Sau gõ latex.

\(\sqrt{x-5}-\left(x-\dfrac{14}{3}+\sqrt{x-5}\right)=3\\ \Leftrightarrow\sqrt{x-5}-x+\dfrac{14}{3}-\sqrt{x-5}=3\\ \Leftrightarrow-x=3-\dfrac{14}{3}=-\dfrac{5}{3}\\ \Rightarrow x=-\dfrac{5}{3}:\left(-1\right)=\dfrac{5}{3}\)

\(a.17+8x=10-6x\\\Leftrightarrow 8x+6x=-17+10\\\Leftrightarrow 2x=-7\\ \Leftrightarrow x=-\frac{7}{2}\)

Vậy nghiệm của phương trình trên là \(-\frac{7}{2}\)

\(b.3\left(x+5\right)+7=19-5\left(x-2\right)\\\Leftrightarrow 3x+15+7=19-5x+10\\ \Leftrightarrow3x+5x=-15-7+19+10\\ \Leftrightarrow8x=7\\\Leftrightarrow x=\frac{7}{8}\)

Vậy nghiệm của phương trình trên là \(\frac{7}{8}\)

\(c.3x-4\left(x+2\right)\left(x+3\right)=14-4\left(x^2-3x\right)\\ \Leftrightarrow3x-4\left(x^2+5x+6\right)=14-4x^2+12x\\ \Leftrightarrow4x^2-4x^2+3x-5x-12x=24+14\\ \Leftrightarrow-14x=38\\ \Leftrightarrow x=-\frac{19}{7}\)

Vậy nghiệm của phương trình trên là \(-\frac{19}{7}\)

\(d.x+\frac{3}{4}+3x+2=\frac{x}{3}-3x-\frac{2}{6}\\ \Leftrightarrow\frac{12x}{12}+\frac{9}{12}+\frac{36x}{12}+\frac{24}{12}=\frac{4x}{12}-\frac{36x}{12}-\frac{4}{12}\\ \Leftrightarrow12x+9+36x+24=4x-36x-4\\ \Leftrightarrow12x+36x+36x-4x=-24-9-4\\ \Leftrightarrow80x=-37\\ \Leftrightarrow x=-\frac{37}{80}\)

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

Lời giải:
Đặt $\frac{1}{x-y+2}=a;\frac{1}{x+y-1}=b$ thì HPT trở thành cơ bản:
\(\left\{\begin{matrix} 14a-10b=9\\ 3a+2b=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 14a-10b=9\\ 15a+10b=20\end{matrix}\right.\)

$\Rightarrow (14a-10b)+(15a+10b)=9+20$

$\Leftrightarrow 29a=29\Leftrightarrow a=1$.

$b=\frac{4-3a}{2}=\frac{1}{2}$

Vậy: \(\left\{\begin{matrix} \frac{1}{x-y+2}=1\\ \frac{1}{x+y-1}=\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x-y+2=1\\ x+y-1=2\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x-y=-1\\ x+y=3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=2\end{matrix}\right.\)

Lam lai nha , nay cau tha qua :(

\(\sqrt{x-5}-\dfrac{x-14}{3+\sqrt{x-5}}=3\) ( x ≥ 5 )

Dat : \(\sqrt{x-5}=a\) ( x ≥ 0 ) , ta co :

\(\sqrt{a}-\dfrac{a-9}{3+\sqrt{a}}=3\)

⇔ \(\sqrt{a}-\sqrt{a}+3=3\)

⇔ \(3=3\left(Luon-dung\right)\)

KL........

