a/ (x-1)² - (x+1)² = 3

<=> (x-1-x-1)(x-1+x+1) = 3

<=> - 2 . 2x = 3

<=> -4x = 3 <=> x = -3/4

KL: x = -3/4

b/ (x-2)(x+2) - (x-3) ^2 = 5x - 1

<=> x² - 4 - x² + 6x - 9 = 5x - 1

<=> 6x - 5x = -1 + 4 + 9

<=> x = 12

KL: x= 12

c/ x(x-2) - (x+3)² - x = 4

<=> x² - 2x - x² - 6x - 9 - x = 4

<=> -9x = 4 + 9

<=> x = -13/9

KL: x= -13/9

d/ 6x(x-1) - 5(x-2)² - x(x - 3) = 4

<=> 6x² - 6x - 5(x² - 4x + 4) - x² - 3x = 4

<=> 6x² - 6x - 5x² + 20x - 20 -x² - 3x = 4

<=> 11x = 24

<=> x = 24/11

KL: x = 24/11

2x(x-3) + 2x - 6 = 0

<=> 2x(x-3) + 2(x - 3) = 0

<=> (x-3)(2x+2)=0

<=> \(\left[{}\begin{matrix}x-3=0\\2x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

vậy............

Ta có:

\(3^{12}\equiv1\left(mod10\right)\)

\(\Rightarrow\left(3^{12}\right)^{167}\equiv1^{167}\equiv1\left(mod10\right)\)

=> \(3^{2004}\cdot3^6\equiv1\cdot9\equiv9\left(mod10\right)\)

Vậy chữ số tận cùng của 3²⁰¹⁰ là 9

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