A = -x^2 - 6x + 1

= -(x^2 + 6x + 9) + 10

= -(x + 3)^2 + 10 ≤ 10

GTLN của A là 10 khi x = -3.

Ta có: \(A=-x^2-6x+1\)

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

\(=-\left(x^2+6x+9-10\right)\)

\(=-\left(x+3\right)^2+10\le10\forall x\)

Dấu '=' xảy ra khi x+3=0

=>x=-3

x=2

Ta có: \(x^3-3x^2+3x-2=0\)

=>\(x^3-3x^2+3x-1-1=0\)

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

=>x-1=1

=>x=2

Ta có: \(8x^3+36x^2+54x+27=0\)

=>\(\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3=0\)

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

=>2x+3=0

=>2x=-3

=>\(x=-\frac32\)

8x^3 + 36x^2 + 54x + 27 = 0

→ (2x)^3 + 3(2x)^2 * 3 + 3(2x) * 3^2 + 3^3 = 0

→ (2x + 3)^3 = 0

→ 2x + 3 = 0

→ x = -3/2

Vậy x = -3/2.

Ta có: \(x^3-3x^2+3x-1=\left(3x+5\right)^3\)

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

=>3x+5=x-1

=>3x-x=-1-5

=>2x=-6

=>x=-3

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

→ (x - 1)^3 = (3x + 5)^3

→ x - 1 = 3x + 5

→ -2x = 6

→ x = -3

Vậy x = -3.

Ta có: \(x^3-6x^2+12x-8=\left(2x+1\right)^3\)

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

=>2x+1=x-2

=>2x-x=-2-1

=>x=-3

x^3 - 6x^2 + 12x - 8 = (2x + 1)^3

→ (x - 2)^3 = (2x + 1)^3

→ x - 2 = 2x + 1

→ -x = 3

→ x = -3

Vậy x = -3.

\(x^2-2x-120=0\)

\(x^2-2x+1-121=0\)

\(\left(x-1\right)^2-11^2\) =0

\(\left(x-1+11\right)\left(x-1-11\right)=0\)

\(\left(x+10\right)\left(x-12\right)=0\)

\(\left[\begin{array}{l}x+10=0\Rightarrow x=-10\\ x-12=0\Rightarrow x=12\end{array}\right.\)

Vậy x = -10; x = 12

\(x^2-2x-120=0\)

=>\(x^2-12x+10x-120=0\)

=>x(x-12)+10(x-12)=0

=>(x-12)(x+10)=0

=>\(\left[\begin{array}{l}x-12=0\\ x+10=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=12\\ x=-10\end{array}\right.\)

a: \(\left(2x-1\right)^2-x\left(4x-3\right)\)

\(=4x^2-4x+1-4x^2+3x\)

=-x+1

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

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

\(=x^2-3x+2-x^2+6x-9=3x-7\)

c: \(\left(x-2\right)^3-x\left(x-1\right)\left(x-2\right)\)

\(=x^3-6x^2+12x-8-x\left(x^2-3x+2\right)\)

\(=x^3-6x^2+12x-8-x^3+3x^2-2x\)

\(=-3x^2+10x-8\)

d: \(\frac12x\left(x-2\right)-\left(2x-3\right)^2\)

\(=\frac12x^2-x-\left(4x^2-12x+9\right)\)

\(=\frac12x^2-x-4x^2+12x-9=-\frac72x^2+11x-9\)

xét tứ giác `KJVT` có :

\(\hat{K}+\hat{J}+\hat{V}+\hat{T}=360^0\)

`=> 133^0 + 85^0 + x + x = 360^0`

`=> 218^0 + 2x = 360^0`

`=> 2x =360^0 - 218^0`

`=> 2x = 142^0`

`=> x = 142^0 : 2`

`=> x = 71^0`

Vậy `x = 71^0`

Giải:

\(x+x\) + 133\(^0\) + 85\(^0\) = 360\(^0\)

2\(x\) + (133\(^0\) + 85\(^0\)) = 360\(^0\) (tống bốn góc của tứ giác luôn bằng 360\(^0\))

2\(x\) + 218\(^0\) = 360\(^0\)

2\(x\) = 360\(^0\) - 218\(^0\)

2\(x\) = 142\(^0\)

\(x\) = 142\(^0\) : 2

\(x\) = 71\(^0\)

(-7\(x^2\).2y\(^4\).z\(^3\)) + 20\(x^2\).y\(^4\).z\(^3\)

= (-7 + 20)\(x^2y^4z^3\)

= 13\(x^2y^4z^3\)

a; -(3 - 2\(x\)).(\(x+1\))

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

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

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

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

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

b; 2\(x\).(3 - \(x\)) - (\(x+1)^2\)

= 6\(x\) - 2\(x^2\) - (\(x^2\) + 2\(x\) + 1)

= 6\(x\) - 2\(x^2\) - \(x^2\) - 2\(x\) - 1

= (6\(x\) - 2\(x\)) - (2\(x^2\) + \(x^2\)) - 1

= 4\(x\) - 3\(x^2\) - 1

= - 3\(x^2\) + 4\(x\) - 1

c; (\(x+4\))(2 - \(x\)) - (\(x^2\) - 2)

= \(x\).(2 - \(x\)) + 4.(2 - \(x\)) - \(x^2\) + 2

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

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

= - 2\(x\) - 2\(x^2\) + 10

= - \(2x\) - 2\(x^2\) + 10

= - 2\(x^2\) - 2\(x\) + 10

d; - 2.(\(x-1\)).3\(x\) - 3\(x\).(\(x+2\))

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

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

= - 9\(x^2\) + 0

= -9\(x^2\)