\(\sqrt{x-5}-\dfrac{x-14}{3+\sqrt{x-5}}=3\)

\(\sqrt{x-5}-\dfrac{x-14}{3+\sqrt{x-5}}=3,x\in\left[5,+\infty\right]\)

\(\sqrt{x-5}-\dfrac{x-14}{3+\sqrt{x-5}}-3=0\)

\(\dfrac{\left(3+\sqrt{x-5}\right)\sqrt{x-5}-\left(x-14\right)-3\left(3+\sqrt{x-5}\right)}{3+\sqrt{x-5}}=0\)

\(3\sqrt{x-5}\sqrt{x-5}-\left(x-14\right)-3\left(3+\sqrt{x-5}\right)=0\)

\(3\sqrt{x-5}+x-5-\left(x-14\right)-9-3\sqrt{x-5}=0\)

\(x-5-x+14-9=0\)

\(0=0\)

\(x\in R,x\in\left[5,+\infty\right]\)

Cach khac :3

\(\sqrt{x-5}-\dfrac{x-14}{3+\sqrt{x-5}}=3\) ( x ≥ 5)

Dat : \(x-5=a\) ( a ≥ 0 ), ta co :

\(\sqrt{a}+\dfrac{a-9}{3+\sqrt{a}}=3\)

⇔ \(\sqrt{a}+\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}{\sqrt{a}+3}=3\)

\(\text{⇔}2\sqrt{a}=6\)

⇔ \(\sqrt{a}=3\text{⇔}a=9\) ( TM )

Khi do : \(x-5=9\text{⇔}x=14\left(TM\right)\)

KL..............

=>x^2+8x+15=x^2+6x+8

=>8x+15=6x+8

=>2x=-7

=>x=-7/2

Đề: 

`=> x^2 + 3x + 5x + 15 = 2x + 8 + x^2 + 4x`

`=> x^2 + 8x + 15 = x^2 + 6x + 8`

`=> x^2 - x^2 + 8x - 6x + 15 - 8 = 0`

`=> 2x + 7 = 0`

`=> 2x = -7`

`=> x = -7/2`

Vậy `x = -7/2`

\(\left(x+3\right)\left(x+5\right)=\left(x+4\right)\left(2+x\right)\)

\(\Leftrightarrow\left(x+3\right)\left(x+5\right)\left(4+x\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\\4+x=0\\x+2=0\end{matrix}\right.\)                 \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\\x=-4\\x=-2\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{-3;-5;-4;-2\right\}\)

\(\dfrac{2}{x-14}-\dfrac{5}{x-13}=\dfrac{2}{x-9}-\dfrac{5}{x-11}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne9;11;13;14\\\left(\dfrac{2}{x-14}-\dfrac{2}{3}\right)-\left(\dfrac{5}{x-13}-\dfrac{5}{4}\right)=\left(\dfrac{2}{x-9}-\dfrac{1}{4}\right)-\left(\dfrac{5}{x-11}-\dfrac{5}{6}\right)\end{matrix}\right.\)

\(\Leftrightarrow2\left(\dfrac{x-17}{3\left(x-14\right)}\right)-5\left(\dfrac{x-17}{4\left(x-13\right)}\right)=\left(\dfrac{x-17}{4\left(x-9\right)}\right)-5\left(\dfrac{x-17}{6\left(x-11\right)}\right)\)

\(\left(x-17\right)\left[\dfrac{2}{3\left(x-14\right)}-\dfrac{5}{4\left(x-13\right)}+\dfrac{5}{6\left(x-11\right)}-\dfrac{1}{4\left(x-9\right)}\right]=0\)

[..] vô nghiệm

x=17

Lời giải:

Bài của bạn ngonhuminh cơ bản không đúng do không có cơ sở khẳng định biểu thức trong ngoặc vuông vô nghiệm.

ĐKXĐ: \(x\neq \left\{9;11;13;14\right\}\)

\(\frac{2}{x-14}-\frac{5}{x-13}=\frac{2}{x-9}-\frac{5}{x-11}\)

\(\Leftrightarrow 2\left(\frac{1}{x-14}-\frac{1}{x-9}\right)=5\left(\frac{1}{x-13}-\frac{1}{x-11}\right)\)

\(\Leftrightarrow \frac{10}{(x-14)(x-9)}=\frac{10}{(x-13)(x-11)}\)

\(\Rightarrow (x-14)(x-9)=(x-13)(x-11)\)

\(\Leftrightarrow x^2-23x+126=x^2-24x+143\)

\(\Leftrightarrow x-17=0\Leftrightarrow x=17\)

Thử lại thấy thỏa mãn.

Vậy \(x=17\)

không vô nghiệm thì tìm đi