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 \(\left(x;y;z\right)=\left(1;1;1\right)\)

a, \(A=\dfrac{5n-4-4n+5}{n-3}=\dfrac{n+1}{n-3}=\dfrac{n-3+4}{n-3}=1+\dfrac{4}{n-3}\Rightarrow n-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

a.\(A=\dfrac{2n+1}{n-3}+\dfrac{3n-5}{n-3}-\dfrac{4n-5}{n-3}\)

\(A=\dfrac{2n+1+3n-5-4n+5}{n-3}\)

\(A=\dfrac{n+1}{n-3}\)

\(A=\dfrac{n-3}{n-3}+\dfrac{4}{n-3}\)

\(A=1+\dfrac{4}{n-3}\)

Để A nguyên thì \(\dfrac{4}{n-3}\in Z\) hay \(n-3\in U\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

n-3=1 --> n=4

n-3=-1 --> n=2

n-3=2 --> n=5

n-3=-2 --> n=1

n-3=4 --> n=7

n-3=-4 --> n=-1

Vậy \(n=\left\{4;2;5;7;1;-1\right\}\) thì A nhận giá trị nguyên

b.hemm bt lèm:vv

cc

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) \(\Rightarrow4=64^a+64^b+64^c\le64^{a+b+c}+2\)

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

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

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

1. \(\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).0\)

\(=0\)

Bài 1: 

Ta có: \(A=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)

=0

a: \(A=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

Ai lm đc câu nào thì giúp mk với , cảm ơn !!

\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\\ A_{min}=\dfrac{1}{9}\Leftrightarrow x=\dfrac{3}{5}\\ B=\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\le\dfrac{2009}{2008}\\ B_{max}=\dfrac{2009}{2008}\Leftrightarrow x=\dfrac{3}{5}\\ C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\le1\dfrac{2}{3}\\ C_{max}=1\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=-4\Leftrightarrow x=-12\)

a: \(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{5}\)

a, A = \(\frac{3n-1}{n-2}=\frac{3n-6+5}{n-2}=\frac{3\left(n-2\right)+5}{n-2}=3+\frac{5}{n-2}\)

Để A thuộc Z <=> n - 2 thuộc Ư(5) = {1;-1;5;-5}

Ta có: n - 2 = 1 => n = 3

          n - 2 = -1 => n = 1

          n - 2 = 5 => n = 7

          n - 2 = -5 => n = -3

Vậy n = {3;1;7;-3}

b, A = \(\frac{3n-1}{n-2}=\frac{3n-6+5}{n-2}=\frac{3\left(n-2\right)+5}{n-2}=3+\frac{5}{n-2}\)

Để A đạt giá trị nhỏ nhất <=> \(\frac{5}{n-2}\) đạt giá trị nhỏ nhất

=> n - 2 đạt giá trị lớn nhất  (n - 2 \(\ne\)0 ; n - 2 < 0)

=> n - 2 = -1 => n = 1

Vậy để A có giá trị nhỏ nhất thì n = 1

c, \(\frac{3n-1}{n-2}=\frac{3n-6+5}{n-2}=\frac{3\left(n-2\right)+5}{n-2}=3+\frac{5}{n-2}\)

Để A đạt giá trị lớn nhất <=> \(\frac{5}{n-2}\)đạt giá trị lớn nhất

=> n - 2 đạt giá trị nhỏ nhất (n - 2 \(\ne\)0 ; n - 2 > 0)

=> n - 2 = 1 => n = 3

Vậy để A đạt giá trị lớn nhất thì n = 3