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A x 2 = (1+2+4+8+16+32+....+1204) x 2
A x 2 = 2 + 4 + 16 + 32 + 64+...+2408
A x 2 - A = (2 + 4 + 16 + 32 + 64+...+2408) - (1+2+4+8+16+32+....+1204)
A = 2408 - 1
A = 2407
Chúc bạn hok tốt nha!@ ^ ^
1, \(x=-\dfrac{7}{3}-\dfrac{1}{3}=-\dfrac{8}{3}\)
2, \(x=\dfrac{1}{8}-\dfrac{3}{8}=-\dfrac{2}{8}=-\dfrac{1}{4}\)
3, \(x=\dfrac{6}{5}+\dfrac{4}{5}=\dfrac{10}{5}=2\)
4, \(x=\dfrac{7}{3}-\dfrac{2}{3}=\dfrac{5}{3}\)
5, \(x+\dfrac{7}{3}=\dfrac{5}{3}\Leftrightarrow x=\dfrac{5}{3}-\dfrac{7}{3}=-\dfrac{2}{3}\)
\(=x+\dfrac{1}{3}=\dfrac{-7}{3}\Leftrightarrow x=\dfrac{-8}{3}\)
\(=\dfrac{1}{8}-x=\dfrac{3}{8}\Leftrightarrow x=\dfrac{-1}{4}\)
\(=x-\dfrac{4}{5}=\dfrac{6}{5}\Leftrightarrow x=2\)
\(=\dfrac{2}{3}+x=\dfrac{7}{3}\Leftrightarrow x=\dfrac{5}{3}\)
\(=x-\dfrac{-7}{3}=\dfrac{5}{3}\Leftrightarrow x=\dfrac{-2}{3}\)
\(\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\)\(\frac{65}{64}-7\)
\(=\frac{96}{64}+\frac{80}{64}+\frac{72}{64}+\frac{66}{64}+\frac{65}{64}-7\)
\(=\frac{96+80+72+66+64}{64}-7\)
\(=\frac{378}{64}-\frac{7}{1}\)
\(=\frac{189}{32}-\frac{224}{32}\)
\(=\frac{-35}{32}\)
\(S=1-3+3^2-3^3+...+3^{98}-3^{99}=\left(1-3+3^2-3^3\right)+3^4\left(1-3+3^3-3^3\right)+...+3^{96}\left(1-3+3^2-3^3\right)=\left(-20\right)+3^4.\left(-20\right)+...+3^{96}.\left(-20\right)=\left(-20\right)\left(1+3^4+...+3^{96}\right)⋮20\)
Ta có: \(S=1-3+3^2-3^3+...+3^{98}-3^{99}\)
\(=\left(1-3+3^2-3^3\right)+...+3^{96}\left(1-3+3^2-3^3\right)\)
\(=-20\cdot\left(1+...+3^{96}\right)⋮20\)
3/2+5/4+9/8/+17/16+33/32-6+x-1/x+1=31/32-2/2015
=(1+1/2)+(1+1/4)+(1+1/8)+(1+1/16)+(1+1/32-6+x-1/x+1=31/32-2/2015
=(1/2+1/4+1/8+1/16+1/32)+(1+1+1+1+1)-6+x-1/x+1=31/32-2/2015
=31/32+5-6+x-1/x+1=31/32-2/2015
=5-6+x-1/x+1=31/32-2/2015-31/32
=-1+x-1/x+1=-2/2015
=x-1/x+1=-2/2015- -1
=x-1/x+1=2013/2015
=>x=2014
1 > 33 ( 17 - 5 ) - 17 ( 33 - 5 )
= 33 . 17 - 33.5 - 17.33 + 17.5
=-33.5 + 17.5
= 5 ( - 33 + 17 )
= 5 . ( -16 )
= - 80
2 > | -100 | + ( 135 - 58 ) - ( 142 - 856 )
= 100 + 135 - 58 - 142 + 856
Bạn tự làm nha , đừng công thức chuyển vế thôi
# chúc bạn học tốt ạ #
\(=3^{2\cdot8}:3^{3\cdot5}=3^{16}:3^{15}=3^{16-15}=3^1=3\)