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\(\frac{2}{7}< \frac{x}{3}< \frac{11}{4};x\inℕ\)
=>\(\frac{12.2}{84}< \frac{28x}{84}< \frac{11.21}{84}\)
=>\(\frac{24}{84}< \frac{28x}{84}< \frac{231}{84}\)
=>24<28x<231
=>28x\(\in\){25;26;27;28;.............................;230}
=>Các số chia hết cho 28 là:28;56;84;112;140;168;196;224
=>x (thỏa mãn)\(\in\){1;2;3;4;5;6;7;8}
Vậy x\(\in\) {1;2;3;4;5;6;7;8}
\(\left(4,5m-\frac{3}{4}.5\frac{1}{3}\right).\frac{1}{12}+\frac{1}{2}x=1\frac{1}{2}\)
\(\left(4,5m-\frac{3}{4}.\frac{16}{3}\right).\frac{1}{2}.\frac{1}{6}+\frac{1}{2}x=\frac{3}{2}\)
\(\left(4,5m-\frac{48}{12}\right).\frac{1}{2}.\left(\frac{1}{6}+x\right)=\frac{3}{2}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{3}{2}:\frac{1}{2}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{3}{2}.\frac{2}{1}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{6}{2}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=3\)
=>3\(⋮\)\(\frac{1}{6}+x\)
=>\(\frac{1}{6}+x\)\(\in\)Ư(3)={\(\pm\)1;\(\pm\)3}
Ta có bảng:
| \(\frac{1}{6}+x\) | -1 | 1 | -3 | 3 |
| x | \(-1\frac{1}{6}\) | \(1\frac{1}{6}\) | \(-3\frac{1}{6}\) | 3\(\frac{1}{6}\) |
Vậy x\(\in\){\(-1\frac{1}{6}\);\(1\frac{1}{6}\);\(-3\frac{1}{6}\);\(\frac{1}{6}\)}
Chúc bn học tốt
5A=\(\frac{1}{5}+\frac{2}{5^2}...+\frac{n}{5^n}...+\frac{11}{5^{11}}\)
=>4A=5A-A=\(\frac{1}{5}+\frac{1}{5^2}...+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
=>20A=\(1+\frac{1}{5}+...+\frac{1}{5^{10}}-\frac{11}{5^{11}}\)
=>16A=20A-4A=\(1-\frac{1}{5^{11}}+\frac{11}{5^{12}}-\frac{11}{5^{11}}\)
Mà \(1-\frac{1}{5^{11}}< 1\),\(\frac{11}{5^{12}}-\frac{11}{5^{11}}< 0\)
=>16A<1
Do đó: A<1/16(đpcm)
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3n+2}\right]=\frac{1}{3}\left[\frac{3n+2}{2\left(3n+2\right)}-\frac{2}{2\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\cdot\frac{3n}{6n+4}=\frac{n}{6n+4}=VP\)
b) Ta có: \(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{4n+3}{12n+9}-\frac{3}{12n+9}\right)\)
\(=\frac{5}{4}.\frac{4n}{12n+9}\)
\(=\frac{5n}{12n+9}\)
( sai đề )
Bài làm:
Ta thấy: \(\frac{1}{2^2}< \frac{1}{1.2}\) ; \(\frac{1}{3^2}< \frac{1}{2.3}\) ; ... ; \(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
=> \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=2-\frac{1}{n}< 2\) (vì n là STN)
=> đpcm
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\)
\(=\frac{1}{2}-\frac{1}{3n+2}=\frac{3n+2}{2\cdot\left(3n+2\right)}-\frac{2}{2\cdot\left(3n+2\right)}\)
\(=\frac{3n+2-2}{6n+4}=\frac{3n}{6n+4}=VP\)
a) \(\frac{3n-2}{4n-3}\)
gọi \(\text{Ư}CLN_{\left(3n-2;4n-3\right)}=d\)
\(\Rightarrow\hept{\begin{cases}3n-2⋮d\\4n-3⋮d\end{cases}\Rightarrow\hept{\begin{cases}4\left(3n-2\right)⋮d\\3\left(4n-3\right)⋮d\end{cases}\Rightarrow}\hept{\begin{cases}12n-8⋮d\\12n-9⋮d\end{cases}}}\)
\(\Rightarrow12n-8-12n+9⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
vậy phân số \(\frac{3n-2}{4n-3}\) là phân số tối giản
b) \(\frac{4n+1}{6n+1}\)
gọi \(\text{Ư}CLN_{\left(4n+1;6n+1\right)}=d\)
\(\Rightarrow\hept{\begin{cases}4n+1⋮d\\6n+1⋮d\end{cases}}\Rightarrow\hept{\begin{cases}3\left(4n+1\right)⋮d\\2\left(6n+1\right)⋮d\end{cases}\Rightarrow\hept{\begin{cases}12n+3⋮d\\12n+2⋮d\end{cases}}}\)
\(\Rightarrow12n+3-12n-2⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
vậy phân số \(\frac{4n+1}{6n+1}\) là phân số tối giản
a) Số số hạng là : \(\frac{\left(4n-4\right)}{4}+1=\frac{4\left(n-1\right)}{4}+1=n-1+1=n\)
Tổng của dãy trên là : \(\frac{\left(4+4n\right)\cdot n}{2}=2n\left(n+1\right)\)
Ta có : \(2n\left(n+1\right)=2964\)
=> \(n\left(n+1\right)=2964:2=1482=38\cdot39\)
=> n = 38
b) \(\frac{1}{2}+1+\frac{3}{2}+...+\frac{n}{2}=33\)
=> \(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+...+\frac{n}{2}=33\)
=> \(\frac{1+2+3+...+n}{2}=33\)
=> \(1+2+3+...+n=66\)
Số số hạng là : \(\left(n-1\right):1+1=n\)
Tổng : \(\frac{\left(1+n\right)\cdot n}{2}=\frac{n\left(n+1\right)}{2}\)
=> \(\frac{n\left(n+1\right)}{2}=66\)
=> \(n\left(n+1\right)=66\cdot2=132=11\cdot12\)
=> n = 11
P/S : K bt có đúng k nx