a. \(A=\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{x^3-1}\left(ĐKXĐ:x\ne1;x\ne-3\right)\)

\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{x+3}{x-1}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{\left(x+3\right)^2}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+3\right)}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{2-3x+x^2+6x+9-x^2+1}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}.\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{3x+12}=\dfrac{x^2+x+1}{x+3}\)

\(M=A.B=\dfrac{x^2+x+1}{x+3}.\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x-2}{x+3}\)

b. -Để M thuộc Z thì:

\(\left(x^2+x-2\right)⋮\left(x+3\right)\)

\(\Rightarrow\left(x^2+3x-2x-6+4\right)⋮\left(x+3\right)\)

\(\Rightarrow\left[x\left(x+3\right)-2\left(x+3\right)+4\right]⋮\left(x+3\right)\)

\(\Rightarrow4⋮\left(x+3\right)\)

\(\Rightarrow x+3\in\left\{1;2;4;-1;-2;-4\right\}\)

\(\Rightarrow x\in\left\{-2;-1;1;-4;-5;-7\right\}\)

c. \(A^{-1}-B=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{x^3-1}\)

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

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

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

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

\(=\dfrac{1}{x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)

\(Max=\dfrac{4}{3}\Leftrightarrow x=\dfrac{-1}{2}\)

a: \(A=\dfrac{15}{x+2}+\dfrac{42}{3\left(x+2\right)}=\dfrac{45+42}{3\left(x+2\right)}=\dfrac{29}{x+2}\)

b: Để A là số nguyên thì \(x+2\in\left\{1;-1;29;-29\right\}\)

hay \(x\in\left\{-1;-3;27;-31\right\}\)

\(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

\(\Rightarrow\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\le\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=2\)

Lại có \(\dfrac{1}{2x+y+z}=\dfrac{1}{x+y+x+z}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\)

Tương tự \(\dfrac{1}{x+2y+z}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}\right)\)

\(\dfrac{1}{x+y+2z}\le\dfrac{1}{4}\left(\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)\)

Cộng vế với vế: \(P\le\dfrac{1}{2}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)=\dfrac{1}{2}.2=1\)

\(\Rightarrow P_{max}=1\) khi \(x=y=z=\dfrac{3}{4}\)

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

Với mọi giá trị của \(x\in R\) ta có:

\(\left(n-1\right)^2\ge0\Rightarrow2.\left(n-1\right)^2\ge0\Rightarrow2.\left(n-1\right)^2+3\ge3\)

\(\Rightarrow\dfrac{1}{2\left(n-1\right)^2+3}\ge\dfrac{1}{3}\) với mọi giá trị của \(x\in R\)

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

\(\Rightarrow2\left(n-1\right)^2=0\Rightarrow\left(n-1\right)^2=0\Rightarrow n-1=0\Rightarrow n=1\)

Vậy GTNN của biểu thức là \(\dfrac{1}{3}\) đạt được khi và chỉ khi \(n=1\)

Chúc bạn học tốt!!!

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

Với mọi giá trị của \(n\in R\) ta có:

\(\left(n-1\right)^2\ge0\Rightarrow2\left(n-1\right)^2\ge0\Rightarrow2\left(n-1\right)^2+3\ge3\Rightarrow\dfrac{1}{2\left(n-1\right)^2+3}\le\dfrac{1}{3}\)

Hay \(B\le\dfrac{1}{3}\) với mọi giá trị của \(n\in R\).

Để \(B=\dfrac{1}{3}\) thì \(\dfrac{1}{2\left(n-1\right)^2+3}=\dfrac{1}{3}\)

\(\Rightarrow2\left(n-1\right)^2+3=3\Rightarrow2\left(n-1\right)^2=0\Rightarrow\left(n-1\right)^2=0\Rightarrow n-1=0\Rightarrow n=1\)

Vậy GTLN của biểu thức B là \(\dfrac{1}{3}\) đạt được khi và chỉ khi \(n=1\)

Chúc bạn học tốt!!!

Ta có :

\(2\left(n-1\right)^2\ge0\Rightarrow2\left(n-1\right)^2+3\ge3\)

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

hay \(B\le\dfrac{1}{3}\)

Dấu " = " xảy ra \(\Leftrightarrow n-1=0\Rightarrow n=1\)

Vậy \(B\) _max = \(\dfrac{1}{3}\) khi n = 1

m = 5 

n = -1

mình nhầm câu trên

Lời giải:

$D=\frac{2(3n+1)-5}{3n+1}=2-\frac{5}{3n+1}$

Để $D$ max thì $\frac{5}{3n+1}$ min 

$\Rightarrow 3n+1$ max

$\Rightarrow n$ max

Với $n$ nguyên thì không có giá trị $n$ max. Nên không tồn tại $n$ nguyên để $D$ max.