ĐK : x \(\ne\) 1
a) D = \(\left(1+\frac{x}{x^2+1}\right):\left(\frac{1}{x-1}-\frac{2x}{x^3+x-x^2-1}\right)=\left(\frac{x^2+1}{x^2+1}+\frac{x}{x^2+1}\right):\left(\frac{x^2+1}{\left(X^2+1\right)\left(x-1\right)}-\frac{2x}{x^2\left(x-1\right)+\left(x-1\right)}\right)\)

\(=\frac{x^2+x+1}{x^2+1}:\frac{x^2-2x+1}{\left(x-1\right)\left(x^2+1\right)}=\frac{x^2+x+1}{x^2+1}\cdot\frac{\left(x-1\right)\left(X^2+1\right)}{\left(x-1\right)^2}=\frac{x^2+x+1}{x^2+1}\cdot\frac{x^2+1}{x-1}=\frac{x^2+x+1}{x-1}\)

b)

D <1

=> \(x^2+x+1< x-1\Rightarrow x^2+x+1-x+1< 0\Rightarrow x^2+2< 0\) ( vô lí )

Vậy D > 1, không có x thỏa mãn

c) D thuộc Z

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

Vì x thuộc Z nên D thuộc Z khi

\(x-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

* x -1 = 1 => x= 2 (tm)

* x-1 = -1 => x = 0 (tm)

* x-1 =3 => x = 4 (tm)

* x-1 = -3 => x = -2 ( tm )

\(ĐKXD:x\ne1\)

\(a,D=\left(1+\frac{x}{x^2+1}\right):\left(\frac{1}{x-1}-\frac{2x}{x^3+x-x^2-1}\right)=\frac{x^2+x+1}{x^2+1}:\left(\frac{1}{\left(x-1\right)}-\frac{2x}{\left(x-1\right)\left(x^2+1\right)}\right)=\frac{x^2+x+1}{x^2+1}:\left(\frac{x^2+1}{\left(x-1\right)\left(x^2+1\right)}-\frac{2x}{\left(x-1\right)\left(x^2+1\right)}\right)=\frac{x^2+x+1}{x^2+1}:\left(\frac{x^2-2x+1}{\left(x-1\right)\left(x^2+1\right)}\right)=\frac{x^2+x+1}{x^2+1}:\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+1\right)}=\frac{x^2+x+1}{x^2+1}:\frac{x-1}{x^2+1}=\frac{\left(x^2+x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x-1\right)}=\frac{x^2+x+1}{x-1}\)

\(D< 1\Leftrightarrow x^2+x+1< x-1\Leftrightarrow\left(x-1\right)-\left(x^2+x+1\right)>0\Leftrightarrow x-1-x^2-x-1>0\Leftrightarrow-\left(x^2+2\right)>0\left(\text{ vô lí}\right).\text{ Nên không tìm được x thỏa mãn}\)

\(ĐểDnguyênthì:x^2+x+1⋮x-1\Leftrightarrow x\left(x-1\right)+2x+1⋮x-1\Leftrightarrow\left(x+2\right)\left(x-1\right)+3⋮x-1\Leftrightarrow3⋮x-1\left(\text{ vì: (x+2)(x-1) chia hết cho x-1}\right)\Leftrightarrow x-1\in\left\{-1;1;-3;3\right\}\Leftrightarrow x\in\left\{0;2;-2;4\right\}.Vậy:x\in\left\{0;2;-2;4\right\}thìDnguyên\)

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

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

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

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

b: Thay x=1/2 vào D, ta được:

\(D=\left(4\cdot\dfrac{1}{2}\right):\left[3\cdot\left(\dfrac{1}{4}-\dfrac{1}{2}+1\right)\right]\)

\(=2:\left[3\cdot\dfrac{1-2+4}{4}\right]\)

\(=2:\left[3\cdot\dfrac{3}{4}\right]=2:\dfrac{9}{4}=\dfrac{8}{9}\)

c: Ta có: D=8/9

nên \(\dfrac{4x}{3\left(x^2-x+1\right)}=\dfrac{8}{9}\)

\(\Leftrightarrow24\left(x^2-x+1\right)=36x\)

\(\Leftrightarrow2x^2-2x+2-3x=0\)

\(\Leftrightarrow2x^2-5x+2=0\)

=>(x-2)(2x+1)=0

=>x=2 hoặc x=-1/2

a,ĐKXĐ của biểu thức D là :

x³+x²+x+1\(\ne0\)

\(\Leftrightarrow\)x²(x+1)+(x+1)\(\ne\)0

\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ne0\\x^2+1\ne0\left(vôlí\right)\end{matrix}\right.\)

\(\Leftrightarrow x\ne-1\)

Ta có : D=\(\frac{3\left(x+1\right)}{x^3+x^2+x+1}=\frac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}=\frac{3.\left(x+1\right)}{\left(x+1\right)\left(x^2+1\right)}=\frac{3}{x^2+1}\)

b,Để D nguyên thì \(\frac{3}{x^2+1}\)(đkxđ: x\(\ne-1\)) nguyên

\(\Leftrightarrow\)x²+1\(\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

ta có bảng giá trị sau :

vậy x\(\in\left\{0;\sqrt{2}\right\}\)thì D nguyên

c, Ta có : D=\(\frac{3}{x^2+1}\left(đkxđ:x\ne-1\right)\)\(\le3\)

Dấu = xảy ra khi : x=0 \(\Leftrightarrow\)D=3

Vậy Max D=3 \(\Leftrightarrow x=0\)

Chọn B

Đáp án: D

Ta có:

⇒ d': x - 2y + 5 = 0

Ta thấy:

 ⇒ d cắt d’

Mà (-1;-2).(1;-2) = -1 + 4 = 3 ≠ 0 nên d cắt d’ nhưng không vuông

a) D (ĐKXĐ: x\(\ge0,x\ne1\))

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

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

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

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

=\(-\sqrt{x}\left(\sqrt{x}-1\right)=\sqrt{x}-x\)

b) \(\sqrt{x}-x=3\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)=3\)

=\(\sqrt{x}-x-3=0\Leftrightarrow\left(x-2.\dfrac{1}{2}x+\dfrac{1}{4}\right)-\dfrac{13}{4}=0\)

\(\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{13}{4}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-\dfrac{1}{2}=\dfrac{13}{4}\\\sqrt{x}-\dfrac{1}{2}=-\dfrac{13}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{225}{16}\\x=\dfrac{121}{16}\end{matrix}\right.\)

a: ĐKXĐ: x>0; x<>9

b: \(D=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\right):\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)

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

c: Để D<-1 thì D+1<0

\(\Leftrightarrow-3\sqrt{x}+2\sqrt{x}+4< 0\)

\(\Leftrightarrow4-\sqrt{x}< 0\)

hay x>16