cho mọi số nguyên dương n>2 cmr \(\dfrac{1}{3}\)\(\dfrac{ }{ }\). \(\dfrac{4}{6}.\dfrac{7}{9}.\dfrac{10}{12}........\dfrac{3n-2}{3n}.\dfrac{3n+1}{3n+3}< \dfrac{1}{3\sqrt{n+1}}\)
CMR":
\(\dfrac{1}{3}\cdot\dfrac{4}{6}\cdot\dfrac{7}{9}\cdot.......\cdot\dfrac{\left(3n-2\right)}{3n}\cdot\dfrac{\left(3n+1\right)}{3n+3}< \dfrac{1}{3\sqrt{n+1}}\)
Tìm n ϵ Z sao cho n là số nguyên
\(\dfrac{2n-1}{n-1};\dfrac{3n+5}{n+1};\dfrac{4n-2}{n+3};\dfrac{6n-4}{3n+4};\dfrac{n+3}{2n-1};\dfrac{6n-4}{3n-2};\dfrac{2n+3}{3n-1};\dfrac{4n+3}{3n+2}\)
C/m:
\(\dfrac{1}{4}.\dfrac{4}{7}.\dfrac{7}{9}.....\dfrac{3n-2}{3n}.\dfrac{3n+1}{3n+3}< \dfrac{1}{\sqrt{3n+1}}\)
Tính :6/ lim\(\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}\)
7/ lim \(\dfrac{\sqrt{n^3-2n+5}}{3+5n}\)
10/ lim\(\dfrac{1+3+5+...+\left(2n+1\right)}{3n^3+4}\)
CMR các phân số sau là phân số tối giản
a) \(A=\dfrac{n+1}{n+2}\)
b) \(B=\dfrac{n+1}{3n+4}\)
c) \(C=\dfrac{3n+2}{5n+3}\)
d) \(D=\dfrac{12n+1}{30n+2}\)
a) Gọi d là ƯCLN(n + 1; n + 2)
\(\Rightarrow n+1⋮d\)
\(n+2⋮d\)
\(\Rightarrow\left[\left(n+2\right)-\left(n+1\right)\right]⋮d\)
\(\Rightarrow\left(n+2-n-1\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
Vậy \(\dfrac{n+1}{n+2}\) là phân số tối giản
b) Gọi d là ƯCLN(n + 1; 3n + 4)
\(\Rightarrow n+1⋮d\) và \(3n+4⋮d\)
Do \(n+1⋮d\Rightarrow3n+3⋮d\)
\(\Rightarrow\left[\left(3n+4\right)-\left(3n+3\right)\right]⋮d\)
\(\Rightarrow\left(3n+4-3n-3\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
Vậy \(\dfrac{n+1}{3n+4}\) là phân số tối giản
c) Gọi d là ƯCLN(3n + 2; 5n + 3)
\(\Rightarrow3n+2⋮d\) và \(5n+3⋮d\)
Do \(3n+2⋮d\)
\(\Rightarrow5\left(3n+2\right)⋮d\)
\(\Rightarrow15n+10⋮d\) (1)
Do \(5n+3⋮d\)
\(\Rightarrow3\left(5n+3\right)⋮d\)
\(\Rightarrow15n+9⋮d\) (2)
Từ (1) và (2) \(\Rightarrow\left[\left(15n+10\right)-\left(15n+9\right)\right]⋮d\)
\(\Rightarrow\left(15n+10-15n-9\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
Vậy \(\dfrac{3n+2}{5n+3}\) là phân số tối giản
d) Gọi d là ƯCLN(12n + 1; 30n + 2)
\(\Rightarrow12n+1⋮d\) và \(30n+2⋮d\)
Do \(12n+1⋮d\)
\(\Rightarrow5\left(12n+1\right)⋮d\)
\(\Rightarrow60n+5⋮d\) (3)
Do \(30n+2⋮d\)
\(\Rightarrow2\left(30n+2\right)⋮d\)
\(\Rightarrow60n+4⋮2\) (4)
Từ (3 và (4) \(\Rightarrow\left[\left(60n+5\right)-\left(60n+4\right)\right]⋮d\)
\(\Rightarrow\left(60n+5-60n-4\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
Vậy \(\dfrac{12n+1}{30n+2}\) là phân số tối giản
a: Gọi d=ƯCLN(n+1;n+2)
=>n+2-n-1 chia hết cho d
=>1 chia hết cho d
=>d=1
=>PSTG
b: Gọi d=ƯCLN(3n+4;n+1)
=>3n+4-3n-3 chia hết cho d
=>1 chia hết cho d
=>d=1
=>PSTG
c: Gọi d=ƯCLN(3n+2;5n+3)
