\(\left\{{}\begin{matrix}x+y=5\\\dfrac{3}{5}+\dfrac{2}{x-y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=5\\\dfrac{2}{x-y}=\dfrac{12}{5}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+y=5\\6\left(x-y\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=5\\x-y=\dfrac{5}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{35}{6}\\y=x-\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{35}{12}\\y=\dfrac{25}{12}\end{matrix}\right.\)

TA có: \(\begin{cases} (a+b)(a+c) = 3 \quad (1) \\ (a+b)(b+c) = 4 \quad (2) \\ (a+c)(b+c) = 5 \quad (3) \end{cases}\)

=>\([(a+b)(b+c)(a+c)]^2 = 3 \times 4 \times 5 = 60\)

=>\((a+b)(b+c)(a+c) = \pm\sqrt{60} = \pm 2\sqrt{15}\)

Vì (a+b)(a+c)=3>0; (a+b)(b+c)=4>0; (a+c)(b+c)=5>0

nen các tổng a+b; a+c; b+c phải cùng dương hoặc cùng âm

TH1: a+b>0; a+c>0; b+c>0

Do đó, ta có: \(\left(a+b\right)\left(b+c\right)\left(a+c\right)=2\sqrt{15}\)

=>\(\begin{cases}b+c=\frac{2\sqrt{15}}{3}\\ a+c=\frac{2\sqrt{15}}{4}=\frac{\sqrt{15}}{2}\\ a+b=\frac{2\sqrt{15}}{5}\end{cases}\)

=>\((a+b) + (b+c) + (a+c) = \frac{2\sqrt{15}}{3} + \frac{\sqrt{15}}{2} + \frac{2\sqrt{15}}{5}\)

=>\(2(a+b+c) = \sqrt{15}\left(\frac{2}{3} + \frac{1}{2} + \frac{2}{5}\right) = \sqrt{15} \cdot \frac{47}{30} = \frac{47\sqrt{15}}{30}\)

\(\implies a+b+c = \frac{47\sqrt{15}}{60}\)

Do đó: \(\begin{cases}a=(a+b+c)-(b+c)=\frac{47\sqrt{15}}{60}-\frac{2\sqrt{15}}{3}=\frac{7\sqrt{15}}{60}\\ b=(a+b+c)-(a+c)=\frac{47\sqrt{15}}{60}-\frac{\sqrt{15}}{2}=\frac{17\sqrt{15}}{60}\\ c=(a+b+c)-(a+b)=\frac{47\sqrt{15}}{60}-\frac{2\sqrt{15}}{5}=\frac{23\sqrt{15}}{60}\end{cases}\)

TH2: a+b<0; a+c<0; b+c<0

Do đó, ta có: \(\left(a+b\right)\left(b+c\right)\left(a+c\right)=-2\sqrt{15}\)

=>\(\begin{cases}b+c=\frac{-2\sqrt{15}}{3}\\ a+c=\frac{2\sqrt{15}}{4}=\frac{-\sqrt{15}}{2}\\ a+b=\frac{-2\sqrt{15}}{5}\end{cases}\)

=>\((a+b)+(b+c)+(a+c)=\frac{-2\sqrt{15}}{3}+\frac{-\sqrt{15}}{2}+\frac{-2\sqrt{15}}{5}\)

=>\(2(a+b+c)=-\sqrt{15}\left(\frac{2}{3}+\frac{1}{2}+\frac{2}{5}\right)=\sqrt{15}\cdot\frac{-47}{30}=\frac{-47\sqrt{15}}{30}\)

=>\(a+b+c=-\frac{47\sqrt{15}}{60}\)

=>\(\begin{cases}a=(a+b+c)-(b+c)=-\frac{47\sqrt{15}}{60}-\left(-\frac{2\sqrt{15}}{3}\right)=-\frac{47\sqrt{15}}{60}+\frac{40\sqrt{15}}{60}=-\frac{7\sqrt{15}}{60}\\ b=(a+b+c)-(a+c)=-\frac{47\sqrt{15}}{60}-\left(-\frac{\sqrt{15}}{2}\right)=-\frac{47\sqrt{15}}{60}+\frac{30\sqrt{15}}{60}=-\frac{17\sqrt{15}}{60}\\ c=(a+b+c)-(a+b)=-\frac{47\sqrt{15}}{60}-\left(-\frac{2\sqrt{15}}{5}\right)=-\frac{47\sqrt{15}}{60}+\frac{24\sqrt{15}}{60}=-\frac{23\sqrt{15}}{60}\end{cases}\)

\(\Leftrightarrow x\left(x-2\right)\left(x^2+x-6\right)\le0\)

\(\Leftrightarrow x\left(x-2\right)^2\left(x+3\right)\le0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\-3\le x\le0\end{matrix}\right.\)

\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)

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

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

\(\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\)

Cho mình sửa lại nhé:

\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)

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

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

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

1) ĐKXĐ: \(x\ge-5\)

\(pt\Leftrightarrow x+5=9\Leftrightarrow x=9-5=4\left(tm\right)\)

2) ĐKXĐ: \(x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\sqrt{x-3}=6\)

\(\Leftrightarrow2\sqrt{x-3}=6\Leftrightarrow\sqrt{x-3}=3\)

\(\Leftrightarrow x-3=9\Leftrightarrow x=12\left(tm\right)\)

3) ĐKXĐ: \(x\ge-1\)

\(pt\Leftrightarrow\sqrt{\left(x+1\right)^2}-2\sqrt{x+1}=0\)

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

\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x+1}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+1=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=3\left(tm\right)\end{matrix}\right.\)

tui uk.......u...a