a) Ta có: \(\left(x-2\right).\left(x^2+2x+4\right)+\left(x-2\right)^3-\left(x-2\right).\left(x+2\right)\) 

\(=\left(x^3-8\right)+\left(x-2\right)^3-\left(x^2-4\right)\)  

\(=x^3-8+x^3-6x^2+12x-8-x^2+4\) 

\(=2x^3-7x^2+12x-12\) 

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

\(=9-12x+4x^2-x^2-6x-9-4x^2+1\)  

\(=3x^2-18x+1\)

\(=2x^3-7x^2+12x-12\)\(a.\left(x-2\right).\left(x^2+2x+4\right)+\left(x-2\right)^3-\left(x-2.\left(x+2\right)\right)\)

\(=\left(x^3-8\right)+\left(x-2\right)^3-\left(x^2-4\right)\)

~còn nữa~

\(b.=9-12x+4x^2-x^2-6x-9-4x+1=3x^2-18x+1\)

~Study well~ :)

1) (x + 1)² + (x - 1)(x² + x + 1) + (x - 1)³

= x² + 2x + 1 + x³ - 1 + x³ - 3x² + 3x - 1

= 2x³ - 2x² + 5x + 1

2) (x - 2)² + (2x + 1)² + (x + 1)³

= x² - 4x + 4 + 4x² + 4x + 1 + x³ + 3x² + 3x + 1

= x³ + 8x² + 3x + 6

3) (x + 1)(x² - x + 1) - (x - 3)²

= x³ + 1 - x² + 6x - 9

= x³ - x² + 6x - 8 

4) (3x + 2)² + (2x - 1)² - (x + 3)²

= 9x² + 12x + 4 + 4x² - 4x + 1 - x² - 6x - 9

= 12x² + 2x - 4

a. \(\frac{3}{4}x-\frac{4}{5}.x=\frac{-2}{3}\)

\(\left(\frac{3}{4}-\frac{4}{5}\right)\) \(.x\) = \(\frac{-2}{3}\)

\(\frac{-1}{20}.x=\frac{-2}{3}\)

\(x=\frac{-2}{3}:\frac{-1}{20}\)

x =\(\frac{-40}{3}\)

a) \(\frac{3}{4}.x\) - \(\frac{4}{5}.x\) = \(\frac{-2}{3}\)

=> \(x.\left(\frac{3}{4}-\frac{4}{5}\right)\) = \(\frac{-2}{3}\)

=> \(x.\frac{-1}{20}\) = \(\frac{-2}{3}\)

=> \(x=\frac{-2}{3}:\frac{-1}{20}\)

=> \(x=\frac{40}{3}\)

1) đặt 2x+1 = a => \(a^4-3a^2+2=\left(a^2-1\right)\left(a^2-2\right)=\)\(\left(a-1\right)\left(a+1\right)\left(a-\sqrt{2}\right)\left(a+\sqrt{2}\right)\)

=(2x+1-1)(2x+1+1)(2x+1-\(\sqrt{2}\))(2x+1+\(\sqrt{2}\)) = 4x(x+1)(2x+1-\(\sqrt{2}\))(2x+1+\(\sqrt{2}\))

2) =(x²-x)(x²-x-2)-3

đặt x²-x = b => b(b-2)-3 = b²-2b-3 = (b+1)(b-3) = (x²-x+1)(x²-x-3)

3) đặt x²+2x-1 = c => c²-3xc+2x² = (c-x)(c-2x) = (x²+2x-1-x)(x²+2x-1-2x) = (x²+x-1)(x²-1) = (x²+x-1)(x-1)(x+1)

tìm x

x³-8 +(x-2)(x+1)=0 <=> (x-2)(x²+2x+4)+(x-2)(x+1)=0 <=>(x-2)(x²+2x+4+x+1)=0 <=> x=2 (vì x²+3x+5= (x+\(\frac{3}{2}\))² +\(\frac{11}{4}\)>0)

vậy x=2 

2) \(x\left(x-1\right)\left(x+1\right)\left(x-2\right)-3\)

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

Đặt \(x^2-x=t\)

\(\Rightarrow\left(1\right)=t\left(t-2\right)-3=t^2-2t+1-4\)

\(=\left(t-1\right)^2-4\)

\(=\left(t+3\right)\left(t-5\right)\)

Thay \(x^2-x=t\), ta được:

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

\(x^3-8+\left(x-2\right)\left(x+1\right)=0\)

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

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

\(TH1:x-2=0\Leftrightarrow x=2\)

\(TH2:x^2-x+5=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{19}{4}=0\)

Mà \(\left(x-\frac{1}{2}\right)^2+\frac{19}{4}>0\)nên loại TH2

Vậy x = 2

a) \(\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\)

\(=x\left(x^2-5x+1\right)-2\left(x^2-5x+1\right)-x\left(x^2+11\right)\)

\(=x^3-5x^2+x-2x^2+10x-2-x^3-11x\)

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

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

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

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

\(=2x^3-3\)

c) \(\left(x-y\right)\left(x+y\right)-2x\left(x-y\right)\)

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

\(=x^2+xy-yx-y^2-2x^2+2xy\)

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

a, \(\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\)

\(=x^3-7x^2+11x-2-x^3-11x=-7x^2-2\)

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

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

c, \(\left(x-y\right)\left(x+y\right)-2x\left(x-y\right)\)

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

Áp dụng CT căn phức tạp : \(\sqrt{A\pm\sqrt{B}}=\sqrt{\frac{A+\sqrt{A^2-B}}{2}}\pm\sqrt{\frac{A-\sqrt{A^2-B}}{2}}\)

ĐKXĐ : \(-1\le x\le1\)

Áp dụng CT căn phức tạp , ta được : \(\sqrt{1+\sqrt{1-x^2}}=\sqrt{\frac{1+\sqrt{1-1+x^2}}{2}}+\sqrt{\frac{1-\sqrt{1-1+x^2}}{2}}\)

\(=\sqrt{\frac{1+\left|x\right|}{2}}+\sqrt{\frac{1-\left|x\right|}{2}}=\hept{\begin{cases}\frac{1}{\sqrt{2}}\left(\sqrt{1+x}+\sqrt{1-x}\right)\text{ nếu x }\ge0\\\frac{1}{\sqrt{2}}\left(\sqrt{1-x}+\sqrt{1+x}\right)\text{ nếu x }< 0\end{cases}}\)( kết quả như nhau )

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

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

\(\Rightarrow M=\frac{1}{\sqrt{2}}.\frac{\left(\sqrt{1+x}+\sqrt{1-x}\right)\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(2+\sqrt{1-x^2}\right)}{2+\sqrt{1-x^2}}\)

\(=\frac{1}{\sqrt{2}}.\left[\left(1+x\right)-\left(1-x\right)\right]=x\sqrt{2}\)