quy đồng r khử mẫu là ok

\(ĐK:x\ne0;2\)

\(\Leftrightarrow\dfrac{x+2}{x}-\dfrac{2x+3}{2\left(x-2\right)}=0\)

\(\Leftrightarrow\dfrac{2\left(x-2\right)\left(x+2\right)-x\left(2x+3\right)}{2x\left(x-2\right)}=0\)

\(\Leftrightarrow2\left(x^2-4\right)-x\left(2x+3\right)=0\)

\(\Leftrightarrow2x^2-8-2x^2-3x=0\)

\(\Leftrightarrow-3x=8\Leftrightarrow x=-\dfrac{8}{3}\left(tm\right)\)

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

\(\Leftrightarrow\) \(2x^2-8=2x^2+3x\)

\(\Leftrightarrow-3x=8\\ \Leftrightarrow x=-\dfrac{8}{3}\)

Vậy S = \(\left\{-\dfrac{8}{3}\right\}\)

\(\Leftrightarrow\dfrac{3\left(2x-1\right)-36}{12}=\dfrac{2\left(1+x\right)}{12}\)

\(\Leftrightarrow\dfrac{6x-3-36}{12}=\dfrac{2+2x}{12}\)

\(\Leftrightarrow6x-39=2+2x\)

\(\Leftrightarrow4x=41\Leftrightarrow x=\dfrac{41}{4}\)

\(\left(2-x\right)^2=\left(2x-1\right)\left(2x+1\right)\\ \Leftrightarrow4-4x+x^2=4x^2-1\\ \Leftrightarrow4x^2-1-x^2+4x-4=0\\ \Leftrightarrow3x^2+4x-5=0\)

Đến đây mik thấy nghiệm rất xấu bạn xem đề đúng chx nhé

\(3.\left(2-x\right)^2=\left(2x-1\right)\left(2x+1\right)\)

\(\Leftrightarrow3.\left(2-x\right)=2x+1\)

\(\Leftrightarrow6-3x=2x+1\)

\(\Leftrightarrow-3x-2x=1-6\)

\(\Leftrightarrow-5x=-5\)

\(\Leftrightarrow x=1\)

Vậy: S = {1}

\(3\left(2-x\right)^2=\left(2x-1\right)\left(2x+1\right)\\ \Leftrightarrow3\left(4-4x+x^2\right)=4x^2-1\\ \Leftrightarrow12-12x+3x^2=4x^2-1\\ \Leftrightarrow4x^2-1-3x^2+12x-12=0\\ \Leftrightarrow x^2+12x-13=0\\ \Leftrightarrow x^2+13x-x-13=0\\ \Leftrightarrow x\left(x+13\right)-\left(x+13\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+13\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-13\end{matrix}\right.\)

\(\left(2x-5\right)\left(x-7\right)-\left(x-7\right)\left(5+x\right)=0\\ \left(x-7\right)\left(2x-5-5-x\right)=0\\ \left(x-7\right)\left(x-10\right)=0\\ \left\{{}\begin{matrix}x-7=0\\x-10=0\end{matrix}\right.\left\{{}\begin{matrix}x=7\\x=10\end{matrix}\right.\)

\(\left(2x-5\right)\left(x-7\right)=\left(x-7\right)\left(5+x\right)\\ \Leftrightarrow\left(2x-5\right)\left(x-7\right)-\left(x-7\right)\left(5+x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(2x-5-5-x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-10=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=10\end{matrix}\right.\)

\(Đk:x\ne0;3\)

\(\Leftrightarrow\dfrac{x+3}{x-3}=\dfrac{18}{x\left(x-3\right)}+\dfrac{8}{x}\)

\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x-3\right)}=\dfrac{18+8\left(x-3\right)}{x\left(x-3\right)}\)

\(\Leftrightarrow x^2+3x=18+8x-24\)

\(\Leftrightarrow x^2-5x+6=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

x+3/x-3=18/x²-3x+8/x

x(x+3)/x(x-3)=18/x(x-3)+8(x-3)/x(x-3)

