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

⇔ x³ + 6x² + 12x + 8 - ( 4x² + 12x + 9 ) + 4x² - 9 = x³ - 8 - 6x² - 12x

⇔ x³ + 10x² + 12x - 1 - 4x² - 12x - 9 = x³ - 6x² - 12x - 8

⇔ x³ + 6x² - 10 = x³ - 6x² - 12x - 8

⇔ x³ + 6x² - 10 - x³ + 6x² + 12x + 8 = 0

⇔ 12x² + 12x - 2 = 0 

⇔ 2( 6x² + 6x - 1 ) = 0

⇔ 6x² + 6x - 1 = 0 (*)

Δ = b² - 4ac = 6² - 4.6.(-1) = 60

Δ > 0 nên (*) có hai nghiệm phân biệt

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-6+\sqrt{60}}{12}=\frac{-3+\sqrt{15}}{6}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-6-\sqrt{60}}{12}=\frac{-3-\sqrt{15}}{6}\end{cases}}\)

Vậy ...

Sửa đề: Tìm GTNN của \(C=x^2-3x+2017\)

Ta có:

\(C=x^2-3x+2017\)

\(C=\left(x^2-3x+\frac{9}{4}\right)+\frac{3}{4}+2014\)

\(C=\left(x-\frac{3}{2}\right)^2+2014\frac{3}{4}\ge2014\frac{3}{4}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-\frac{3}{2}\right)^2=0\Rightarrow x=\frac{3}{2}\)

Vậy \(Min_C=2014\frac{3}{4}\Leftrightarrow x=\frac{3}{2}\)

\(\frac{x^2+4x+3}{2x+6}\)

\(=\frac{x^2+3x+x+3}{2\left(x+3\right)}\)

\(=\frac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)

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

\(=\frac{x+1}{2}\)

\(\frac{x^2+4x+3}{2x+6}\)

ĐKXĐ : x ≠ -3

\(=\frac{x^2+3x+x+3}{2\left(x+3\right)}\)

\(=\frac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)

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

\(=\frac{x+1}{2}\)

\(\frac{x+y}{\left(x+3\right)^2+\left(y-2\right)^2}\)

ĐKXĐ : \(\left(x+3\right)^2+\left(y-2\right)^2\ne0\)

⇔ \(\hept{\begin{cases}\left(x+3\right)^2\ne0\\\left(y-2\right)^2\ne0\end{cases}}\)

⇔ \(\hept{\begin{cases}x+3\ne0\\y-2\ne0\end{cases}}\)

⇔ \(\hept{\begin{cases}x\ne-3\\y\ne2\end{cases}}\)

Ta có : x² + x + 1

= ( x² + x + 1/4 ) + 3/4

= ( x + 1/2 )² + 3/4 ≥ 3/4 ∀ x

Dấu "=" xảy ra khi x = -1/2

=> GTNN của biểu thức = 3/4 <=> x = -1/2

Đặt \(A=x^2+x+1\)  , ta có :

\(A=x^2+x+1\)

\(=x^2+x+\frac{1}{4}+\frac{3}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(\Rightarrow minA=\frac{3}{4}\) khi và chỉ khi  \(x=\frac{-1}{2}\)

a) (x+3)² + (4+x)(4-x) = 10

x² + 6x + 9 + 16- x² = 10

6x + 25 = 10

6x = -15

x = -15/6

b) 9(x+1)² - (3x-2)(3x+2) = 10

9x² + 18x + 9 - 9x² + 4 =10

18x + 13 = 10

18x = -3

x = -1/6

a) ( x + 3 )² + ( 4 + x )( 4 - x ) = 10

⇔ x² + 6x + 9 + 16 - x² = 10

⇔ 6x + 25 = 10

⇔ 6x = -15

⇔ x = -15/6 = -5/2

b) 9( x + 1 )² - ( 3x - 2 )( 3x + 2 ) = 10

⇔ 9( x² + 2x + 1 ) - ( 9x² - 4 ) = 10

⇔ 9x² + 18x + 9 - 9x² + 4 = 10

⇔ 18x + 13 = 10

⇔ 18x = -3

⇔ x = -3/18 = -1/6

x² + xy + 5x + 5y = ( x² + xy ) + ( 5x + 5y ) = x( x + y ) + 5( x + y ) = ( x + y )( x + 5 )

x² - y² + 3x - 3y = ( x² - y² ) + ( 3x - 3y ) = ( x - y )( x + y ) + 3( x - y ) = ( x - y )( x + y + 3 )

x² + xy + 5x + 5y 

= (x²+ xy) + ( 5x+5y)

= x(x+y) + 5(x+y)

= (x+y)(x+5)

x² - y² + 3x - 3y

= (x² - y²) + ( 3x -3y)

= (x-y)(x+y) + 3(x-y)

= (x-y)(x+y+3)

chúc bạn học tốt ^^

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

x^2-2x+1-2x^2+2+x^2+4x+4=1

2x+7=1

2x=-6

x=-3

vậy x=-3

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

⇔ x² - 2x + 1 - 2( x² - 1 ) + x² + 4x + 4 = 1

⇔ 2x² + 2x + 5 - 2x² + 2 = 1

⇔ 2x + 7 = 1

⇔ 2x = -6

⇔ x = -3

1) x³ + y³ + z³ - 3xyz

= ( x + y )³ - 3xy( x + y ) + z³ - 3xyz

= [ ( x + y )³ + z³ ) - [ 3xy( x + y ) + 3xyz ]

= ( x + y + z )[ ( x + y )² - ( x + y )z + z² ] - 3xy( x + y + z )

= ( x + y + z )( x² + y² + z² + 2xy - xz - yz - 3xy )

= ( x + y + z )( x² + y² + z² - xy - yz - xz )

2) Tạm thời đang bí chưa làm được :(

3) ( x² - 2x )²( x² - 2x - 1 ) - 6 ( đề có vấn đề -- )

4) x⁴ - 7x³ + 14x² - 7x + 1

= x⁴ - 3x² - 4x² + x² + 12x² + x² - 4x - 3x + 1

= ( x⁴ - 3x² + x² ) - ( 4x³ - 12x² + 4x ) + ( x² - 3x + 1 )

= x²( x² - 3x + 1 ) - 4x( x² - 3x + 1 ) + ( x² - 3x + 1 )

= ( x² - 3x + 1 )( x² - 4x + 1 )