Câu 2: pt đã cho \(\Leftrightarrow x^3-3x^2+3x-1+x^3+x^3+3x^2+3x+1=x^3+6x^2+12x+8\)

\(\Leftrightarrow2x^3-6x^2-6x-8=0\)

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

\(\Leftrightarrow\left(x-1\right)^3-6\left(x-1\right)-9=0\) (*)

Đặt  \(x-1=t\) thì (*) trở thành \(t^3-6t-9=0\) 

\(\Leftrightarrow t^3-9t+3t-9=0\)

\(\Leftrightarrow t\left(t^2-9\right)+3\left(t-3\right)=0\)

\(\Leftrightarrow\left(t-3\right)\left(t^2+3t\right)+3\left(t-3\right)=0\)

\(\Leftrightarrow\left(t-3\right)\left(t^3+3t+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=3\\t^2+3t+3=0\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow x-1=3\) 

\(\Leftrightarrow x=4\)

Vậy pt đã cho có nghiệm \(x=4\)

bài đấy thì em làm được rồi á. Chỉ là em đăng lên xem còn cách nào giải hay hơn thôi ạ...

b: 

ĐKXĐ: \(x\notin\left\{0;2;-2\right\}\)

\(\left(\dfrac{4}{x^3-4x}+\dfrac{1}{x+2}\right):\left(\dfrac{x-2}{x^2+2x}-\dfrac{x}{2x+4}\right)\)

\(=\left(\dfrac{4}{x\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right):\left(\dfrac{x-2}{x\left(x+2\right)}-\dfrac{x}{2\left(x+2\right)}\right)\)

\(=\dfrac{4+x\left(x-2\right)}{x\left(x-2\right)\cdot\left(x+2\right)}:\dfrac{2\left(x-2\right)-x^2}{x\left(x+2\right)\cdot2}\)

\(=\dfrac{x^2-2x+4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{2x\left(x+2\right)}{-\left(x^2-2x+4\right)}\)

\(=\dfrac{-2}{x-2}\)

c:ĐKXĐ: x<>0

\(\left(x-\dfrac{3}{x}\right):\left(\dfrac{x^2+2x+1}{x}-\dfrac{2x+4}{x}\right)\)

\(=\dfrac{x^2-3}{x}:\dfrac{x^2+2x+1-2x-4}{x}\)

\(=\dfrac{x^2-3}{x}\cdot\dfrac{x}{x^2-3}\)

=1

Bài làm:

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

\(\Leftrightarrow\left(\frac{x}{3}+\frac{x}{4}\right)=2+\frac{1}{3}+\frac{3}{4}\)

\(\Leftrightarrow\frac{7}{12}x=\frac{37}{12}\)

\(\Leftrightarrow x=\frac{37}{12}\div\frac{7}{12}\)

\(\Rightarrow x=\frac{37}{7}\)

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

\(\Leftrightarrow\frac{4\left(x-1\right)}{12}+\frac{3\left(x-3\right)}{12}=\frac{24}{12}\)

\(\Leftrightarrow4\left(x-1\right)+3\left(x-3\right)=24\)

\(\Leftrightarrow4x-4+3x-9=24\)

\(\Leftrightarrow7x-13=24\)

\(\Leftrightarrow7x=37\)

\(\Leftrightarrow x=\frac{37}{7}\)

\(\frac{x-1}{3}+\frac{x-3}{4}=2\Leftrightarrow\frac{4\left(x-1\right)}{12}+\frac{3\left(x-3\right)}{12}=\frac{24}{12}\)

Khử mẫu : \(4x-4+3x-9=24\)

\(\Leftrightarrow7x-13=24\Leftrightarrow7x=37\)

\(\Leftrightarrow x=\frac{37}{7}\)Vậy pt có nghiệm là x = 37/7 

a) Điều kiện: \(x\ne\pm1\)

 \(B=\frac{x-1}{x+1}-\frac{x+1}{x-1}-\frac{4}{1-x^2}\)

\(B=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}-\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{-4}{\left(x-1\right).\left(x+1\right)}\)

\(B=\frac{x^2-x-x+1-x^2-x-x-1+4}{\left(x-1\right).\left(x+1\right)}\)

\(B=\frac{-4x+4}{\left(x-1\right).\left(x+1\right)}=\frac{-4.\left(x-1\right)}{\left(x-1\right).\left(x+1\right)}=\frac{-4}{x+1}\)

b) \(x^2-x=0\Leftrightarrow x.\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Khi  \(x=0\Leftrightarrow\frac{-4}{0-1}=\frac{-4}{-1}=4\)

Khi \(x=1\Leftrightarrow\frac{-4}{1-1}=0\)

c) \(\frac{-4}{x+1}=-3\Leftrightarrow-3.\left(x+1\right)=-4\Leftrightarrow x+1=\frac{4}{3}\Leftrightarrow x=\frac{1}{3}\)

\(x-5=\frac{1}{3\left(x+2\right)}\left(đkxđ:x\ne-2\right)\)

\(< =>3\left(x-5\right)\left(x+2\right)=1\)

\(< =>3\left(x^2-3x-10\right)=1\)

\(< =>x^2-3x-10=\frac{1}{3}\)

\(< =>x^2-3x-\frac{31}{3}=0\)

giải pt bậc 2 dễ r

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

\(< =>\frac{4x+3x}{12}=\frac{6x-5x}{30}\)

\(< =>\frac{7x}{12}=\frac{x}{30}< =>12x=210x\)

\(< =>x\left(210-12\right)=0< =>x=0\)

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

\(< =>6x^2-9x-9-6x^2=0\)

\(< =>-9x-9=0< =>9x+9=0\)

\(< =>x=-\frac{9}{9}=-1\)

b) 3x = 2y

=>  x/2 = y/3      (1)

7y = 5z

=> y/5 = z/7       (2)

Từ (1) và (2), có:

     \(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{7}\Rightarrow\)\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)

áp dụng tính chất của dãy tỉ số bằng nhau, có:

     \(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)

x/10 = 2            => x = 2 x 10 =20

y/15 = 2            => y = 2 x 15 = 30

z/21 = 2            => z = 2 x 21 = 42