\(a)=\frac{-2\left(x+3\right)}{x\left(1-3x\right)}.\frac{1-3x}{x\left(x+3\right)}\)

\(=\frac{-2}{x^2}\)

\(b)=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)

\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}\)

\(=x\left(x-3\right)\)

\(c)=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\left(x+1\right)}\)

\(=\frac{\left(x+3\right).x}{x\left(x-1\right)\left(x+1\right)}-\frac{1.\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x\left(x+3\right)-\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x+3}{x+1}\)

# Sắp ik ngủ nên làm vậy hoi, ko chắc phần kq câu b và c đâu nha

Ta thấy \(\left(x-3\right)\left(2x+3\right)=2x^2-3x-9.\)

\(\left(1\right)\Leftrightarrow\frac{x}{x-3}-\frac{2x^2+9}{\left(x-3\right)\left(2x+3\right)}=\frac{1}{2x+3}\)

ĐK: \(x\ne3\)và \(x\ne-\frac{3}{2}\)

\(\Rightarrow x\left(2x+3\right)-2x^2-9=x-3\)

\(\Leftrightarrow2x^2+3x-2x^2-9=x-3\Leftrightarrow2x=6\Leftrightarrow x=2\)

Thỏa mãn ĐK

Các trường hợp khác làm tương tự

a/ Do \(x=0\) không phải nghiệm, pt tương đương:

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

Đặt \(x+\frac{3}{x}-3=a\) ta được:

\(\frac{3}{a+2}-\frac{2}{a}=-1\)

\(\Leftrightarrow3a-2\left(a+2\right)=-a\left(a+2\right)\)

\(\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{3}{x}-3=1\\x+\frac{3}{x}-3=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-4x+3=0\\x^2+x+3=0\end{matrix}\right.\)

b/ Đặt \(x^2+2x+\frac{5}{2}=a>0\)

Phương trình trở thành:

\(\frac{1}{\left(a-\frac{1}{2}\right)^2}+\frac{1}{\left(a+\frac{1}{2}\right)^2}=\frac{5}{4}\)

\(\Leftrightarrow4\left(a+\frac{1}{2}\right)^2+4\left(a-\frac{1}{2}\right)^2=5\left(a^2-\frac{1}{4}\right)^2\)

\(\Leftrightarrow8a^2+2=5\left(a^4-\frac{1}{2}a^2+\frac{1}{16}\right)\)

\(\Leftrightarrow5a^4-\frac{21}{2}a^2-\frac{27}{16}=0\Rightarrow\left[{}\begin{matrix}a^2=\frac{9}{4}\\a^2=-\frac{3}{20}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+2x+\frac{5}{2}=\frac{3}{2}\\x^2+2x+\frac{5}{2}=-\frac{3}{2}\end{matrix}\right.\)

c/ ĐKXĐ: \(x\ne\pm1\)

\(\Leftrightarrow\left(\frac{x}{x+1}\right)^2+\left(\frac{x}{x-1}\right)^2+\frac{2x^2}{x^2-1}-\frac{2x^2}{x^2-1}-\frac{10}{9}=0\)

\(\Leftrightarrow\left(\frac{x}{x+1}+\frac{x}{x-1}\right)^2-\frac{2x^2}{x^2-1}-\frac{10}{9}=0\)

\(\Leftrightarrow\left(\frac{2x^2}{x^2-1}\right)^2-\frac{2x^2}{x^2-1}-\frac{10}{9}=0\)

Đặt \(\frac{2x^2}{x^2-1}=a\)

\(\Rightarrow a^2-a-\frac{10}{9}=0\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{5}{3}\\a=-\frac{2}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{2x^2}{x^2-1}=\frac{5}{3}\\\frac{2x^2}{x^2-1}=-\frac{2}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=-5\left(l\right)\\x^2=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow x=\pm\frac{1}{2}\)

d/ĐKXĐ: ...

\(\Leftrightarrow\left(x^2+\frac{36}{x^2}\right)-13\left(x-\frac{6}{x}\right)=0\)

Đặt \(x-\frac{6}{x}=a\Rightarrow x+\frac{36}{x^2}=a^2+12\)

\(\Rightarrow a^2-13a+12=0\Rightarrow\left[{}\begin{matrix}a=1\\a=12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-\frac{6}{x}=1\\x-\frac{6}{x}=12\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-x-6=0\\x^2-12x-6=0\end{matrix}\right.\)

a, 2(4x - 7 ) = 3(x + 1) + 18

⇌ 8x -14 = 3x + 3 + 18

⇌ 5x = 35 ⇌ x = 7

→ S = \(\left\{7\right\}\)

b, ( 2x - 1 )² - 4x ( x - 3 ) = -11

⇌ 4x² - 2x + 1 - 4x² + 12 = -11

⇌ 10x = -12

⇌ x = \(-\frac{12}{10}\)

→ S = \(\left\{-\frac{12}{10}\right\}\)

c, ( 2x - 5 )² - ( x + 2 )² = 0

⇌ ( 2x - 5 -x + 2 )² = 0

⇌ ( x - 3 )² = 0

⇌ x - 3 = 0 ⇌ x = 3

→ S = \(\left\{3\right\}\)

d, ( x - 6 ) ( x + 1 ) = 2(x + 1)

⇌ ( x - 6 - 2 ) ( x+ 1) = 0

⇌ x² - 7x - 8 =0

⇌ ( x - 8 ) ( x + 1 ) = 0

⇒\(\left\{{}\begin{matrix}x-8=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-1\end{matrix}\right.\)

→ S = \(\left\{8;-1\right\}\)

e, \(\frac{x-3}{2}=2-\frac{1-2x}{5}\)

⇌ 5( x - 3) = 20 - 2(1 - 2x)

⇌ 5x - 4x = 15 + 20 + 2

⇌ x = 37

→ S = \(\left\{37\right\}\)

g, \(\frac{3x+2}{2}+\frac{5-2x}{3}=\frac{11}{6}\)

⇌ 3(3x + 2) + 2(5 - 2x) = 11

⇌ 6x + 6 + 10 - 4x = 11

⇌ 2x = -5

⇌ x = \(-\frac{5}{2}\)

→ S = \(\left\{-\frac{5}{2}\right\}\)

h, \(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{9x-66}{x^2-4}\)

⇌ (x - 2)² - 3(x - 2) = 9x - 66

⇌ x² - 4x + 4 - 3x - 6 = 9x - 66

⇌ x² -16 + 64 = 0

⇌ (x - 8)² = 0

⇌ x - 8 = 0

⇌ x = 8

→ S = \(\left\{8\right\}\)