uses crt;

var s:real;

a,n:integer;

begin

clrscr;

write('Nhap a='); readln(a);

n:=0;

s:=0;

while s<=a do 

  begin

n:=n+1;

s:=s+1/n;

end;

writeln(n);

readln;

end.

Đặt \(\left(\dfrac{x}{6};\dfrac{y}{3};\dfrac{z}{2}\right)=\left(a;b;c\right)\Rightarrow2^{6a}+4^{3b}+8^{2c}=4\)

\(\Leftrightarrow64^a+64^b+64^c=4\)

Áp dụng BĐT Cô-si:

\(4=64^a+64^b+64^c\ge3\sqrt[3]{64^{a+b+c}}\Rightarrow64^{a+b+c}\le\dfrac{64}{27}\)

\(\Rightarrow a+b+c\le log_{64}\left(\dfrac{64}{27}\right)\Rightarrow M=log_{64}\left(\dfrac{64}{27}\right)\)

Lại có: \(x;y;z\ge0\Rightarrow a;b;c\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}64^a\ge1\\64^b\ge1\\64^c\ge1\end{matrix}\right.\) \(\Rightarrow\left(64^b-1\right)\left(64^c-1\right)\ge0\)

\(\Rightarrow64^{b+c}+1\ge64^b+64^c\) (1)

Lại có: \(b+c\ge0\Rightarrow64^{b+c}\ge1\Rightarrow\left(64^a-1\right)\left(64^{b+c}-1\right)\ge0\)

\(\Rightarrow64^{a+b+c}+1\ge64^a+64^{b+c}\) (2)

Cộng vế (1);(2) \(\Rightarrow4=64^a+64^b+64^c\le64^{a+b+c}+2\)

\(\Rightarrow64^{a+b+c}\ge2\Rightarrow a+b+c\ge log_{64}2\)

\(\Rightarrow N=log_{64}2\)

\(\Rightarrow T=2log_{64}\left(\dfrac{64}{27}\right)+6log_{64}\left(2\right)\approx1,4\)

\(u_{n+1}^2=\dfrac{u_n^2}{1+u_n^2}\Rightarrow\dfrac{1}{u_{n+1}^2}=\dfrac{1}{u_n^2}+1\)

Đặt \(\dfrac{1}{u_n^2}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=\dfrac{1}{2018^2}\\v_{n+1}=v_n+1\end{matrix}\right.\)

\(v_n\) là cấp số cộng với công sai d=1 \(\Rightarrow v_n=\dfrac{1}{2018^2}+n-1\)

\(\Rightarrow u_n^2=\dfrac{1}{v_n}=\dfrac{1}{n+\dfrac{1}{2018^2}-1}\)

\(u_n^2< \dfrac{1}{2018^2}\Rightarrow\dfrac{1}{n+\dfrac{1}{2018^2}-1}< \dfrac{1}{2018^2}\Rightarrow n...\)

Lời giải:
$\frac{1}{m}+\frac{n}{6}=12$

$\Rightarrow 6+mn=72m$

$\Leftrightarrow 6=m(72-n)$

Vì $m,72-n$ là số nguyên với mọi $m,n$ nguyên nên xét các TH:

$m=1; 72-n=6\Rightarrow (m,n)=(1,66)$

$m=6, 72-n=1\Rightarrow (m,n)=(6,71)$

$m=-1, 72-n=-6\Rightarrow (m,n)=(-1,78)$

$m=-6, 72-n=-1\Rightarrow (m,n)=(-6,73)$

$m=-2, 72-n=-3\Rightarrow (m,n)=(-2,75)$

$m=-3, 72-n=-2\Rightarrow (m,n)=(-3,74)$

$m=2, 72-n=3\Rightarrow (m,n)=(2,69)$

$m=3, 72-n=2\Rightarrow (m,n)=(3,70)$

ta có : 1/2<6/x<3/4

       Hay 6/12<6/x<6/8

=> xϵ{11;10;9}

1.

Gọi \(d=ƯC\left(2n^2+3n+1;3n+1\right)\)

\(\Rightarrow2n^2+3n+1-\left(3n+1\right)⋮d\)

\(\Rightarrow2n^2⋮d\Rightarrow2n\left(3n+1\right)-3.2n^2⋮d\)

\(\Rightarrow2n⋮d\Rightarrow2\left(3n+1\right)-3.2n⋮d\Rightarrow2⋮d\Rightarrow\left[{}\begin{matrix}d=1\\d=2\end{matrix}\right.\)

\(d=2\Rightarrow3n+1=2k\Rightarrow n=2m+1\)

\(\Rightarrow n\) lẻ thì A không tối giản

\(\Rightarrow n\) chẵn thì A tối giản

2.

Giả thiết tương đương:

\(xy^2+\dfrac{x^2}{z}+\dfrac{y}{z^2}=3\)

Đặt \(\left(x;y;\dfrac{1}{z}\right)=\left(a;b;c\right)\Rightarrow a^2c+b^2a+c^2b=3\)

Ta có: \(9=\left(a^2c+b^2a+c^2b\right)^2\le\left(a^4+b^4+c^4\right)\left(c^2+a^2+b^2\right)\)

\(\Rightarrow9\le\left(a^4+b^4+c^4\right)\sqrt{3\left(a^4+b^4+c^4\right)}\)

\(\Rightarrow3\left(a^4+b^4+c^4\right)^3\ge81\Rightarrow a^4+b^4+c^4\ge3\)

\(\Rightarrow M=\dfrac{1}{a^4+b^4+c^4}\le\dfrac{1}{3}\)

\(M_{max}=\dfrac{1}{3}\) khi \(\left(a;b;c\right)=\left(1;1;1\right)\) hay \(\left(x;y;z\right)=\left(1;1;1\right)\)

\(N=\dfrac{\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3}{\left(ab\right)\left(bc\right)\left(ca\right)}\)

Đặt \(\left(ab;bc;ca\right)=\left(x;y;z\right)\Rightarrow x+y+z=0\Rightarrow N=\dfrac{x^3+y^3+z^3}{xyz}\)

\(N=\dfrac{x^3+y^3+z^3-3xyz+3xyz}{xyz}=\dfrac{\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]+3xyz}{xyz}=\dfrac{3xyz}{xyz}=3\)

a) \(\dfrac{32}{\left(-2\right)^n}=4\)

\(\Rightarrow\left(-2\right)^n=8=\left(-2\right)^3\)

=> n = 3

b) \(\dfrac{8}{2^n}=2\)

\(\Rightarrow2^n=4=2^2\)

=> n = 2

c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)

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

=> 2n - 1 = 3

=> 2n = 4

=> n = 2

Giải:

a) \(\dfrac{32}{\left(-2\right)^n}=4\) 

\(\Rightarrow\left(-2\right)^n=32:4=8\) 

\(\Rightarrow\left(-2\right)^n=8\) 

Vì \(\left(-2\right)^n=2^3\) là ko thể nên n ∈ ∅

b) \(\dfrac{8}{2^n}=2\)

\(\Rightarrow2^n=8:2=4\) 

\(\Rightarrow2^n=4\) 

\(\Rightarrow2^n=2^2\) 

\(\Rightarrow n=2\) 

c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\) 

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

\(\Rightarrow2n-1=3\rightarrow n=2\)