\(\Leftrightarrow4x^2+x-8x-2-4x^2-27x=1\)

=>-34x=3

hay x=-3/34

\(\left(x-2\right)\left(4x+1\right)-4x\left(x+7\right)=1\\ \Rightarrow4x^2-8x+x-2-4x^2-28x=1\\ \Rightarrow-35x=3\\ \Rightarrow x=\dfrac{-3}{35}\)

`a)`

`A(x) + B(x) = 2x - 4x^2 + 1 + x^3 - 4x^2 + 5 - 2x`

                  `= x^3 - ( 4x^2 + 4x^2 ) + ( 2x - 2x ) + ( 1+ 5 )`

                  `= x^3 - 8x^2 + 6`

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`b)`

    `P(x) + B(x) = A(x)`

`=>P(x) = A(x) - B(x)`

`=>P(x) = 2x - 4x^2 + 1 + x^3 + 4x^2 - 5 + 2x`

`=>P(x) = x^3 + ( -4x^2 + 4x^2 ) + ( 2x + 2x ) + ( 1 - 5 )`

`=>P(x) = x^3 + 4x - 4`

a: \(\dfrac{x}{6}=\dfrac{8}{3}\)

=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)

b: \(\dfrac{5}{x}=\dfrac{4}{9}\)

=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)

c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)

=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)

=>x=-1-3=-4

d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)

=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)

=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)

=>\(x=-\dfrac{69}{8}\)

f: ĐKXĐ: x<>1

\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)

=>\(\left(x-1\right)^2=3\cdot27=81\)

=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)

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

\(\Rightarrow x+2-2=0\)

\(\Rightarrow x=0\)

\(\left(x+3\right)+1=7\)

\(\Rightarrow x+3+1=7\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=3\)

\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)

\(\Rightarrow3x=12\)

\(\Rightarrow x=4\)

\(\left(5x+4\right)-1=13\)

\(\Rightarrow5x+4-1=13\)

\(\Rightarrow5x+3=13\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

\(\left(4x-8\right)-3=5\)

\(\Rightarrow4x-8-3=5\)

\(\Rightarrow4x-11=5\)

\(\Rightarrow4x=16\)

\(\Rightarrow x=4\)

\(8-\left(2x+4\right)=2\)

\(\Rightarrow8-2x-4=2\)

\(\Rightarrow4-2x=2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

\(7+\left(5x+2\right)=14\)

\(\Rightarrow7+5x+2=14\)

\(\Rightarrow9+5x=14\)

\(\Rightarrow5x=5\)

\(\Rightarrow x=1\)

\(5-\left(3x-11\right)=1\)

\(\Rightarrow5-3x+11=1\)

\(\Rightarrow16-3x=1\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

b/ (x² + x + 1)(x⁶ - x⁴ + x³ - x + 1)

c/ (x - 1)(2x³ + 1)

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

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

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

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

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

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

Ta có : 

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

\(\Leftrightarrow\)\(4x^2-4x-3x^2+15=x-3-x-4\)

\(\Leftrightarrow\)\(x^2-4x+15=-7\)

\(\Leftrightarrow\)\(\left(x^2-2.x.2+2^2\right)+11=-7\)

\(\Leftrightarrow\)\(\left(x-2\right)^2=-18\)

Mà \(\left(x-2\right)^2\ge0\) \(\left(\forall x\inℝ\right)\)

\(\Rightarrow\)\(x\in\left\{\varnothing\right\}\)

Vậy không có giá trị nào của x thoã mãn đề bài 

Chúc bạn học tốt ~ 

a.

\(\left(x^2-x+1\right)\left(x^2-x+2\right)=12\)

Đặt \(x^2-x+1=y\) ta được:

\(y\left(y+1\right)=12\)

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-2=0\\x^2-x+5=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

b.

\(3y^3-7y^2-7y+3=0\)

\(\Leftrightarrow3\left(y^3+1\right)-7y\left(y+1\right)=0\)

\(\Leftrightarrow3\left(y+1\right)\left(y^2-y+1\right)-7y\left(y+1\right)=0\)

\(\Leftrightarrow\left(y+1\right)\left(3y^2-3y+3-7y\right)=0\)

\(\Leftrightarrow\left(y+1\right)\left(3y^2-10y+3\right)=0\)

\(\Leftrightarrow\left(y+1\right)\left(3y-1\right)\left(y-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=-1\\y=\dfrac{1}{3}\\y=3\end{matrix}\right.\)