Bài 2: 

a: Ta có: \(x\left(2x-1\right)-2x+1=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)

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

Ta có: M=A+B

\(=\dfrac{x-\sqrt[3]{x}}{x-1}+\dfrac{1}{\sqrt[3]{x}-1}+\dfrac{1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\)

\(=\dfrac{x-\sqrt[3]{x}}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}+\dfrac{\sqrt[3]{x^2}+\sqrt[3]{x}+1+\sqrt[3]{x}-1}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}\)

\(=\dfrac{x+\sqrt[3]{x}+\sqrt[3]{x^2}}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}\)

\(=\dfrac{\sqrt[3]{x}}{\sqrt[3]{x}-1}\)

Câu 2:

a) \(A=\left(x+5\right)\left(2x-3\right)-2x\left(x+3\right)-\left(x-15\right)\)

\(=x\left(2x-3\right)+5\left(2x-3\right)-2x^2-6x-x+15\)

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

\(=0\)

b) \(B=2\left(x-5\right)\left(x+1\right)+\left(x+3\right)-\left(x-15\right)\)

\(=2\left[x\left(x+1\right)-5\left(x+1\right)\right]+x+3-x+15\)

\(=2.\left[\left(x^2+x\right)-\left(5x+5\right)\right]+x+3-x+15\)

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

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

\(=2x^2-8x+8\)

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

Thay: \(x=\frac{3}{4}\) vào B ta đc:

\(2.\frac{3}{4}\left(\frac{3}{4}-4\right)+8\)

\(=\frac{3}{2}.\frac{-13}{4}+8\)

\(=\frac{25}{8}\)

c) \(C=5x^2\left(3x-2\right)-\left(4x+7\right)\left(6x^2-x\right)-\left(7x-9x^3\right)\)

\(=5x^23x-5x^22-\left[4x\left(6x^2-x\right)+7\left(6x^2-x\right)\right]-7x+9x^3\)

\(=15x^3-10x^2-\left[4x6x^2-4x^2+42x^2-7x\right]-7x+9x^3\)

\(=15x^3-10x^2-24x^3+4x^2-42x^2+7x-7x+9x^3\)

\(=-48x^2\)

P/s: Ko chắc!

\(A=x^2+4x-21-x^2-4x+5=-16\\ B=-2\left(4x^2+20x+25\right)-\left(1-16x^2\right)\\ B=-8x^2-40x-50-1+16x^2=8x^2-40x-51\\ C=x^2\left(x^2-16\right)-\left(x^4-1\right)=x^4-16x^2-x^4+1=1-16x^2\\ D=x^3+1-\left(x^3-1\right)=2\\ E=x^3-3x^2+3x-1-x^3+1-9x^2+1=-12x^2+3x+1\)

a: \(=2x\left(4x^2-4x+1\right)-3x^2-9x-4x^2-4x\)

\(=8x^3-8x^2+2x-7x^2-13x\)

\(=8x^3-15x^2-11x\)

c: \(=5x^3-5x^2-5x^3+5x^2-15=-15\)

d: \(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)

\(=x^2+10x+25-16x^3-48x^2-36x-\left(2x-1\right)\left(x^2-9\right)\)

\(=-16x^3-47x^2-26x+25-2x^3+18x+x^2-9\)

\(=-18x^3-46x^2-8x+16\)

a)

$(2x+1)^2-(2x+1)(2x-1)=(2x+1)[(2x+1)-(2x-1)]$

$=2(2x+1)$

b)

$(4x+3)(x-1)-2x(2x+1)=4x^2-x-3-4x^2-2x=-3x-3=-3(x+1)$

c)

$(2x+3)^2-(4x+1)(x+5)=(4x^2+12x+9)-(4x^2+21x+5)$

$=-9x+4$

d)

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

$=6x^2+12x+9$

e)

$(x+2)(x^2-2x+1)-(x+3)(x-3)=(x^3-3x+2)-(x^2-9)$

$=x^3-x^2-3x+11$

f)

$(x+3)(x^2-3x+9)-(x^2+2x+4)(x-2)$

$=x^3+3^3-(x^3-2^3)=3^3+2^3=35$