Rút Gọn :\frac{4^5\cdot 9^4-2\cdot 6^9}{2^{10}\cdot 3^8+6^8\cdot 20}
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A=\(\frac{1.2.3.4...8.9}{2.3.4.5...9.10}\)
A=\(\frac{1}{10}\)
mình làm đc 1 câu thôi. Bạn thông cảm nhé
\(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{6}+\frac{1}{6}\cdot\frac{1}{7}+\frac{1}{7}\cdot\frac{1}{8}+\frac{1}{8}\cdot\frac{1}{9}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)
\(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+...+\frac{1}{8}\cdot\frac{1}{9}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}\)
* LÀM NỐT *
#Louis
a/ \(\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\left(\frac{-9}{2}\right)+\frac{1}{7}.\left(-\frac{9}{10}\right)\)
= \(-\frac{9}{10}.\left(\frac{5}{14}+\frac{1}{7}\right)+\frac{1}{10}.\left(-\frac{9}{2}\right)\)
= \(-\frac{9}{10}.\frac{1}{2}+\frac{1}{10}.\left(-\frac{9}{2}\right)\)
= \(\frac{-9}{20}+\left(-\frac{9}{20}\right)=\frac{-18}{20}=\frac{-9}{10}\)
b/ \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{11}\right).132\)
\(=\left(\frac{1}{2}.132\right)+\left(\frac{1}{3}.132\right)+\left(\frac{1}{4}.132\right)+\left(\frac{1}{6}.132\right)\)\(+\left(\frac{1}{11}.132\right)\)
\(=66+44+33+22+12=177\)
c/ \(-\frac{2}{3}.\left(\frac{8}{9}.\frac{8}{13}-\frac{8}{27}.\frac{8}{13}+\frac{4}{3}.\frac{22}{39}\right)\)
= \(-\frac{2}{3}.\left[\frac{8}{13}\left(\frac{8}{9}-\frac{8}{27}\right)+\frac{88}{117}\right]\)
= \(-\frac{2}{3}.\left(\frac{8}{13}.\frac{16}{27}+\frac{88}{117}\right)\)
= còn lại làm nốt nha! bận ròy
a.4^7
b.8^5
c.cho x mk sẻ tính kết quả nhưng tìm xmk ko tính đâu
1/4.2/6.3/8.4/10.........30/62.31/64=4x
=1/2.1/2.1/2.1/2.............1/2.1/64=4^x
=1/2^30.1/2^6=4^x
=1/2^36=4^x
=1/4^18=4^x
=>x=-18
b) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=\frac{4^5.\left(1+1+1+1\right)}{3^5.\left(1+1+1\right)}.\frac{6^5.\left(1+1+1+1+1+1\right)}{2^5.\left(1+1\right)}\)
\(=\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=\frac{4^6}{3^6}.\frac{6^6}{2^6}=\frac{2^{12}.2^6.3^6}{3^6.2^6}=2^{12}\)
Ta có: \(2^{12}=\left(2^3\right)^4=8^4\)
Vậy x= 4
\(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}.....\dfrac{30}{62}.\dfrac{31}{64}=2^x\)
\(\Leftrightarrow\dfrac{1}{2.2}.\dfrac{2}{2.3}.\dfrac{3}{2.4}.\dfrac{4}{2.5}.\dfrac{5}{2.6}.....\dfrac{30}{2.31}.\dfrac{31}{2.32}=2^x\)
\(\Leftrightarrow\dfrac{1.2.3.4.5.....30.31}{2.2.2.3.2.4.2.5.2.6.....2.31.2.32}=2^x\)
\(\Leftrightarrow\dfrac{2.3.4.5.....30.31}{2^{31}.32.\left(2.3.4.5.....31\right)}=2^x\)
\(\Leftrightarrow\dfrac{1}{2^{31}.2^5}=2^x\)
\(\Leftrightarrow\dfrac{1}{2^{36}}=2^x\)
\(\Leftrightarrow2^{-36}=2^x\)
\(\Leftrightarrow x=-36\)