=>x(x+3)=18+8(x-3)

x²+3x=18+8x-24

x²+3x-8x=18-24

x²-5x=-6

x²-5x+6=0

x²-2x-3x+6=0

x(x-2)-3(x-2)=0

(x-3)(x-2)=0

=>x-3=0 hoặc x-2=0

=>x=3 hoặc x=2

a. ĐKXĐ: $x\in\mathbb{R}$

PT \(\Rightarrow \left\{\begin{matrix} 2-x\geq 0\\ x^2+x+2=(3-x)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq 2\\ x^2+x+2=x^2-6x+9\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq 2\\ 7x=7\end{matrix}\right.\Leftrightarrow x=1\)

b. ĐKXĐ: $x\geq -1$

PT $\Leftrightarrow (x^2-1)+\sqrt{x+1}=0$

$\Leftrightarrow (x-1)(x+1)+\sqrt{x+1}=0$

$\Leftrightarrow \sqrt{x+1}[(x-1)\sqrt{x+1}+1]=0$

$\Leftrightarrow \sqrt{x+1}=0$ hoặc $(x-1)\sqrt{x+1}+1=0$

Nếu $\sqrt{x+1}=0$

$\Leftrightarrow x=-1$ (tm)

Nếu $(x-1)\sqrt{x+1}+1=0$

$\Leftrightarrow (x-1)\sqrt{x+1}=-1$

$\Rightarrow (x-1)^2(x+1)=1$

$\Leftrightarrow x^3-x^2-x=0$

$\Leftrightarrow x(x^2-x-1)=0$

$\Leftrightarrow x=0$ hoặc $x^2-x-1=0$

$\Leftrightarrow x=0$ hoặc $x=\frac{1\pm \sqrt{5}}{2}$

Kết hợp đkxđ suy ra $x=0; -1; \frac{1\pm \sqrt{5}}{2}$

c. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{(x-2)(x+2)}-2\sqrt{x-2}=0$

$\Leftrightarrow \sqrt{x-2}(\sqrt{x+2}-2)=0$

$\Leftrightarrow \sqrt{x-2}=0$ hoặc $\sqrt{x+2}-2=0$

$\Leftrightarrow x=2$  (thỏa mãn)

d. ĐKXĐ: $x\geq 3$ hoặc $x\leq -4$

PT \(\Rightarrow \left\{\begin{matrix} 8-x\geq 0\\ x^2+x-12=(8-x)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq 8\\ x^2+x-12=x^2-16x+64\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq 8\\ 17x=76\end{matrix}\right.\Leftrightarrow x=\frac{76}{17}\) (tm)

e. ĐKXĐ: $x\geq \frac{-3}{2}$
PT $\Leftrightarrow x=\sqrt{2x+3}$

\(\Rightarrow \left\{\begin{matrix} x\geq 0\\ x^2=2x+3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ (x-3)(x+1)=0\end{matrix}\right.\Leftrightarrow x=3\) (tm)

f. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow \sqrt{x+2}=\sqrt{x}+\sqrt{x+1}$

$\Leftrightarrow x+2=2x+1+2\sqrt{x(x+1)}$ (bp hai vế)

$\Leftrightarrow 1-x=2\sqrt{x(x+1)}$

\(\Rightarrow \left\{\begin{matrix} 1-x\geq 0\\ (1-x)^2=4x(x+1)\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq 1\\ 3x^2+6x-1=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq 1\\ x=\frac{-3\pm 2\sqrt{3}}{3}\end{matrix}\right.\)

Kết hợp cả đkxđ suy ra  $x=\frac{-3+2\sqrt{3}}{3}$

lỗi r 

...

11 wasn't invited

12 were paiting

13 hasn't had

14 will be sent

15 was equipped

1. had already prepared (một hành động xảy ra trc một hành động khác trong quá khứ với dấu hiệu already trong câu => dùng thì QKHT)

2. will have lived (By the time: dấu hiệu của thì TLHT)

3. encourages (HTĐ)

4. will have (TLĐ)

5. is shopping (HTTD) - will be (TLĐ)