(7 + 33 : 32) . 4 - 3
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\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
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, \(x+\dfrac{7}{3}=\dfrac{5}{3}\Leftrightarrow x=\dfrac{5}{3}-\dfrac{7}{3}=-\dfrac{2}{3}\)
\(=x+\dfrac{1}{3}=\dfrac{-7}{3}\Leftrightarrow x=\dfrac{-8}{3}\)
\(=\dfrac{1}{8}-x=\dfrac{3}{8}\Leftrightarrow x=\dfrac{-1}{4}\)
\(=x-\dfrac{4}{5}=\dfrac{6}{5}\Leftrightarrow x=2\)
\(=\dfrac{2}{3}+x=\dfrac{7}{3}\Leftrightarrow x=\dfrac{5}{3}\)
\(=x-\dfrac{-7}{3}=\dfrac{5}{3}\Leftrightarrow x=\dfrac{-2}{3}\)
1.\(\dfrac{8}{11}+\dfrac{8}{33}\times\dfrac{3}{4}=\dfrac{8}{11}+\dfrac{8\times3}{33\times4}=\dfrac{8\times12}{11\times12}+\dfrac{8\times3}{33\times4}=\dfrac{8\times15}{132}=\dfrac{120}{132}=\dfrac{10}{11}\)
2.\(\dfrac{7}{9}\times\dfrac{3}{14}:\dfrac{5}{8}=\dfrac{7}{9}\times\dfrac{3}{14}\times\dfrac{8}{5}=\dfrac{4}{15}\)
3.\(\dfrac{5}{12}-\dfrac{7}{32}:\dfrac{21}{16}=\dfrac{5}{12}-\dfrac{7}{32}\times\dfrac{16}{21}=\dfrac{5}{12}-\dfrac{1}{6}=\dfrac{5-2}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)
`@V.Tr.V`
\(\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\)\(\frac{65}{64}-7\)
\(=\frac{96}{64}+\frac{80}{64}+\frac{72}{64}+\frac{66}{64}+\frac{65}{64}-7\)
\(=\frac{96+80+72+66+64}{64}-7\)
\(=\frac{378}{64}-\frac{7}{1}\)
\(=\frac{189}{32}-\frac{224}{32}\)
\(=\frac{-35}{32}\)
A = 8⁸ + 2²⁰
= (2³)⁸ + 2²⁰
= 2²⁴ + 2²⁰
= 2²⁰.(2⁴ + 1)
= 2²⁰.17 ⋮ 17
Vậy A ⋮ 17
Mọi người tk mình đi mình đang bị âm nè!!!!!!
Ai tk mình mình tk lại nha !!!
\(S=1+\frac{1}{2}+1+\frac{1}{4}+1+\frac{1}{8}+1+\frac{1}{16}+1+\frac{1}{32}+1+\frac{1}{64}-7\)
\(S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}-1\)
Ta đặt: \(P=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
=> \(2P=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
=> \(2P-P=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
=> \(P=1-\frac{1}{64}\)
Mà \(S=P-1\)
=> \(S=1-\frac{1}{64}-1=-\frac{1}{64}\)
Vậy \(S=-\frac{1}{64}\)
a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)
b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)
c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
\(\left(7+3^3\div3^2\right).4-3\text{ }\)
\(=\left(7+3\right).4-3\)
\(=10.4-3\)
\(=40-3\)
\(=37\)