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

 x²-2x+3

=x²-x-x+1+2

=x.(x-1)-(x-1)+2

=(x-1)(x-1)+2

Để x²-2x+3 chia hết cho x-1 thì:

(x-1)(x-1)+2 chia hết cho x-1

=>2 chia hết cho x-1

=>x-1 thuộc Ư(2)={1;-1;2;-2}

Ta có bàng sau:

Vậy x={2;0;3;-1}

Ta có : 3x - 7/3 - 2x - 1/2 = 7 .

=>       x ( 3 - 2 ) - ( 7/3 + 1/2 ) = 7 .

=>       x - ( 14/6 + 3/6 ) = 7 .

=>            x - 17/6         = 7 .

=>                x                = 7 + 17/6 .

=>                x                 = 59/6 .

vậy x = 59/6 .

\(3x-\frac{7}{3}-2x-\frac{1}{2}=7\)

\(\Leftrightarrow\left(3x-2x\right)-\left(\frac{7}{3}+\frac{1}{2}\right)=7\)

\(\Leftrightarrow x-\frac{17}{6}=7\)

\(\Leftrightarrow x=7+\frac{17}{6}\)

\(\Leftrightarrow x=\frac{59}{6}\)

\(\frac{3x-7}{3}\)  -    \(\frac{2x-1}{2}\) \(=\) \(7\)

\(\frac{3x-7}{3}\) = \(7\) +  \(\frac{2x-1}{2}\)

\(\frac{3x-7}{3}\)  = \(\frac{14}{2}\) +   \(\frac{2x-1}{2}\)

\(\frac{3x-7}{3}\)  =   \(\frac{2x+13}{2}\)

<=> \(\left(3x-7\right)\)  x  \(2\) =    \(\left(2x+13\right)\)x   \(3\)

<=> \(6x-14=6x+39\) <=> 6x - 6x = 14 + 39

                                                       <=> 0x = 53 => x ko tốn tại

vậy x ko tồn tại