Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:

\(\frac{1}{x+1+\frac{1}{x}}+\frac{2}{x+2+\frac{1}{x}}=\frac{8}{15}\)

Đặt \(x+1+\frac{1}{x}=a\)

\(\frac{1}{a}+\frac{2}{a+1}=\frac{8}{15}\)

\(\Leftrightarrow a+1+2a=\frac{8}{15}a\left(a+1\right)\)

\(\Leftrightarrow8a^2-37a-15=0\Rightarrow\left[{}\begin{matrix}a=5\\a=-\frac{3}{8}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+1+\frac{1}{x}=5\\x+1+\frac{1}{x}=-\frac{3}{8}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+1=0\\x^2+\frac{11}{8}x+1=0\end{matrix}\right.\)

Câu 6 :

a, Ta có : \(x+\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\)

=> \(\frac{15x}{15}+\frac{5\left(2x+\frac{x-1}{5}\right)}{15}=\frac{15}{15}-\frac{3\left(3x-\frac{1-2x}{3}\right)}{15}\)

=> \(15x+5\left(2x+\frac{x-1}{5}\right)=15-3\left(3x-\frac{1-2x}{3}\right)\)

=> \(15x+10x+\frac{5\left(x-1\right)}{5}=15-9x+\frac{3\left(1-2x\right)}{3}\)

=> \(15x+10x+x-1=15-9x+1-2x\)

=> \(15x+10x+x-1-15+9x-1+2x=0\)

=> \(37x-17=0\)

=> \(x=\frac{17}{37}\)

Vậy phương trình trên có nghiệm là \(S=\left\{\frac{17}{37}\right\}\)

Bài 7 :

a, Ta có : \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)

=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)

=> \(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)

=> \(x-23=0\)

=> \(x=23\)

Vậy phương trình trên có nghiệm là \(S=\left\{23\right\}\)

c, Ta có : \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)

=> \(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)

=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)

=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\)

=> \(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)

=> \(x+2005=0\)

=> \(x=-2005\)

Vậy phương trình trên có nghiệm là \(S=\left\{-2005\right\}\)

e, Ta có : \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)

=> \(\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)

=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)

=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)

=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)

=> \(x-100=0\)

Vậy phương trình trên có nghiệm là \(S=\left\{100\right\}\)

Theo bài ra , ta có :

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{\left(x-3\right)\left(x+3\right)}\)

ĐKXĐ : \(x\ne3,x\ne-3,x\ne-\frac{7}{2}\)

Quy đồng và khử mẫu phương trình ta đk :

\(13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow\left(x+3\right)\left(13+x-3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow\left(x+3\right)\left(x+10\right)=12x+42\)

\(\Leftrightarrow x^2+13x+30=12x+42\)

\(\Leftrightarrow x^2+13x-12x+30-42=0\)

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

\(\Leftrightarrow x^2-3x+4x-12=0\)

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

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

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

Kết hợp với ĐKXĐ ta có : x = -4

Vậy \(S=\left\{-4\right\}\)

Chúc bạn học tốt =))

ĐKXĐ: x\(\ne\)3;-7/2;-3

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\Leftrightarrow\frac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\frac{\left(x-3\right)\left(x+3\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\frac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)

\(\Leftrightarrow13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow13x+39+x^2-9=12x+42\\ \Leftrightarrow x^2+x=12\)

\(\Leftrightarrow x^2+x-12=0\Leftrightarrow x^2-3x+4x-12=0\\ \Leftrightarrow x\left(x-3\right)+4\left(x-3\right)=0\)

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

Nhận thấy x=3 không thỏa mãn ĐKXĐ nên pt có 1 nghiệm duy nhất là x=-4

Đk: x khác 3; -3 và -7/2

Từ đề suy ra:

13/(x-3)(2x+7) + 1/2x+7 - 6/(x-3)(x+3) = 0

<=> \(\frac{13\left(x+3\right)+\left(x-3\right)\left(x+3\right)-6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}=0\)

<=> 13(x+3) + (x-3)(x+3) + 6(2x+7) = 0

<=> 13x + 39 + x² - 9 + 12x + 42 = 0

<=> x² + 25x + 72 = 0

<=> (x + 25/2)² = 337/4

<=> \(\left[\begin{matrix}x+\frac{25}{2}=\sqrt{\frac{337}{4}}\\x+\frac{25}{2}=-\sqrt{\frac{337}{4}}\end{matrix}\right.\)<=> \(\left[\begin{matrix}x=\frac{-25+\sqrt{337}}{2}\\x=\frac{25-\sqrt{337}}{2}\end{matrix}\right.\)(TM)

Vây ...

