\(\frac{1}{6}-\frac{1}{2}\) ; \(\frac{-7}{8}-(-1)\) ; \(\frac{2}{5}-\frac{5}{6};\frac{-1}{15}-\frac{1}{16};\frac{7}{24}-\frac{-5}{36};\frac{-7}{9}-\frac{-7}{12}\)
các bạn giải ra hộ mình nha cảm ơn ^_^ các bn lm đúng hộ nha
bài 1: tính A:=\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}-\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{2}{3}-\frac{1}{2}\)
Bài 2: Cho B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+.....+\frac{1}{49}-\frac{1}{50}\)
Chứng minh rằng: \(\frac{7}{12}< A< \frac{5}{6}\)
So sánh:\(\frac{\frac{\frac{1}{2}}{\frac{3}{4}}}{\frac{\frac{5}{6}}{\frac{7}{8}}}+\frac{\frac{\frac{8}{7}}{\frac{6}{5}}}{\frac{\frac{4}{3}}{\frac{2}{1}}}\) và\(\frac{\frac{\frac{1}{2}}{\frac{3}{4}}+\frac{\frac{8}{7}}{\frac{6}{5}}}{\frac{\frac{5}{6}}{\frac{7}{8}}+\frac{\frac{4}{3}}{\frac{2}{1}}}\)và \(\frac{\frac{\frac{1}{2}+\frac{8}{7}}{\frac{3}{4}+\frac{6}{5}}}{\frac{\frac{5}{6}+\frac{4}{3}}{\frac{7}{8}+\frac{2}{1}}}\)và\(\frac{\frac{\frac{1+8}{2+7}}{\frac{3+6}{4+5}}}{\frac{5+4}{\frac{6+3}{2+1}}}\)
\(3\frac{1}{2}-4\frac{2}{3}+\left[\frac{3}{4}-2\frac{1}{3}\right]-\left(\frac{5}{6}-\frac{7}{4}\right)+5\frac{1}{2}-3\)
\(2\frac{2}{3}-1\frac{2}{5}+1\frac{3}{10}-\left(\frac{2}{5}-\frac{5}{6}\right)+\frac{4}{15}-1\frac{1}{3}\)
\(\left[2\frac{1}{3}-1\frac{4}{3}\right]-\left(\frac{5}{4}-\frac{7}{12}+\frac{-11}{6}\right)+\frac{4}{3}-\frac{3}{4}\)
\(-3\frac{3}{2}+5\frac{4}{3}-\left(\frac{7}{6}-1\frac{3}{4}\right)+\left[\frac{2}{3}-2\frac{1}{4}\right]\)
\(2\frac{2}{3}-\frac{5}{12}-\left(1\frac{3}{4}-2\frac{1}{4}\right)-\left[1-1\frac{1}{6}\right]+\left[\frac{-5}{3}\right]\)
\(1\frac{1}{3}-5\frac{1}{2}-\left[\frac{5}{6}-2\frac{2}{3}\right]+\left[\frac{7}{12}-\frac{5}{6}\right]\)
\(\frac{8}{15}-\left(\frac{2}{5}-3\frac{1}{3}+\left[\frac{-5}{6}\right]\right)+\left[\frac{1}{2}-\frac{4}{5}\right]-\left(\frac{1}{6}-1\frac{1}{3}\right)\)
\(-2\frac{3}{2}+\left[\frac{5}{6}-1\frac{1}{3}\right]-\left(\frac{5}{12}-\frac{7}{6}\right)+\left[\frac{4}{3}-3\frac{1}{4}\right]\)
\(\frac{9}{10}-1\frac{2}{5}-\left(\frac{5}{6}-3\frac{1}{2}\right)-\left[2\frac{1}{4}-5\frac{2}{36}\right]-\left[1-2\frac{1}{15}\right]\)
\(\frac{5}{7}-\frac{5}{21}+1\frac{2}{3}-\left(1\frac{1}{2}-\frac{5}{14}-\frac{1}{3}\right)+\left[\frac{1}{6}-\frac{4}{3}\right]\)
\(\frac{5}{7}-\frac{5}{21}+1\frac{2}{3}-\left(1\frac{1}{2}-\frac{5}{14}-\frac{1}{3}\right)+\left[\frac{1}{6}-\frac{4}{3}\right]\)
\(1\frac{1}{5}-\left(\frac{-9}{10}-2\frac{1}{2}+\frac{3}{4}\right)+\left[\frac{1}{5}-2\frac{1}{2}\right]+\frac{7}{10}-\left(\frac{1}{2}-\frac{1}{4}\right)\)
\(2\frac{1}{3}-\left(5\frac{1}{2}-2\frac{2}{3}\right)+\left[1\frac{1}{6}-2\frac{1}{2}\right]-\frac{5}{12}+\left(\frac{1}{4}-\frac{1}{8}\right)\)
e,\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{42}\right)\)
\(\Rightarrow A=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}=4-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(\Rightarrow A=4-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=4-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(\Rightarrow A=4-\left(\frac{1}{1}-\frac{1}{7}\right)=4-\frac{6}{7}=3\frac{1}{7}\)
BN mún hỏi j vậy, đây k phải câu hỏi, mà có thì phải là toán lớp 6
Tính nhanh:
a) \(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+\frac{3^2}{13.16}+...+\frac{3^2}{97.100}\)
b)\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{940}\)
c) A= \(\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}+\frac{6}{208}\)
d) M= \((1-\frac{1000}{2016}).(1-\frac{1001}{2016}).(1-\frac{1002}{2016})...(1-\frac{2017}{2016})\)
e) A= \(8400.(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25})\)
f) T= \((\frac{1}{2}+1).(\frac{1}{3}+1).(\frac{1}{4}+1)...(\frac{1}{98}+1).(\frac{1}{99}+1)\)
h) A=\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)phần \(\frac{1}{5}+\frac{5}{3}+\frac{5}{6}+\frac{1}{2}+...+\frac{1}{9}\)
c) \(A=\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}+\frac{6}{208}\)
\(=\frac{6}{1.4}+\frac{6}{4.7}+\frac{6}{7.10}+\frac{6}{10.13}+\frac{6}{13.16}\)
\(=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)
\(=2\left(1-\frac{1}{16}\right)\)
\(=2.\frac{15}{16}\)
\(=\frac{15}{8}\)
Vậy A=\(\frac{15}{8}\)
a) \(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)
\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}=\frac{297}{100}\)
\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{10}{20}-\frac{1}{20}\)
\(=\frac{9}{20}\)
Tính.
