h, \(27x^3-8=\left(3x-2\right)\left(9x^2+6x+4\right)\)

\(\Rightarrow\left(27x^3-8\right):\left(3x-2\right)\\ =\left(3x-2\right)\left(9x^2+6x+4\right):\left(3x-2\right)\\ =9x^2+6x+4\)

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

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

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

Dấu '=' xảy ra khi x=-1

2: \(=-\left(4x^2-12x-10\right)\)

\(=-\left(4x^2-12x+9-19\right)\)

\(=-\left(2x-3\right)^2+19< =19\)

Dấu '=' xảy ra khi x=3/2

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

Dấu '=' xảy ra khi x=-2

Giải:

5) \(-x^2+x-\dfrac{1}{2}\)

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

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

\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\le\dfrac{3}{4}\)

\(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy ...

6) \(-\dfrac{1}{4}x^2+x-2\)

\(=-\dfrac{1}{4}x^2+x-1-1\)

\(=-\left(\dfrac{1}{4}x^2-x+1\right)-1\)

\(=-\left(\dfrac{1}{2}x-1\right)^2-1\le-1\)

\(\Leftrightarrow\dfrac{1}{2}x-1=0\Leftrightarrow x=2\)

Vậy ...

7) \(-\dfrac{1}{9}x^2-\dfrac{1}{3}x+1\)

\(=-\dfrac{1}{9}x^2-\dfrac{1}{3}x-\dfrac{1}{4}+\dfrac{5}{4}\)

\(=-\left(\dfrac{1}{9}x^2+\dfrac{1}{3}x+\dfrac{1}{4}\right)+\dfrac{5}{4}\)

\(=-\left(\dfrac{1}{3}x+\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{1}{3}x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{3}{2}\)

Vậy ...

8) \(-2x^2+2xy-2y^2+2x+2y-8\)

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y-1=0\end{matrix}\right.\Leftrightarrow x=y=1\)

Vậy ...

a) ( 5x - y )( 25x² + 5xy + y² ) = ( 5x )³ - y³ = 125x³ - y³

b) ( x - 3 )( x² + 3x + 9 ) - ( 54 + x³ ) = x³ - 3³ - 54 - x³ = -27 - 54 = -81

c) ( 2x + y )( 4x² - 2xy + y² ) - ( 2x - y )( 4x² + 2xy + y² ) = ( 2x )³ + y³ - [ ( 2x )³ - y³ ]= 8x³ + y³ - 8x³ + y³ = 2y³

d) ( x + y )² + ( x - y )² + ( x + y )( x - y ) - 3x² = x² + 2xy + y² + x² - 2xy + y² + x² - y² - 3x² = y²

e) ( x - 3 )³ - ( x - 3 )( x² + 3x + 9 ) + 6( x + 1 )²

= x³ - 9x² + 27x - 27 - ( x³ - 3³ ) + 6( x² + 2x + 1 )

= x³ - 9x² + 27x - 27 - x³ + 27 + 6x² + 12x + 6

= -3x² + 39x + 6

= -3( x² - 13x - 2 )

f) ( x + y )( x² - xy + y² ) + ( x - y )( x² + xy + y² ) - 2x³

= x³ + y³ + x³ - y³ - 2x³

= 0

g) x² + 2x( y + 1 ) + y² + 2y + 1

= x² + 2x( y + 1 ) + ( y² + 2y + 1 )

= x² + 2x( y + 1 ) + ( y + 1 )²

= ( x + y + 1 )²

= [ ( x + y ) + 1 ]²

= ( x + y )² + 2( x + y ) + 1

= x² + 2xy + y² + 2x + 2y + 1

méo hiểu đề bạn à

a) A=x^3 + 3x^2*5 + 3x*5^2 + 5^3

       =(x+5)^3

Thay x = -10 vào biểu thức A ta được:

A  = (-10+5)^3

    =(-5)^3

    =-75

Làm tương tự nhé

\(a)\)

\(21\left(x+3\right)^3:\left(3x+9\right)^2\)

\(=[21\left(x+3\right)^3]:[3^2\left(x+3\right)^2]\)

\(=7\left(x+3\right):3\)

Thay vào ta được: \(7.\frac{\left(-6+3\right)}{3}=7.\left(-3\right):3=-7\)

\(b)\)

Thay vào ta được:

\(\left(2.2^2-5.2+3\right)^4:[\left(2.2-3\right)^3:\left(2-1\right)^2]\)

\(=\left(2.4-10+3\right)^4:[\left(4-3\right)^31^2]\)

\(=1^4:\left(1^3.1\right)\)

\(=1:1\)

\(=1\)

\(c)\)

Thay vào ta được:

\(36.10^4.7^3:\left(-6.10^3.7^2\right)\)

\(=-6.10.7\)

\(=-420\)

B1 : a, M = x³-3xy(x-y)-y³-x²+2xy-y²

= ( x³-y³)-3xy(x-y) -(x²-2xy+y²)

= (x-y)(x²+xy+y²)-3xy(x-y)-(x-y)²

= (x-y) [(x²+xy+y²-3xy-(x-y)]

= (x-y)[(x²-2xy+y²)-(x-y)

= (x-y)[(x-y)²-(x-y)]

= (x-y)(x-y)(x-y-1)

= (x-y)²(x-y-1)

= 7²(7-1) = 49 . 6= 294

N = x²(x+1)-y²(y-1)+xy-3xy(x-y+1)-95

= x³+x²-(y³-y²)+xy-(3x²y-3xy²+3xy)-95

= x³+x²-y³+y²+xy-3x²y+3xy²-3xy-95

= (x³-y³)+(x²-2xy+y²)-(3x²y+y²)-(3x²y-3xy²)-95

=(x-y)(x²+xy+y²)+(x-y)²-3xy(x-y)-95

= (x-y)(x²+xy+y²+x-y-3xy)-95

= (x-y)[(x²-2xy+y²)+(x-y)]-95

= (x-y)[(x-y)²+(x-y)]-95

=(x-y)(x-y)(x-y+1)-95

= (x-y)²(x-y+1)-95

= 7²(7+1)-95=297