a ) \(\left(x^2+x\right)\left(x^2+x+1\right)=6\)

\(x^2.\left(x^2+x+1\right)+x.\left(x^2+x+1\right)=6\)

\(x^2.x^2+x^2.x+x^2.x+x.x^2+x.x+x.1=_{ }6\)

\(x^4+x^3+x^3+x^3+x^2+x=6\)

\(x^4+3x^3+x^2+x=6\)

Tới đây .........

Đặt x^2+x=a

Theo bài ra ta có:a.(a+1)=6

\(\Rightarrow\)a=2

thay a=x^2+x ta có

x^2+x=2

\(\Rightarrow\)x.(x+1)=2

\(\Rightarrow\)=1

a)2+4+6+..+2x=210

=>2*1+2*2+...+2*x=210

=>2(1+2+3+...+x)=210

=>2[x(x+1)/2]=210

=>x*(x+1)=210

hay x*(x+1)=14(14+1)

vậy x=14

c)x+(x+1)+(x+2)+...+(x+2010)

=>(x+x+x+...+x)+(1+2+3+...+2010)=2029099

=>2011x           + 2021055         =2029099

     2011x                                     =2029099-2021055

       2011x                                   =8044

         x                                          =8044 /2011

          x                                          =4

k cho mình nhé

=> (x - 1)² = 64 => (x - 1)² = 8² => x - 1 = 8 => x = 9

b/ => x² - x = 0 

=> x(x - 1) = 0 

=> x = 0 hoặc x - 1 = 0 => x = 1

a) 2 . (x - 1)² = 128

(x - 1)² = 64 = + 8

=> x = 9 hoặc x = -7

b) x² = x

<=> x \(\in\) {0; -1; 1}

\(6\left(x+1\right)^2-2\left(x+1\right)^3-2\left(x-1\right)\left(x^2+x+1\right)=1\)

\(\Leftrightarrow6\left(x^2+2x+1\right)-2\left(x^3+3x^2+3x+1\right)-2\left(x^3+x^2+x-x^2-x-1\right)=1\)

\(\Leftrightarrow6x^2+12x+6-2x^3-6x^2-6x-2-2x^3-2x^2-2x+2x^2+2x+2=1\)

\(\Leftrightarrow-4x^3+6x+5=0\)

\(\Leftrightarrow x=1.5233401602\)