( 2x + 5 )( 2x - 5 ) - ( 4x⁵ - 2x⁴ ) : ( -x³ ) = 15

<=> 4x² - 25 - [ ( 4x⁵ : ( -x³ ) - ( 2x⁴ : ( -x³ ) ] = 15

<=> 4x² - 25 - ( -4x² + 2x ) = 15

<=> 4x² - 25 + 4x² - 2x - 15 = 0

<=> 8x² - 2x - 40 = 0

Δ = b² - 4ac = (-2)² - 4.8.(-40) = 1284

Δ > 0 nên phương trình có hai nghiệm phân biệt

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\text{Δ}}}{2a}=\frac{2+\sqrt{1284}}{16}=\frac{1+\sqrt{321}}{8}\\x_2=\frac{-b-\sqrt{\text{Δ}}}{2a}=\frac{2-\sqrt{1284}}{16}=\frac{1-\sqrt{321}}{8}\end{cases}}\)

Vậy ...

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

\(\Leftrightarrow\left(x+3\right)\left[x\left(x-1\right)-x^2\right]=-4\)

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

\(\Leftrightarrow-x\left(x+3\right)=-4\)

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

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

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

\(\Leftrightarrow x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)

vậy........

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

⇔ ( x + 3 )[ x( x - 1 ) - x² ] + 4 = 0

⇔ ( x + 3 )( x² - x - x² ) + 4 = 0

⇔ ( x + 3 ).(-x) + 4 = 0

⇔ -x² - 3x + 4 = 0

⇔ -( x² + 3x - 4 ) = 0

⇔ -( x² - x + 4x - 4 ) = 0

⇔ -[ x( x - 1 ) + 4( x - 1 ) ] = 0

⇔ -( x - 1 )( x + 4 ) = 0

⇔ x - 1 = 0 hoặc x + 4 = 0

⇔ x = 1 hoặc x = -4

a) x( x + y ) - 7x - 7y

= x( x + y ) - 7( x + y )

= ( x + y )( x - 7 )

b) 16x - 5x² - 3

= -5x² + 15x + x - 3

= -5x( x - 3 ) + ( x - 3 )

= ( x - 3 )( 1 - 5x )

c) x⁴ - y⁴ - x² + y²

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

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

= ( x - y )( x + y )( x² + y² - 1 )

1. a) 2x² - 8x

= 2x(x - 4)

b) x² - xy + x - y

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

= (x + 1)(x - y)

2. a) Ta có M = x² + 5y² + 4xy + 4y + 11

= (x² + 4xy + 4y²) + (y² + 4y + 4) + 7

= (x + 2y)² + (y + 2)² + 7 \(\ge7\forall x;y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+2y=0\\y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2y\\y=-2\end{cases}}\Rightarrow\hept{\begin{cases}x=4\\y=-2\end{cases}}\)

Vậy Min M = 7 <=> x = 4 ; y = -2

a) x² +2x +1 -2x +6 

=x² + 7

b) 2x² +x +x²-1 -x

= 3x² -1

\(a)x^2+\left(2a-2\right)x+a+6\)

\(b)x^2+2x\)