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1.a) (2 + 1)(2² + 1)((2⁴ + 1)(2⁸ + 1) = (2² - 1)(2² + 1)(2⁴ + 1)(2⁸ + 1) = (2⁴ - 1)(2⁴ + 1)(2⁸ + 1)

= (2⁸ - 1)(2⁸ + 1) = 2¹⁶ - 1

b) 7(2³ + 1)(2⁶ + 1)(2¹² + 1)(2²⁴ + 1) = (2³ - 1)(2³ + 1)(2⁶ + 1)(2¹² + 1)(2²⁴ + 1)

= (2⁶ - 1)(2⁶ + 1)(2¹² + 1)(2²⁴ + 1) = (2¹² - 1)(2¹² + 1)(2²⁴ + 1) = (2²⁴ - 1)(2²⁴ + 1) = 2⁴⁸ - 1

c) (x² - x + 1)(x² + x + 1)(x² - 1) = [(x² - x + 1)(x + 1)][(x² + x + 1)(x - 1)] = (x³ + 1)(x³ - 1) = x⁶ - 1

2. Đặt A = 4x - x² - 1 = -(x^2 - 4x + 4) + 3 = -(x - 2)² + 3 \(\le\)3 \(\forall\)x

Dấu "=" xảy ra <=> x - 2 = 0 <=> x = 2

Vậy MaxA = 3 khi x = 2

2. 4x - x² - 1

= -x² + 4x - 1

= -x² + 4x - 4 + 3

= -( x² - 4x + 4 ) + 3

= -( x - 2 )² + 3

( x - 2 )² ≥ 0 ∀ x => -( x - 2 )²  ≤ 0 ∀ x 

=> -( x - 2 )² + 3  ≤ 3 ∀ x 

Dấu bằng xảy ra <=> x - 2 = 0 => x = 2

Vậy GTLN của biểu thức = 3 khi x = 2

Bài 1 : 12 - 12 + 11 + 10 - 9 + 8 - 7 + 5 - 4 + 3 + 2 - 1 

= ( 12 - 12 ) + ( 11 - 1 ) + ( 10 - 9 ) + ( 8 - 7 ) + ( 5 - 4 ) + ( 3 + 2 )

= 0 + 10 + 1 + 1 + 1 + 5 

= 18

Bài 2 :

3x + 27 = 9

        3x = 9 - 27

        3x = - 18

          x = - 6

2x + 12 = 3( x - 7 )

2x + 12 = 3x - 21

 3x - 2x = 12 + 21 

         x = 33

2x² - 1 = 49

     2x² = 49 + 1 

     2x² = 50

       x² = 50 : 2 

       x² = 25

=> x = 5 hoặc x = - 5

- | 9 - x | - 5 = 12

     - | 9 - x | = 12 + 5 

     - | 9 - x | = 17

TH1 : 9 - x >= 0 <=> x <= 9

=> - ( 9 - x ) = 17 

=> x = 26 ( loại )

TH2 : 9 - x < 0 <=> x > 9 

=> - ( 9 - x ) = -17 

=> x = - 8 ( loại ) 

=> ko có giá trị nào thõa mãn

Bài 3 a,: A = ( - a - b + c ) - ( - a - b - c )

              = - a - b + c + a + b + c

              = 2c

b, thay c = - 2 vào biểu thức A = 2c

Ta được : A = 2 x ( -2 ) = - 4

\(2\left(x-2\right)\left(x+3\right)-x^2+4=0\)

\(2\left(x^2+3x-2x-6\right)-x^2+4=0\)

\(2x^2+6x-4x-12-x^2+4=0\)

\(x^2+2x-8=0\)

\(x^2+4x-2x-8=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x+4=0\rightarrow x=\left(-4\right)\\x-2=0\rightarrow x=2\end{cases}}\)

3/ 

a/ \(2\left(x+1\right)^2-3\left(x-1\right)^2+\left(x+2\right)\left(5-x\right)\)

\(=2\left(x^2+2x+1\right)-3\left(x^2-2x+1\right)+\left(5x-x^2+10-2x\right)\)

