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: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

b: Ta có: \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

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

\(=\dfrac{2}{x+\sqrt{x}+1}\)

c: Ta có: \(x+\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ

\(\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}>0\forall x\)

Bạn tham khảo lời giải tại đây:

Tìm GTLN của biểu thức: \(A=\left(\dfrac{x^2}{x^2-3x 2} \dfrac{x^2}{x^2-5x 6}\right):\dfrac{x^4 x^2 1}{x^2-4x 3}\) - Hoc24

Lời giải:

ĐK: $x\neq 1;2;3$

\(A=x^2\left[\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}\right].\frac{(x-1)(x-3)}{x^4+x^2+1}\)

\(=x^2.\frac{x-3+x-1}{(x-1)(x-2)(x-3)}.\frac{(x-1)(x-3)}{x^4+x^2+1}=x^2.\frac{2(x-2)}{(x-1)(x-2)(x-3)}.\frac{(x-1)(x-3)}{x^4+x^2+1}=\frac{2x^2}{x^4+x^2+1}\)

Áp dụng BĐT AM-GM: $x^4+1\geq 2x^2$

$\Rightarrow A\leq \frac{2x^2}{2x^2+x^2}=\frac{2}{3}$

Vậy $A_{\max}=\frac{2}{3}$. Giá trị đạt tại $x^4=1$ hay $x=-1$ (do $x\neq 1$)

`a)M=(x^4+2)/(x^6+1)+(x^2-1)/(x^4-x^2+1)-(x^2+3)/(x^4+4x^2+3)`

`=(x^4+2)/(x^6+1)+(x^2-1)/(x^4-x^2+1)-(x^2+3)/((x^2+1)(x^2+3))`

`=(x^4+2)/(x^6+1)+((x^2-1)(x^2+1))/(x^6+1)-1/(x^2+1)`

`=(x^4+2+x^4-1-x^4+x^2-1)/(x^2+1)`

`=(x^4+x^2)/(x^2+1)`

`=(x^2(x^2+1))/(x^2+1)`

`=x^2`

`b)` tìm gtnn chứ?

`M=x^2>=0`

Dấu '=" `<=>x=0`

 Bài này chỉ tìm được GTLN thôi nhé bạn.

 Ta thấy \(A=-\dfrac{1}{3}x^2+2x\) 

\(A=-\dfrac{1}{3}\left(x^2-6x\right)\)

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

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

 Vì \(\left(x-3\right)^2\ge0\) nên \(A\le3\) (dấu "=" xảy ra khi \(x-3=0\Leftrightarrow x=3\)). Như vậy GTLN của A là 3, đạt được khi \(x=3\).

1:

\(A=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

3: A nguyên

=>-5căn x-15+17 chia hết cho căn x+3

=>căn x+3 thuộc Ư(17)

=>căn x+3=17

=>x=196

a: ĐKXĐ: x<>1; x<>2; x<>3

\(K=\left(\dfrac{x^2}{\left(x-2\right)\left(x-3\right)}+\dfrac{x^2}{\left(x-1\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+2x^2+1-x^2}\)

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

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

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

b: