Tính
a) A= \(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
b) B = \(\frac{\left(7^4-7^3\right)^2}{49^3}\)
tính
a, A=\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
b, B=\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)...\left(\frac{1}{100}-1\right)\left(\frac{1}{121}-1\right)\)
c,C=\(\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+\frac{9}{16.25}+\frac{11}{25.36}\)
a)
\(A=\left(\frac{1}{9}-\frac{1}{10}\right)-\left(\frac{1}{8}-\frac{1}{9}\right)-....-\left(1-\frac{1}{2}\right)=\frac{1}{9}-\frac{1}{10}-\frac{1}{8}+\frac{1}{9}-....-1+\frac{1}{2}\)
\(A=-\left(\frac{1}{10}+1\right)=-\frac{11}{10}\)
a)\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\\ \Rightarrow A=-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}\\ \Rightarrow A=-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)Đặt \(B=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}+\frac{1}{90}\)
\(\Rightarrow B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow B=1-\frac{1}{10}=\frac{9}{10}\)
Ta có : \(A=-B\)
\(\Rightarrow A=-\frac{9}{10}\)
a) A=\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
A=\(-\left(\frac{1}{90}+\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
A=\(-\left(\frac{1}{9.10}+\frac{1}{8.9}+\frac{1}{7.8}+\frac{1}{6.7}+\frac{1}{5.6}+\frac{1}{4.5}+\frac{1}{3.4}+\frac{1}{2.3}+\frac{1}{1.2}\right)\)
A=-\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+........+\frac{1}{9}-\frac{1}{10}\right)\)
A=-(\(1-\frac{1}{10}\))
A=-\(\frac{9}{10}\)
Tính nhanh:
A=\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
B=\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}\)
A = \(\frac{-79}{90}\)
B = \(\frac{8}{9}\)
Tính
A)\(\left[\left(\frac{1}{9}:\frac{8}{27}\right):\frac{-1}{3}\right]:\frac{81}{128}\)
b)\(\left(\frac{-7}{15}\right).\frac{5}{8}.\frac{15}{-7}.\left(-32\right)\)
c)\(0,5+\frac{1}{3}+0,4+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}\)
d)\(\frac{8}{9}-\frac{1}{12}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
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
\(1\frac{1}{2}+2\frac{1}{6}+3\frac{1}{12}+4\frac{1}{20}+5\frac{1}{30}+6\frac{1}{42}+7\frac{1}{56}+8\frac{1}{72}+9\frac{1}{90}+\frac{1}{10}\)\(\frac{1}{10}=?\)
Đầu tiên , ta cộng các phần nguyên lại với nhau trước :
( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 ) + ( \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{8}{72}+\frac{1}{90}+\frac{1}{10}\)
= 45 + \(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{42}+\frac{1}{72}\right)+\left(\frac{1}{10}+\frac{1}{20}\right)+\left(\frac{1}{30}+\frac{1}{90}\right)+\frac{1}{56}\)
= 45 +
tới đây tớ chịu , các cậu giúp với
Đầu tiên , cộng các phần nguyên lại với nhau , ta có :
( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 ) + ( \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\))
= 45 + \(\left(\frac{1}{6}+\frac{1}{30}\right)+\frac{1}{2}+\frac{1}{12}+\frac{1}{20}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
sau khi cộng trong ngoặc , ta được 6 / 30 , rút gọn tối giản còn 1 / 5
= 45 + \(\left(\frac{1}{5}+\frac{1}{20}\right)+\frac{1}{2}+\frac{1}{12}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
sau khi cộng trong ngoặc và rút gọn tối giản , ta được 1 / 4
= 45 + \(\left(\frac{1}{4}+\frac{1}{2}\right)+\frac{1}{12}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
sau khi cộng trong ngoặc rồi rút gọn , ta được 3 / 4
= 45 + \(\left(\frac{3}{4}+\frac{1}{12}\right)+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
rút gọn lại ta được 5 / 6
= 45 + \(\left(\frac{5}{6}+\frac{1}{42}\right)+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
rút gọn tối giản ra 6 / 7
= 45 + \(\left(\frac{6}{7}+\frac{1}{56}\right)+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
sau khi tính trong ngoặc rút gọn được 7 / 8
= 45 + \(\left(\frac{7}{8}+\frac{1}{72}\right)+\frac{1}{90}+\frac{1}{10}\)
tính trong ngoặc rồi rút gọn ra 8 / 9
= 45 + \(\left(\frac{8}{9}+\frac{1}{90}\right)+\frac{1}{10}\)
cũng rút gọn tiếp ta được 9 / 10
= 45 + \(\left(\frac{9}{10}+\frac{1}{10}\right)\)
= 45 + 1
= 46
cái này thì bạn lên làm giảng viên đến nơi rồi
Câu 1. Tính hợp lý giá trị các biểu thức sau :
a. A = ( 689 - 31 ) - ( 269 - 131 )
b. B = \(\left(\frac{1}{2}+\frac{2016}{2017}+\frac{2017}{2018}+1\right)\times\left(\frac{2016}{2017}+\frac{2017}{2018}+\frac{3}{4}\right)-\left(\frac{1}{2}+\frac{2016}{2017}+\frac{2017}{2018}\right)\times\left(\frac{2016}{2017}+\frac{2017}{2018}+\frac{3}{4}+1\right)\)c. C = \(1-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!
\(0,25:\left(10,3-9,8\right)-\frac{3}{4}\\ b,\left(3\frac{4}{5}-2.x\right).1\frac{1}{3}\\ c,\frac{x}{7}=\frac{6}{-21}\\ d,\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{ }{132}\)cái cuối là 1/132
c) x=-2 nha
d) =\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+......+\(\frac{1}{11.12}\)
=\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+.....+\(\frac{1}{11}\)-\(\frac{1}{12}\)
=\(\frac{1}{5}\)-\(\frac{1}{12}\)= \(\frac{7}{60}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}\)
\(=\frac{7}{60}\)
Tính hợp lý.
\(a0.5+\frac{1}{3}+0.4+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}\)
\(b\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
Giai dùm mình hai bài này với.
a) \(0.5+\frac{1}{3}+0.4\frac{5}{7}+\frac{1}{6}-\frac{4}{35}\)
b) \(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)