Lời giải:
PT $\Leftrightarrow (\frac{x+1}{2022}+1)+(\frac{x+2}{2021}+1)+...+(\frac{x+23}{2000}+1)=0$

$\Leftrightarrow \frac{x+2023}{2022}+\frac{x+2023}{2021}+...+\frac{x+2023}{2000}=0$

$\Leftrightarrow (x+2023)(\frac{1}{2022}+\frac{1}{2021}+...+\frac{1}{2000})=0$
Dễ thấy tổng trong () luôn dương 

$\Rightarrow x+2023=0$

$\Leftrightarrow x=-2023$

bạn rút gọn trong ngoặc rồi chi cho x+1

Lời giải:

$\frac{x^3+8}{x^2-2x+1}.\frac{x^2+3x+2}{1-x^2}=\frac{(x^3+8)(x^2+3x+2)}{(x^2-2x+1)(1-x^2)}$

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

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

a) \(6x^2-2x-6x^2+13=0\\ -2x=-13\\ x=\dfrac{13}{2}\)

b: =>2x-2x-1=x-6x

=>-5x=-1

hay x=1/5

Lời giải:
a. 

$2x(3x-1)=6x^2-13$

$\Leftrightarrow 6x^2-2x=6x^2-13$

$\Leftrightarrow 2x=13$

$\Leftrightarrow x=\frac{13}{2}$

b. 

$\frac{x}{3}-\frac{2x+1}{6}=\frac{x}{6}-x$

$\Leftrightarrow \frac{2x-(2x+1)}{6}=\frac{-5}{6}x$

$\Leftrightarrow \frac{-1}{6}=\frac{-5}{6}x$
$\Leftrightarrow x=\frac{-1}{6}: \frac{-5}{6}=\frac{1}{5}$

Để D là số nguyên thì \(2x+4⋮3x-1\)

=>\(6x+12⋮3x-1\)

=>\(6x-2+14⋮3x-1\)

=>\(14⋮3x-1\)

=>\(3x-1\in\left\{1;-1;2;-2;7;-7;14;-14\right\}\)

=>\(3x\in\left\{2;0;3;-1;8;-6;15;-13\right\}\)

=>\(x\in\left\{\dfrac{2}{3};0;1;-\dfrac{1}{3};\dfrac{8}{3};-2;5;-\dfrac{13}{3}\right\}\)

mà x nguyên

nên \(x\in\left\{0;1;-2;5\right\}\)

\(D=\dfrac{2x+4}{3x-1}\\ =>3D=\dfrac{6x+12}{3x-1}=\dfrac{2\left(3x-1\right)+14}{3x-1}=2+\dfrac{14}{3x-1}\)

Để 3D nguyên thì : \(\dfrac{14}{3x-1}\in Z\)

\(=>14⋮\left(3x-1\right)\\ =>3x-1\inƯ\left(14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

\(=>3x\in\left\{2;0;3;-1;8;-6;15;-13\right\}\\ =>x\in\left\{\dfrac{2}{3};0;1;-\dfrac{1}{3};\dfrac{8}{3};-2;5;-\dfrac{13}{3}\right\}\)

Mà x nguyên \(=>x\in\left\{0;1;-2;5\right\}\)

Do những giá trị trên chỉ là 3D nguyên nên chưa chắc D đã nguyên

Vậy thử lại thay từng giá trị x vào bt D

Kết luận : \(x\in\left\{0;1;-2;5\right\}\)

`(2/3 x +1/2) (-2x+3)=0`

\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x+\dfrac{1}{2}=0\\-2x+3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{1}{2}\\-2x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}.\dfrac{3}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(\left(\dfrac{2}{3}x+\dfrac{1}{2}\right)\cdot\left(-2x+3\right)=0\\ =>\left[{}\begin{matrix}\dfrac{2}{3}x+\dfrac{1}{2}=0\\-2x+3=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{1}{2}\\-2x=-3\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\)

=>8(x+1/x)^2+4[(x+1/x)^2-2]^2-4[(x+1/x)^2-2](x+1/x)^2=(x+4)^2

Đặt x+1/x=a(a>=2)

=>8a^2+4[a^2-2]^2-4[a^2-2]*a^2=(x+4)^2

=>8a^2+4a^4-16a^2+16-4a^4+8a^2=(x+4)^2

=>(x+4)^2=16

=>x+4=4 hoặc x+4=-4

=>x=-8;x=0

Điều kiện: \(x\ne0\)

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2\)

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\\ \Leftrightarrow\left(x+4\right)^2=16\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

Vì \(x\ne0\) nên \(S=\left\{-8\right\}\)

Ta có: \(\sqrt{2x+7}-6=x\)

\(\Leftrightarrow\sqrt{2x+7}=x+6\)

\(\Leftrightarrow x^2+12x+36-2x-7=0\)

\(\Leftrightarrow x^2+10x+29=0\)(Vô lý)

Vậy: \(S=\varnothing\)

Điều kiện : x ≥ 0

\(\sqrt{2x+97}-6=x\text{⇔}\sqrt{2x+97}=x+6\\ \text{⇔}2x+97=x^2+12x+36\text{⇔}x^2+10x-61=0\\ \text{⇔}\left[{}\begin{matrix}x=-5+\sqrt{86}\\x=-5-\sqrt{86}\end{matrix}\right.\)

\(\sqrt{2x+97}-6=x\)

\(\Leftrightarrow\sqrt{2x+97}=x+6\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+6\ge0\\2x+97=\left(x+6\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-6\\x^2+10x-61=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-6\\\left[{}\begin{matrix}x=-5+\sqrt{86}\\x=-5-\sqrt{86}\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x=-5+\sqrt{86}\)

Vậy..,