c)\(7^{2n}+7^{2n+2}=2450\)

⇒\(7^{2n}+7^{2n}.7^2=2450\)

⇒\(7^{2n}.50=2450\)

⇒\(7^{2n}=49\)\(=7^2\)

⇒2n=2

⇒n=1

a)\(\left(-\dfrac{1}{5}\right)^n=-\dfrac{1}{125}\)                   b)\(\left(-\dfrac{2}{11}\right)^m=\dfrac{4}{121}\)

\(\left(-\dfrac{1}{5}\right)^n=\left(-\dfrac{1}{5}\right)^3\)                    \(=\left(-\dfrac{2}{11}\right)^m=\left(-\dfrac{2}{11}\right)^2\)

⇒n=3                                          ⇒m=2

\(1-\dfrac{3}{n\left(n+2\right)}=\dfrac{n\left(n+2\right)-3}{n\left(n+2\right)}=\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)

\(\Rightarrow M=\dfrac{1.5}{2.4}.\dfrac{2.6}{3.5}.\dfrac{3.7}{4.6}...\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)

\(=\dfrac{1.2.3...\left(n-1\right)}{2.3.4...n}.\dfrac{5.6.7...\left(n+3\right)}{4.5.6...\left(n+2\right)}\)

\(=\dfrac{1}{n}.\dfrac{n+3}{4}=\dfrac{n+3}{4n}=\dfrac{1}{4}+\dfrac{3}{4n}>\dfrac{1}{4}\) (đpcm)

(-3)^n=(-3)^7

n=7

\(\dfrac{\left(-3\right)^n}{9^2}=\left(-27\right)\)

\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^7\)

hay n=7

len vietjack la co het

$\frac{1.3.5...(2n-1)}{(n+1)(n+2)...(n+n)}=\frac{1}{2^n}(*)$

Với $n=1$ thì $(*)\Leftrightarrow \frac{1}{2}=\frac{1}{2}$

Vậy $(*)$ đúng với $n=1$

Giả sử với $n=k$,$ k\in \mathbb{N^*}$ thì $(*)$ đúng, tức là: 

$\frac{1.3.5...(2k-1)}{(k+1)(k+2)...(k+k)}=\frac{1}{2^k}$

Ta cần chứng minh với $n=k+1$ thì $(*)$ đúng, tức là: 

$\frac{1.3.5...(2k+1)}{(k+2)(k+3)...(2k+2)}=\frac{1}{2^{k+1}}=\frac{1}{2^k}.\frac{1}{2}$

$\Leftrightarrow \frac{1.3.5...(2k+1)}{(k+2)(k+3)...(2k+2)}=\frac{1.3.5...(2k-1)}{2(k+1)(k+2)...(k+k)}$

$\Leftrightarrow \frac{1.3.5...(2k-1)2k(2k+1)}{(k+2)(k+3)...2k(2k+1)(2k+2)}=\frac{1.3.5...(2k-1)}{2(k+1)(k+2)...2k}$

$\Leftrightarrow \frac{2k(2k+1)}{2k(2k+1)(2k+2)}=\frac{1}{2(k+1)}$

$\Leftrightarrow \frac{1}{(2k+2)}=\frac{1}{2(k+1)}$

Do đó với $n=k+1$ thì $(*)$ đúng

$\Rightarrow \frac{1.3.5...(2n-1)}{(n+1)(n+2)...(n+n)}=\frac{1}{2^n}$

Ta có: \(\dfrac{2}{\left(n-1\right)n\left(n+1\right)}=\dfrac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\)

a)

b)

c)

a) $\frac{16}{2^n}=\mathrm{2\; =>\; \backslash (\backslash dfrac\{2^4\}\{2^n\}=2\backslash Rightarrow2^\{4-n\}=2\backslash Rightarrow4-n=1\backslash Rightarrow\; n=3\backslash )}$

b,
\(\dfrac{\left(-3\right)^n}{81}=-27\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^4}=-27\Rightarrow\left(-3\right)^{n-4}=\left(-3\right)^3\Rightarrow n-4=3\Rightarrow n=7\)

c,\(8^n:2^n=4\Rightarrow4^n=4\Rightarrow n=1\)

=> (-3)n-4 = (-3)3

=> n - 4 = 3 => n = 7

c) 8n : 2n = 4

4n = 4.

Với mọi k thuộc N và k > 2 thì ta có :

\(1-\frac{1}{1+2+....+k}=1-\frac{1}{\frac{k\left(k+1\right)}{2}}=1-\frac{2}{k\left(k+1\right)}=\frac{k^2+k-2}{k\left(k+1\right)}=\frac{\left(k+2\right)\left(k-1\right)}{k\left(k+1\right)}\)

Áp dụng vào A ta được :

\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+....+n}\right)\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)

\(=\frac{\left[1.2.3....\left(n-1\right)\right]\left[4.5.6.....\left(n+2\right)\right]}{\left(2.3.4......n\right)\left[3.4.5.....\left(n+1\right)\right]}\)

\(=\frac{n+2}{n.3}=\frac{n+2}{3n}\)

Bài 1:

a, \(\left(x-2\right)^2=9\)

\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)

b, \(\left(3x-1\right)^3=-8\)

\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)

\(\Rightarrow x=-\dfrac{1}{3}\)

c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)

\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)

d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)

\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)

Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)

e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)

Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)

a: f(1)=1

=>\(a\cdot1^2+b\cdot1+1=1\)

=>a+b=0

f(-1)=3

=>\(a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+1=3\)

=>a-b=2

mà a+b=0

nên \(a=\dfrac{2+0}{2}=1;b=2-1=1\)

b: a=1 và b=1 nên \(f\left(x\right)=x^2+x+1\)

\(\Leftrightarrow\dfrac{n}{f\left(n\right)}=\dfrac{n}{n^2+n+1}\)

Gọi d=ƯCLN(n^2+n+1;n)

=>\(\left\{{}\begin{matrix}n^2+n+1⋮d\\n⋮d\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}n^2+n+1⋮d\\n\left(n+1\right)⋮d\end{matrix}\right.\)

=>\(\left(n^2+n+1\right)-n\left(n+1\right)⋮d\)

=>\(1⋮d\)

=>d=1

=>ƯCLN(n^2+n+1;n)=1

=>\(\dfrac{n}{f\left(n\right)}=\dfrac{n}{n^2+n+1}\) là phân số tối giản