Bài 1.

a)

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

b)

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

Bài 2.

a)

\((14x^3+12x^2-14x):2x=(x+2)(3x-4)\\\Leftrightarrow 14x^3:2x+12x^2:2x-14x:2x=3x^2-4x+6x-8\\ \Leftrightarrow 7x^2+6x-7=3x^2+2x-8\\\Leftrightarrow (7x^2-3x^2)+(6x-2x)+(-7+8)=0\\\Leftrightarrow 4x^2+4x+1=0\\\Leftrightarrow (2x)^2+2\cdot 2x\cdot 1+1^2=0\\\Leftrightarrow (2x+1)^2=0\\\Leftrightarrow 2x+1=0\\\Leftrightarrow 2x=-1\\\Leftrightarrow x=\frac{-1}2\)

b)

\((4x-5)(6x+1)-(8x+3)(3x-4)=15\\\Leftrightarrow 24x^2+4x-30x-5-(24x^2-32x+9x-12)=15\\\Leftrightarrow 24x^2-26x-5-(24x^2-23x-12)=15\\\Leftrightarrow 24x^2-26x-5-24x^2+23x+12=15\\\Leftrightarrow -3x+7=15\\\Leftrightarrow -3x=8\\\Leftrightarrow x=\frac{-8}3\\Toru\)

\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)

\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)

Bài 1 :

\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)=2.50=100\)

\(B=27^2-25^2=\left(27-25\right)\left(27+25\right)=2.52=104\)

Vì \(100< 104\Rightarrow A< B\)

Bài 2 :

\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)

\(\Rightarrow4\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)

\(\Rightarrow4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)

\(\Rightarrow4x=-2\)\(\Leftrightarrow x=-\frac{1}{2}\)

a: \(x^3-4x^2-x+4=0\)

=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)

=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

=>\(\left(x+1\right)^3=0\)

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

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

=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)

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

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

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

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)