Bài 2:

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

\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1\)

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

b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

\(=27x-55\)

a) \(\Rightarrow9x^2+24x+16-9x^2+1=49\)

\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)

b) \(\Rightarrow x^2-13x+22=0\)

\(\Rightarrow\left(x-11\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=11\\x=2\end{matrix}\right.\)

c) \(\Rightarrow x^2-3x-10=0\)

\(\Rightarrow\left(x-5\right)\left(x+2\right)=0\)

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

\(a.2x\left(x-1\right)-3\left(x^2+4x\right)+x\left(x+2\right)\) 

\(=2x^2-2x-3x^2-12x+x^2+2x\) 

\(=-12x\) 

\(b.\left(2x-3\right)\left(3x+5\right)-\left(x-1\right)\left(6x+2\right)+3-5x\) 

\(=6x+10x-9x^2-15-6x^2-2x-6x-2+3-5x\) 

\(=-15x^2+3x-14\) 

\(c.\left(x-y\right)\left(x^2+xy+y^2\right)-\left(x+y\right)\left(x^2-y^2\right)\) 

\(=x^3-y^3-x^3+y^3+x^2y-y^3\)

\(=y^3+x^2y\) 

c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)

\(\Leftrightarrow3x^2+26x=0\)

\(\Leftrightarrow x\left(3x+26\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)

\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)

\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)

\(\left(2x+1\right)2-4\left(x+2\right)2=9\)

\(4x+2-8x-16=9\)

\(4x-8x=9+16-2\)

\(-4x=23\)

\(x=-\frac{23}{4}\)

a, \(\left(2x+1\right)2-4\left(x+2\right)2=9\)

\(\Leftrightarrow4x+2-8x-16=0\Leftrightarrow-4x-14=0\Leftrightarrow x=-\frac{7}{2}\)

b, \(\left(x+1\right)3-2x\left(x+3\right)=2\)

\(\Leftrightarrow3x+3-2x^2-6x=2\Leftrightarrow-3x+1-2x^2=0\)

2:

a: \(9x^2-1=\left(3x\right)^2-1=\left(3x-1\right)\left(3x+1\right)\)

b: \(2\left(x-1\right)+x^2-x\)

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

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

c: \(3x^2+14x-5\)

\(=3x^2+15x-x-5\)

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

3: 

a: \(2x\left(x-1\right)-2x^2=4\)

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

=>-2x=4

=>x=-2

b: \(x\left(x-3\right)-\left(x+2\right)\left(x-1\right)=5\)

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

=>\(x^2-3x-x^2-x+2=5\)

=>-4x=3

=>x=-3/4

c: \(4x^2-25+\left(2x+5\right)^2=0\)

=>\(\left(2x-5\right)\left(2x+5\right)+\left(2x+5\right)^2=0\)

=>\(\left(2x+5\right)\left(2x-5+2x+5\right)=0\)

=>4x(2x+5)=0

=>\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)

bruh

\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)

\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)

\(P=x\left(2x+1\right)-x^2\left(x+2\right)+x^3-x+3\\ P=2x^2+x-x^3-2x^2+x^3-x+3\\ P=3\left(đfcm\right)\)

\(a,\Rightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=54\\ \Rightarrow26x=26\Rightarrow x=1\\ b,\Rightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\\ \Rightarrow39x=-39\Rightarrow x=-1\)