ĐKXĐ: \(x>0\)

Ta có:

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

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

\(\frac{\Leftrightarrow1}{2x\sqrt{x}}-\sqrt{x}=2\left(x-\frac{1}{x}+\frac{1}{4x^3}\right)\)

Đặt : \(\frac{1}{2x\sqrt{x}}-\sqrt{x}=a\Rightarrow a^2=x-\frac{1}{x}+\frac{1}{4x^3}\)

Khi đó pt đã cho trở thành:

\(a=2a^2\Leftrightarrow\orbr{\begin{cases}a=0\\a=\frac{1}{2}\end{cases}}\)

+) a = 0\(\Rightarrow x=\frac{1}{\sqrt{2}}\)

Tương tự

\(\frac{2x}{x+1}=\frac{x^2-x+8}{\left(x+1\right)\cdot\left(x-4\right)}\\ \Leftrightarrow\frac{2x^2-8x}{\left(x+1\right)\cdot\left(x-4\right)}-\frac{x^2-x+8}{\left(x+1\right)\cdot\left(x-4\right)}=0\\ \Leftrightarrow\frac{2x^2-8x-x^2+x-8}{\left(x+1\right)\cdot\left(x-4\right)}=0\\ \Leftrightarrow\frac{x^2-7x-8}{\left(x+1\right)\cdot\left(x-4\right)}=0\\ \Leftrightarrow x^2-7x-8=0\\ \Rightarrow\left[{}\begin{matrix}x=\frac{7+\sqrt{17}}{2}\\x=\frac{7-\sqrt{17}}{2}\end{matrix}\right.\)

\(\frac{2x}{x+1}=\frac{x^2-x+8}{\left(x+1\right)\left(x+1\right)}\) (ĐKXĐ: x \(\ne\) 1)

\(\Leftrightarrow\) \(\frac{2x\left(x+1\right)}{\left(x+1\right)\left(x+1\right)}=\frac{x^2-x+8}{\left(x+1\right)\left(x+1\right)}\)

\(\Leftrightarrow\) 2x(x + 1) = x² - x + 8

\(\Leftrightarrow\) 2x² + 2x - x² + x - 8 = 0

\(\Leftrightarrow\) x² + 3x - 8 = 0

\(\Leftrightarrow\) (x + \(\frac{3}{2}\))² - \(\frac{41}{4}\) = 0

\(\Leftrightarrow\) (x + \(\frac{3}{2}\) - \(\frac{\sqrt{41}}{2}\))(x + \(\frac{3}{2}\) + \(\frac{\sqrt{41}}{2}\)) = 0

\(\Leftrightarrow\) (x + \(\frac{3-\sqrt{41}}{2}\))(x + \(\frac{3+\sqrt{41}}{2}\)) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3-\sqrt{41}}{2}=0\\x+\frac{3+\sqrt{41}}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{41}-3}{2}\\x=\frac{-3-\sqrt{41}}{2}\end{matrix}\right.\)

Vậy S = {\(\frac{\sqrt{41}-3}{2}\); \(\frac{-3-\sqrt{41}}{2}\)}

Chúc bn học tốt!!

ĐKXĐ: ...

\(\Leftrightarrow x^3-\frac{1}{x^3}-3\left(x-\frac{1}{x}\right)-1=0\)

Đặt \(x-\frac{1}{x}=a\Rightarrow a^3=x^3-\frac{1}{x^3}-3\left(x-\frac{1}{x}\right)\)

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

Phương trình trở thành:

\(a^3+3a-3a-1=0\Rightarrow a^3=1\Rightarrow a=1\)

\(\Rightarrow x-\frac{1}{x}=1\Rightarrow x^2-x-1=0\)

\(\frac{2}{x^2+1}+\frac{4}{x^2+3}+\frac{6}{x^2+5}=3+\frac{x^2-1}{x^2+6}\)

\(\Leftrightarrow\frac{x^2-1}{x^2+6}+1-\frac{2}{x^2+1}+1-\frac{4}{x^2+3}+1-\frac{6}{x^2+5}=0\)

\(\Leftrightarrow\frac{x^2-1}{x^2+6}+\frac{x^2-1}{x^2+1}+\frac{x^2-1}{x^2+3}+\frac{x^2-1}{x^2+5}=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(\frac{1}{x^2+6}+\frac{1}{x^2+1}+\frac{1}{x^2+3}+\frac{1}{x^2+5}\right)=0\)

\(\Rightarrow x=\pm1\)

a, Ta có : \(\frac{x+1}{2}+\frac{x-2}{4}=1-\frac{2\left(x-1\right)}{3}\)

=> \(\frac{6\left(x+1\right)}{12}+\frac{3\left(x-2\right)}{12}=\frac{12}{12}-\frac{8\left(x-1\right)}{12}\)

=> \(6\left(x+1\right)+3\left(x-2\right)=12-8\left(x-1\right)\)

=> \(6x+6+3x-6=12-8x+8\)

=> \(17x=20\)

=> \(x=\frac{20}{17}\)

b, Ta có : \(\frac{5x-1}{6}+x=\frac{6-x}{4}\)

=> \(\frac{5x-1+6x}{6}=\frac{6-x}{4}\)

=> \(4\left(11x-1\right)=6\left(6-x\right)\)

=> \(44x-4-36+6x=0\)

=> \(\)\(50x=40\)

=> \(x=\frac{4}{5}\)

c, Ta có : \(\frac{5\left(1-2x\right)}{3}+\frac{x}{2}=\frac{3\left(x-5\right)}{4}-2\)

=> \(\frac{20\left(1-2x\right)}{12}+\frac{6x}{12}=\frac{9\left(x-5\right)}{12}-\frac{24}{12}\)

=> \(20\left(1-2x\right)+6x=9\left(x-5\right)-24\)

=> \(20-40x+6x-9x+45+24=0\)

=> \(43x=89\)

=> \(x=\frac{89}{43}\)