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\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(2\times A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2\times A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\)
\(A=1-\frac{1}{128}\)
\(A=\frac{127}{128}\)
\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(2\times B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)
\(B=1-\frac{1}{16}=\frac{15}{16}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\Leftrightarrow4\times x+\frac{15}{16}=1\)
\(\Leftrightarrow4\times x=\frac{1}{16}\)
\(\Leftrightarrow x=\frac{1}{64}\)
1 a) (x+ 1) + (x + 2 ) + (x + 3) + ... + (x + 100) = 205550 (100 cặp)
=> (x + x + ... + x) + (1 + 2 + 3 + ... + 100) = 205 550
100 số hạng x 100 số hạng
=> 100.x + 100 . 101 : 2 = 205 550
=> 100.x + 5050 = 205 550
=> 100 . x = 205 550 - 5050
=> 100 . x = 200500
=> x = 200500 : 100
=> x = 2005
\(\frac{1999x1999}{1995x1995_{ }}=\frac{1999^2}{1995^2}=\left(\frac{1999}{1995}\right)^2\)\(>1^2\)\(=1\)
Bài 3 :
b) Ta có 1+ 2 + 3 +4 + ...+ x =15
Nên \(\frac{x\left(x+1\right)}{2}=15\)
\(x\left(x+1\right)=30\)
=> \(x\left(x+1\right)=5.6\)
=> x = 5
Bài 2:
h; \(\dfrac{2}{3}\)\(x\) + 50% + \(x\) = \(\dfrac{1}{10}\)
\(\dfrac{2}{3}\)\(x\) + \(\dfrac{1}{2}\) + \(x\) = \(\dfrac{1}{10}\)
(\(\dfrac{2}{3}\)\(x\) + \(x\)) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) (\(\dfrac{2}{3}\) + 1) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{1}{10}\) - \(\dfrac{1}{2}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{-2}{5}\)
\(x\) = \(\dfrac{-2}{5}\): \(\dfrac{5}{3}\)
\(x\) = - \(\dfrac{6}{25}\)
Lớp 5 chưa học số âm em nhé.
\(\frac{127}{128}\times x=1\)
\(\Rightarrow x=\frac{128}{127}\)