\(=2x^2+4x+2-3x^2+6x-3+5x-x^2+10-2x\)

\(=-2x^2+13x+9\)

b/ \(\left(3x-1\right)^3+\left(3x-1\right)^3-6x^2+9\)

\(=2\left(3x-1\right)^3-6x^2+9\)

\(=2\left(\left(3x\right)^3-3\left(3x\right)^2\cdot1+3\cdot3x\cdot1-1\right)-6x^2+9\)

\(=2\left(27x^3-27x^2+9x-1\right)-6x^2+9\)

\(=54x^3-54x^2+18x-2-6x^2+9\)

\(=54x^3-60x^2+18x+7\)

Số hơi dài, nên dễ tính sai -,- tính mik hay cẩu thả có j sai ibbb ạ

2) 2.(x - 2).(x + 3) - x² + 4 = 0

<=> x² + 2x - 8 = 0

<=> (x - 2).(x + 4) = 0

        x - 2 = 0 hoặc x + 4 = 0

        x = 0 + 2         x = 0 - 4

        x = 2               x = -4

=> x = 2 hoặc x = -4

3) a) 2.(x + 1)² - 3.(x - 1)² + (x + 2).(5 - x)

= 2.(x² + 2x + 1) - 3.(x² - 2x + 1) + (x + 2).(5 - x)

= 2x² + 4x + 2 - 3x² + 6x - 3 + (x + 2).(5 - x)

= 2x² + 4x + 2 - 3x² + 6x - 3 + 3x - x² + 10

= (2x² - 3x² - x²) + (4x + 6x + 3x) + (2 - 3 + 10)

= -2x² + 13x + 9

b) (3x - 1)³ + (3x - 1)³ - 6x² + 9

= 2.(3x - 1)³ - 6x² + 9

= 2.(27x³ - 27x² + 9x - 1) - 6x² + 9

= 54x³ - 54x² + 18x - 2 - 6x² + 9

= 54x³ + (-54x² - 6x²) + 18x + (-2 + 9)

= 54x³ - 60x + 18x + 7

4) a) A = (x - 5).(x + 2) + 3.(x - 2).(x + 2) - (3x - 1)² + 5x²

A = (x - 5).(x + 2) + 3.(x - 2).(x + 3) - (9x² - 6x + 1) + 5x²

A = x² - 3x - 10 + 3x² - 12 - (9x² - 6x + 1) + 5x²

A = x² - 3x - 10 + 3x² - 12 - 9x² + 6x - 1 + 5x²

A = (x² + 3x² - 9x² + 5x²) + (-3x + 6x) + (-10 - 12 - 1)

A = 3x - 23 (1)

b) Thay x = 1/2 vào (1), ta có:

A = 3x - 23 = 3.(1/2) - 23

                   = 3/2 - 23

                   = -43/2

A khi x = 1/2 là -43/2

2A = 2 + 1 + 1/2 + 1/2^2 + ... + 1/2^2011

A = 1 + 1/2 + .. + 1/2^2011 + 1/2^2012

2A  - A = 2 + 1 + 1/2 + .. + 1/2^2011 - 1 - 1/2 - ... - 1/2^2011 - 1/2^2012

A = 2 - 1/2^2012

A = \(\frac{2^{2012}-2}{2^{2012}}\)

A = 1+1/2+1/2^2+1/2^3+.....+1/2^2012

2A= 2. (1+1/2+1/2^2+1/2^3+.....+1/2^2012)

2A= 2 + 1 + 1/2 + 1/2^2 + 1/2^3 + ...+ 1/2^2011

2A - A= (2 + 1 + 1/2 + 1/2^2 + 1/2^3+ ...+ 1/2^2011) - (1+1/2+1/2^2+1/2^3+.....+1/2^2012)

1A= 2 + 1 + 1/2 + 1/2^2 + 1/2^3 + ...+ 1/2^2011 - 1-1/2-1/2^2+1/2^3+.....+1/2^2012

1A= 2 - 1/2^2012

A= 2-1/2^2012

A= 2 - 1/2^2012