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\(\frac{x}{7}=\frac{x+1}{14}\Leftrightarrow14x=7x+7\Leftrightarrow7x=7\Leftrightarrow x=1\)
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\le x\le\frac{15}{4}+\frac{18}{8}\)
\(\Leftrightarrow1\le x\le6\Leftrightarrow x=1;2;3;4;5;6\)
\(\frac{1}{2}+\frac{-3}{5}+\frac{1}{10}\le x\le\frac{8}{3}+\frac{14}{6}\)
\(\Leftrightarrow\frac{1}{2}-\frac{3}{5}+\frac{1}{10}\le x\le\frac{8}{3}+\frac{14}{6}\)
\(\Leftrightarrow0\le x\le5\Leftrightarrow x=0;1;2;3;4;5\)
\(\frac{x}{7}=\frac{x+1}{14}\)
=> \(\frac{x\cdot2}{7\cdot2}=\frac{x+1}{14}\)
=> \(2x=x+1\)
=> \(2x-x-1=0\)
=> \(1x-1=0\)
=> \(x=1\)
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\le x\le\frac{15}{4}+\frac{18}{8}\)
=> \(1\le x\le6\)
=> \(x=\left\{1;2;3;4;5;6\right\}\)
\(\frac{1}{2}+\frac{-3}{5}+\frac{1}{10}\le x\le\frac{8}{3}+\frac{14}{6}\)
=> \(0\le x\le5\)
=> \(x=\left\{0;1;2;3;4;5\right\}\)
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\(a)\) Ta có :
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}< 1\)
Vậy \(A< 1\)
Chúc bạn học tốt ~
1)
A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{99.101}\)
A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{99}-\frac{1}{101}\)
A = \(\frac{1}{1}-\frac{1}{101}\)
A = \(\frac{100}{101}\)
Vậy A = \(\frac{100}{101}\)
B = \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
B = \(\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}.\frac{100}{101}\)
B = \(\frac{250}{101}\)
Vậy B = \(\frac{250}{101}\)
2)
Gọi ƯCLN ( 2n + 1 ; 3n + 2 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}\Rightarrow\hept{\begin{cases}3\left(2n+1\right)⋮d\\2\left(3n+2\right)⋮d\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d\Rightarrow1⋮d}\)
\(\Rightarrow d=1\)
Vậy \(\frac{2n+1}{3n+2}\)là p/s tối giản
Gọi ƯCLN ( 2n+3 ; 4n+4 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+4⋮d\end{cases}\Rightarrow\hept{\begin{cases}2n+3⋮d\\\left(4n+4\right):2⋮d\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\2n+2⋮d\end{cases}\Rightarrow\left(2n+3\right)-\left(2n+2\right)⋮d}\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
Vậy ...
\(C< \frac{2}{3}.\frac{4}{5}......\frac{80}{81}\Rightarrow C.C< \frac{C.2....80}{3.5....81}=\frac{1.2.3....79.80}{2.3.4....81}=\frac{1}{81}=\left(\frac{1}{9}\right)^2mà:C>0\Rightarrow C< \frac{1}{9}\)
Shitbo ơi em có thể giải theo cách cấp 1 được không?