a: =>9x^2+12x+4-9x^2+12x-4=5x+38

=>24x=5x+38

=>19x=38

=>x=2

e: =>x^3+1-2x=x^3-x

=>-2x+1=-x

=>-x=-1

=>x=1

f: =>x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1

=>12x-9=3x+1

=>9x=10

=>x=10/9

b: \(\Leftrightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)

=>-3x+3=3x-9

=>-6x=-12

=>x=2

   R(x) =           2x² + 3x - 1

^-  M(x) =   -x³ + x² 

                x³ + x² + 3x - 1

Vậy R(x) - M(x) = x³ + x² + 3x - 1

Bài 1:

a: ĐKXĐ: x∉{2;3;-1;-2}

b: \(B=\left(\frac{x-2}{x^2-5x+6}-\frac{x+3}{2-x}-\frac{x+2}{x-3}\right):\left(2-\frac{x}{x+1}\right)\)

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

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

\(=\frac{x-2+x^2-9-x^2+4}{\left(x-2\right)\left(x-3\right)}\cdot\frac{x+1}{x+2}=\frac{\left(x-7\right)\left(x+1\right)}{\left(x-2\right)\left(x-3\right)\left(x+2\right)}\)

c: B=0

=>(x-7)(x+1)=0

=>x=7(nhận) hoặc x=-1(loại)

Bài 2:

a: ĐKXĐ: x∉{1;-1}

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

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

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

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

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

c: \(C=\frac25\)

=>\(\frac{\left(x-1\right)\left(x+1\right)}{2x^2+x+1}=\frac25\)

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

=>\(5x^2-5=4x^2+2x+2\)

=>\(x^2-2x-7=0\)

=>\(x^2-2x+1-8=0\)

=>\(\left(x-1\right)^2=8\)

=>\(\left[\begin{array}{l}x-1=2\sqrt2\\ x-1=-2\sqrt2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\sqrt2+1\left(nhận\right)\\ x=-2\sqrt2+1\left(nhận\right)\end{array}\right.\)

Bài dài quá, lần sau chia nhỏ câu hỏi nhé!!!!!

đúng vậy

a: Ta có: \(\left(x+1\right)^3-\left(x+2\right)\left(x-1\right)^2-3\left(x-3\right)\left(x+3\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x+2\right)\left(x^2-2x+1\right)-3\left(x^2-9\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-2x^2+x+2x^2-4x+2\right)-3\left(x^2-9\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x-2-3x^2+9=5\)

\(\Leftrightarrow6x=-3\)

hay \(x=-\dfrac{1}{2}\)

b: Ta có: \(\left(x+1\right)^3+\left(x-1\right)^3=\left(x+2\right)^3+\left(x-2\right)^3\)

\(\Leftrightarrow x^3+3x^2+3x+1+x^3-3x^2+3x-1=x^3+6x^2+12x+8+x^3-6x^2+12x-8\)

\(\Leftrightarrow2x^3+6x=2x^3+24x\)

\(\Leftrightarrow x=0\)

c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-1=-10\)

\(\Leftrightarrow12x=-11\)

hay \(x=-\dfrac{11}{12}\)