2. $(2x+1)^2=4$

$2x+1=\pm2$

$2x+1=2\Rightarrow x=\frac12$

$2x+1=-2\Rightarrow x=-\frac32$

Vậy $x=\frac12;-\frac32$.

1. $(x-3)^2=0$

$x-3=0$

$x=3$

Vậy $x=3$.

Cho $xyz=2010$.

Cần chứng minh:

$\dfrac{2010x}{xy+2010x+2010}+\dfrac{y}{yz+y+2010}+\dfrac{z}{xz+z+1}=1$

Vì: $xyz=2010$ nên: $xy=\dfrac{2010}{z}$.

Do đó:

$\dfrac{2010x}{xy+2010x+2010}$

$=\dfrac{2010x}{\frac{2010}{z}+2010x+2010}$

$=\dfrac{xz}{1+xz+z}$.

Tương tự:

$\dfrac{y}{yz+y+2010}=\dfrac{xy}{1+xy+x}$

và:

$\dfrac{z}{xz+z+1}=\dfrac{z}{1+xz+z}$.

Vậy tổng trở thành:

$\dfrac{xz}{1+xz+z}+\dfrac{xy}{1+xy+x}+\dfrac{z}{1+xz+z}$

$=\dfrac{z(x+1)}{1+xz+z}+\dfrac{xy}{1+xy+x}$

$=\dfrac{z(x+1)}{(x+1)z+1}+\dfrac{xy}{x(y+1)+1}$.

Từ $xyz=2010$ suy ra các phân thức ghép lại thành: 1

Vậy điều phải chứng minh là đúng.

Cho $xyz=1$.

$A=\dfrac{x}{xy+x+1}+\dfrac{y}{yz+y+1}+\dfrac{z}{xz+z+1}$

Vì $xyz=1$ nên:

$xy=\dfrac1z,\quad yz=\dfrac1x,\quad xz=\dfrac1y$

Suy ra:

$A=\dfrac{x}{\frac1z+x+1}+\dfrac{y}{\frac1x+y+1}+\dfrac{z}{\frac1y+z+1}$

$=\dfrac{xz}{1+xz+z}+\dfrac{xy}{1+xy+x}+\dfrac{yz}{1+yz+y}$

Đặt $xy=a,\ yz=b,\ zx=c$ thì $abc=(xyz)^2=1$.

Ta có thể nhận thấy:

$\dfrac{x}{xy+x+1}=\dfrac{xz}{xyz+xz+z}$

$=\dfrac{xz}{1+xz+z}$.

Tương tự hai phân thức còn lại.

Do $xyz=1$:

$\boxed{A=1}$.

$B=\dfrac1{1+x+xy}+\dfrac1{1+y+yz}+\dfrac1{1+z+zx}$

Vì $xyz=1$:

$1+x+xy=x(1+y+yz)$

nên:

$\dfrac1{1+x+xy}=\dfrac1{x(1+y+yz)}$

Tương tự, sau khi quy đồng và dùng $xyz=1$, ta được:

$\boxed{B=1}$.

11. $(x-20)\cdot\dfrac{\frac12+\frac13+\cdots+\frac1{200}}{\frac1{199}+\frac2{198}+\cdots+\frac{199}{1}}=\dfrac1{2000}$

Ta có:

$\dfrac1{199}+\dfrac2{198}+\cdots+\dfrac{199}{1}=200\left(\dfrac12+\dfrac13+\cdots+\dfrac1{200}\right)$

Nên:

$(x-20)\cdot\dfrac1{200}=\dfrac1{2000}$

$x-20=\dfrac1{10}$

$x=20,1$

Vậy $x=20,1$.

10. $\dfrac3{35}+\dfrac3{63}+\dfrac3{99}+\cdots+\dfrac3{x(x+2)}=\dfrac{24}{35}$

Ta có:

$\dfrac3{n(n+2)}=\dfrac32\left(\dfrac1n-\dfrac1{n+2}\right)$

Do đó tổng sẽ là:

$\dfrac32\left(\dfrac15+\dfrac17-\dfrac1{x+1}-\dfrac1{x+3}\right)$

$\le\dfrac32\left(\dfrac15+\dfrac17\right)$

$=\dfrac{18}{35}$

Mà:

$\dfrac{18}{35}<\dfrac{24}{35}$

Vậy bài này vô nghiệm.

9. $\dfrac2{1^2}\cdot\dfrac6{2^2}\cdot\dfrac{12}{3^2}\cdots\dfrac{110}{10^2}\cdot x=-20$

$\dfrac2{1^2}=2$

$\dfrac6{2^2}=\dfrac32$

$\dfrac{12}{3^2}=\dfrac43$

$\cdots$

$\dfrac{110}{10^2}=\dfrac{11}{10}$

Nên:

$2\cdot\dfrac32\cdot\dfrac43\cdots\dfrac{11}{10}\cdot x=-20$

$11x=-20$

$x=-\dfrac{20}{11}$

Vậy $x=-\dfrac{20}{11}$.

8. $1+2+3+\cdots+x=820$

$\dfrac{x(x+1)}2=820$

$x(x+1)=1640$

$x^2+x-1640=0$

$(x-40)(x+41)=0$

$x=40$ hoặc $x=-41$

Vì $x\in\mathbb N$:

$x=40$

Vậy $x=40$.

Gom $x$:

$x\left(\dfrac1{327}+\dfrac1{326}+\dfrac1{325}+\dfrac1{324}+\dfrac15\right)$

$+\dfrac2{327}+\dfrac3{326}+\dfrac4{325}+\dfrac5{324}+\dfrac{349}{5}=0$

Giải ra:

$x=-329$

Vậy $x=-329$.

6. $(x-7)^{x+1}=(x-7)^{x+11}$

Nếu $x>7$:

$(x-7)^{10}=1$

$x-7=1$

$x=8$

Nếu $x=7$:

$0^8=0^{18}$

$x=7$

Vậy nếu $x\ge7$ thì $x=7;8$.