=>15n+10-15n-9 chia hết cho d
=>1 chia hết cho d
=>d=1
=>PSTG
d: Gọi d=ƯCLN(12n+1;30n+2)
=>60n+5-60n-4 chia hết cho d
=>1 chia hết cho d
=>d=1
=>PSTG
cmr:
\(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}....\dfrac{2n-1}{2n}\le\dfrac{1}{\sqrt{3n+1}}\left(\forall n\ge1\right)\)
Tìm các giới hạn sau:
a) \(lim\dfrac{5n}{n-\sqrt{n^2-n-1}}\)
b) \(lim\dfrac{\sqrt{n+\sqrt{n+1}}}{n-\sqrt{n}}\)
c) \(lim\dfrac{\sqrt{2n^4-n^2+7}}{3n+5}\)
d) \(lim\dfrac{\sqrt{3n^2+2n}-n}{3n-2}\)
\(a=\lim\dfrac{5n\left(n+\sqrt{n^2-n-1}\right)}{n+1}=\lim\dfrac{5\left(n+\sqrt{n^2-n-1}\right)}{1+\dfrac{1}{n}}=\dfrac{+\infty}{1}=+\infty\)
\(b=\lim\dfrac{\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^3}+\dfrac{1}{n^4}}}}{1-\dfrac{1}{\sqrt{n}}}=\dfrac{0}{1}=0\)
\(c=\lim\dfrac{\sqrt{2n^2-1+\dfrac{7}{n^2}}}{3+\dfrac{5}{n}}=\dfrac{+\infty}{3}=+\infty\)
\(d=\lim\dfrac{\sqrt{3+\dfrac{2}{n}}-1}{3-\dfrac{2}{n}}=\dfrac{\sqrt{3}-1}{3}\)
Tính các giới hạn sau:
a) \(\lim\limits\dfrac{\sqrt[3]{n^6-7n^3-5n+8}}{n+12}\)
b) \(\lim\limits\dfrac{1}{\sqrt{3n+2}-\sqrt{2n+1}}\)
c) \(\lim\limits\dfrac{4.3^n+7^{n+1}}{2.5^n+7^n}\)
a.
\(A=\lim\frac{\sqrt[3]{n^6-7n^3-5n+8}}{n+12}=\lim \frac{\sqrt[3]{\frac{n^6-7n^3-5n+8}{n^3}}}{\frac{n+12}{n}}=\lim \frac{\sqrt[3]{n^3-7-\frac{5}{n^2}+\frac{8}{n^3}}}{1+\frac{12}{n}}\)
Ta thấy:
\(\lim\sqrt[3]{n^3-7-\frac{5}{n^2}+\frac{8}{n^3}}=\infty \)
\(\lim (1+\frac{12}{n})=1\)
Suy ra $A=\infty$
b.
\(B=\lim\frac{1}{\sqrt{3n+2}-\sqrt{2n+1}}=\lim \frac{1}{\frac{3n+2-(2n+1)}{\sqrt{3n+2}+\sqrt{2n+1}}}=\lim \frac{\sqrt{3n+2}+\sqrt{2n+1}}{n+1}\)
\(=\lim \frac{\sqrt{\frac{3n+2}{n}}+\sqrt{\frac{2n+1}{n}}}{\frac{n+1}{\sqrt{n}}}=\lim \frac{\sqrt{3+\frac{2}{n}}+\sqrt{2+\frac{1}{n}}}{\sqrt{n}+\frac{1}{\sqrt{n}}}\)
Ta thấy:
\(\lim( \sqrt{3+\frac{2}{n}}+\sqrt{2+\frac{1}{n}})=\sqrt{3}+\sqrt{2}>0\)
\(\lim (\sqrt{n}+\frac{1}{\sqrt{n}})=\infty\)
$\Rightarrow B=\infty$
c.
\(C=\lim \frac{4.3^n+7^{n+1}}{2.5^n+7^n}=\lim \frac{4(\frac{3}{7})^n+7}{2(\frac{5}{7})^n+1}\)
Ta thấy:
\(\lim [4(\frac{3}{7})^n+7]=4.0+7=7\) với $|\frac{3}{7}|<1$
\(\lim [2(\frac{5}{7})^n+1]=2.0+1=1\) với $|\frac{5}{7}|<1$
$\Rightarrow C=\frac{7}{1}=7$
CMR \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{2n-1}{2n}\le\dfrac{1}{\sqrt{3n+1}}\)
\(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}......\dfrac{2n-1}{2n}=\dfrac{1.2.3.....\left(2n-1\right)}{2.3.4.....2n}=\dfrac{1}{2n}\)
Khi đó ta có điều cần chứng minh:
\(\dfrac{1}{2n}\le\dfrac{1}{\sqrt{3n+1}}\left(n>\dfrac{1}{3}\right)\)
Hay
\(\dfrac{\sqrt{3n+1}}{2n\left(\sqrt{3n+1}\right)}\le\dfrac{2n}{2n\left(\sqrt{3n+1}\right)}\)
Hay \(\sqrt{3n+1}\le2n\)(luôn đúng)