{30.2}=\frac{31}{60}\)
a) \(\left(-\frac{3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-9}{4}\right):\frac{3}{7}\)
= \(\left(-\frac{3}{4}+\frac{2}{5}\right)\cdot\frac{7}{3}+\left(\frac{3}{5}+\frac{-9}{4}\right)\cdot\frac{7}{3}\)
= \(\left(-\frac{15}{20}+\frac{8}{20}\right)\cdot\frac{7}{3}+\left(\frac{12}{20}-\frac{45}{20}\right)\cdot\frac{7}{3}\)
= \(-\frac{7}{20}\cdot\frac{7}{3}-\frac{33}{20}\cdot\frac{7}{3}\)
=\(\frac{7}{3}\cdot\left(-\frac{7}{20}-\frac{33}{20}\right)\)
=\(\frac{7}{3}\cdot\left(-2\right)\)
=\(-\frac{14}{3}\)
Câu a:
{3849 - 2.[146 - (2.14 + 6: 2)]} : 32
= {3849 - 2.[146 - (28 + 3)]} : 32
= {3489 - 2.[146 - 31]} : 32
= {3489 - 2.115} : 32
= {3489 - 230} : 32
= 3259 : 32
= 3259/32
Câu b:
134 - 2.{156 - 6[54 - 2.(9 + 6)]}
= 134 - 2.{156 - 6.[54- 2. 15]}
= 134 - 2.{156 - 6.[54 - 30]}
= 134 - 2.{156 - 6.24]
= 134 - 2.12
= 134 - 24
= 110
a: \(\Leftrightarrow-\dfrac{9}{46}+\dfrac{108}{46}-\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=1\)
\(\Leftrightarrow\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=\dfrac{53}{46}\)
\(\Leftrightarrow-\dfrac{5}{3}x+\dfrac{13}{4}=\dfrac{186}{53}\)
=>-5/3x=55/212
hay x=-33/212
c: \(\Leftrightarrow\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{18}{19}\)
\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{18}{19}\)
=>x+3=19
hay x=16
Câu 17:
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{2006}\right).\left(1+\frac{1}{2007}\right)\)
=\(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{2007}{2006}.\frac{2008}{2007}\)
\(=\frac{2008}{2}=1004\)
Câu 18:
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2006}\right).\left(1-\frac{1}{2007}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2005}{2006}.\frac{2006}{2007}\)
\(=\frac{1}{2007}\)
anh ms hok lp 8
xàm quá bạn. thế thì làm bài lớp 6 xem nào
lớp 6 mà
thầy giáo giao đó
bạn xem lại đề xem phải là
\(A=\frac{\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+...+\frac{18}{2}+\frac{19}{1}}{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{20}}\)
đúng thế không?
quên hết rùi(nỗi khổ!)
tui lớp 6 mà có bài nào thế đâu