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Ta có :
\(S=\left(\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\frac{31}{32}+\frac{63}{64}+\frac{127}{128}\right)-6\)
\(S=\left(\frac{64}{128}+\frac{102}{128}+\frac{112}{128}+\frac{120}{128}+\frac{124}{128}+\frac{126}{128}+\frac{127}{128}\right)-6\)
\(S=\frac{64+102+112+120+124+126+127}{128}-6\)
\(S=\frac{775}{128}-6\)
\(S=\frac{775}{128}-\frac{768}{128}\)
\(S=\frac{7}{128}\)
S=1/2+3/4+7/8+15/16+31/32+63/64+127/128 -6
S= 1-1/2 + 1-1/4 + 1-1/8 + 1-1/16 + 1-1/32 + 1-1/64+ 1-1/128 - 6
S= (1+1+1+1+1+1+1-6)- (1/2+1/4+1/8+1/16 + 1/32+1/64+1/128)
S= 1- 111/128
S= 17/128
(Làm lụi nha bn)
\(\dfrac{45^{10}\cdot5^{20}}{75^{15}}=\dfrac{\left(3^2\cdot5\right)^{10}\cdot5^{20}}{\left(3\cdot5^2\right)^{15}}=\dfrac{3^{20}\cdot5^{10}\cdot5^{20}}{3^{15}\cdot5^{30}}=3^5=243\\ \dfrac{6^6+6^3+3^3+3^6}{-73}=\dfrac{46656+216+27+729}{-73}=-\dfrac{47628}{73}\\ \dfrac{27^7+3^{15}}{9^9-27}=\dfrac{\left(3^3\right)^7+3^{15}}{\left(3^2\right)^9-3^3}=\dfrac{3^{21}+3^{15}}{3^{18}-3^3}=\dfrac{3^{15}\left(3^6+1\right)}{3^3\left(3^{15}-1\right)}=\dfrac{3^5\cdot730}{3^{15}-1}\\ \dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)
Q=\(\dfrac{1}{2}+\left(\dfrac{3}{4}+\dfrac{7}{8}\right)+\left(\dfrac{15}{16}+\dfrac{31}{32}\right)+\left(\dfrac{63}{64}+\dfrac{127}{128}\right)-6\)
Q=\(\dfrac{1}{2}+\dfrac{13}{8}+\dfrac{61}{32}+\dfrac{253}{128}\)\(-6\)
Q= \(\dfrac{64}{128}+\dfrac{208}{128}+\dfrac{244}{128}+\dfrac{253}{128}-6\)
Q= \(\dfrac{769}{128}-6\)
Q=\(\dfrac{769}{128}-\dfrac{768}{128}\)
Q= \(\dfrac{1}{128}\)
= [15/32 . 64 . 81 . 25]/225 . 27 . 0,0625 . 64
= 60750/225 . 27 . 0,0625 . 64
= 270 . 27 . 0,0625 . 64
= 29160
a) Ta có: \(\left(x-1\right)^{x+2}-\left(x-1\right)^{x+4}=0\)
\(\Leftrightarrow\left(x-1\right)^x\cdot\left(x-1\right)^2-\left(x-1\right)^x\cdot\left(x-1\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\cdot\left[1-\left(x-1\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
b) Ta có: \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{15}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\)
\(\Leftrightarrow2x=\dfrac{1}{64}\)
hay \(x=\dfrac{1}{128}\)
bằng -55/72
Có phải bạn soạn nhầm 16 thành 6 đúng không :)
\(\frac{\frac{15}{8}-\frac{15}{16}-\frac{15}{32}+\frac{15}{64}}{3-\frac{3}{2}-\frac{3}{4}+\frac{3}{8}}\)
\(=\frac{15.\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}+\frac{1}{16}\right)}{6.\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}+\frac{1}{16}\right)}\)
\(=\frac{15}{6}\)
\(=\frac{5}{2}\)