a) $\frac{1}{6} + \frac{3}{2} + \frac{1}{2}$
b) $\frac{3}{8} + \frac{1}{2} + \frac{1}{8}$
c) $\frac{2}{5} + \frac{6}{{10}} + \frac{3}{5}$
a) $\frac{1}{6} + \frac{3}{2} + \frac{1}{2} = \frac{1}{6} + \left( {\frac{3}{2} + \frac{1}{2}} \right) = \frac{1}{6} + \frac{4}{2} = \frac{1}{6} + \frac{{12}}{6} = \frac{{13}}{6}$
b) $\frac{3}{8} + \frac{1}{2} + \frac{1}{8} = \left( {\frac{3}{8} + \frac{1}{8}} \right) + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1$
c) $\frac{2}{5} + \frac{6}{{10}} + \frac{3}{5} = \frac{2}{5} + \frac{3}{5} + \frac{3}{5} = \frac{{2 + 3 + 3}}{5} = \frac{8}{5}$
Tính:
a)\(1\frac{1}{2} + \frac{1}{5}.\left[ {\left( { - 2\frac{5}{6} + \frac{1}{3}} \right)} \right];\)
b)\(\frac{1}{3}.\left( {\frac{2}{5} - \frac{1}{2}} \right):{\left( {\frac{1}{6} - \frac{1}{5}} \right)^2}.\)
a)
\(\begin{array}{l}1\frac{1}{2} + \frac{1}{5}.\left[ {\left( { - 2\frac{5}{6} + \frac{1}{3}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\left[ {\left( { - \frac{{17}}{6} + \frac{2}{6}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\frac{{ - 15}}{6}\\ = \frac{3}{2} + \frac{{ - 1}}{2}\\ = \frac{2}{2}\\=1\end{array}\)
b)
\(\begin{array}{l}\frac{1}{3}.\left( {\frac{2}{5} - \frac{1}{2}} \right):{\left( {\frac{1}{6} - \frac{1}{5}} \right)^2}\\ = \frac{1}{3}.\left( {\frac{4}{{10}} - \frac{5}{{10}}} \right):{\left( {\frac{5}{{30}} - \frac{6}{{30}}} \right)^2}\\ = \frac{1}{3}.\frac{{ - 1}}{{10}}:{\left( {\frac{{ - 1}}{{30}}} \right)^2}\\ = \frac{{ - 1}}{{30}}:\frac{1}{{{{30}^2}}}\\ = \frac{{ - 1}}{{30}}{.30^2}\\ = - 30\end{array}\)
\(\frac{6-\frac{1}{\frac{1}{2}-\frac{1}{3}}}{6+\frac{1}{\frac{1}{2}-\frac{1}{3}}}\)
\(\frac{6-\frac{1}{\frac{1}{2}-\frac{1}{3}}}{6+\frac{1}{\frac{1}{2}-\frac{1}{3}}}=0\) NHA Nguyễn Giang !
A = \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}-1\right)\times\left(1-\frac{8}{1}-\frac{4}{1}-\frac{2}{1}\right)\)
B = \(\frac{\frac{3}{1}-\frac{6}{3}-\frac{9}{6}-\frac{369}{1}}{\frac{1}{3}+\frac{3}{6}+\frac{6}{9}-\frac{1}{963}}\)
C = \(\frac{1}{1}-\frac{1}{2}+\frac{3}{1}-\frac{1}{4}+\frac{5}{1}-\frac{1}{6}+\frac{7}{1}-\frac{1}{8}+\frac{9}{1}-\frac{1}{10}\)
so sánh các số trên ( A , B , C )
a= 1/2 + 1/4 + 1/8 - 1 x 1 + 8/1 - 4/1 - 2/1=\(1\frac{7}{8}\)=1,875
b=3/1 - 6/3 - 9/6 - 369/1 : 1/3 + 3/6 + 6/9 - 1/963 \(\approx\)186,665628245067
c=1/1 - 1/2 + 3/1 - 1/4 + 5/1 - 1/6 + 7/1 - 1/8 + 9/1 - 1/10=\(\approx\)23,8583333333333
vậy a>b>c
**************************l i k e***********************************8
A = \(\left(-\frac{1}{8}\right)\times\left(-13\right)=\frac{13}{8}\) => 0 < A < 2
B: Tử âm ; mẫu dương => B < 0
C = \(\left(\frac{1}{1}+\frac{3}{1}+\frac{5}{1}+\frac{7}{1}+\frac{9}{1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)
= 25 \(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)
Dễ có: B < A < C
a) \(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}+\frac{12}{\sqrt{6}-3}-\sqrt{6}\)b)\(\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\frac{\sqrt{3}}{2-\sqrt{6}}+\frac{\sqrt{3}}{2+\sqrt{6}}\right)-\frac{1}{\sqrt{2}}\)c) \(\left(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right)\frac{1}{\sqrt{3}+5}\)d) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)