\(\left(2x-1\right)^2-\left(2x-4\right)\left(2x-5\right)=-13\)

\(\Leftrightarrow4x^2-4x+1-\left(4x^2-18x+20\right)=-13\)

\(\Leftrightarrow14x-19=-13\Leftrightarrow14x=6\Leftrightarrow x=\frac{3}{7}\)

\(\left(2x-1\right)^2-\left(2x-4\right).\left(2x-5\right)=-13\)

\(4x^2-4+1-\left(4x^2-10x-8x+20\right)=-13\) 

\(4x^2-4+1-4x^2+10x+8x-20=-13\) 

\(18x-3=-13\)

\(18x=-10\)

\(x=\frac{-5}{9}\)

a) ( a - b - c )² - ( a - b + c )²

= [ ( a - b - c ) - ( a - b + c ) ][ ( a - b - c ) + ( a - b + c ) ]

= ( a - b - c - a + b - c )( a - b - c + a - b + c )

= -2c( 2a - 2b )

= -2c.2( a - b )

= -4c( a - b )

b) ( a - x - y )³ - ( a + x - y )³

= [ ( a - x - y ) - ( a + x - y ) ][ ( a - x - y )² + ( a - x - y )( a + x - y ) + ( a + x - y )² ]

= ( a - x - y - a - x + y ){ [ ( a - x ) - y ]² + [ ( a - y ) - x ][ ( a - y ) + x ] + [ ( a + x ) - y ] ² }

= -2x{ [ ( a - x )² - 2( a - x )y + y² ] + [ ( a - y )² - x² ] + [ ( a + x )² - 2( a + x )y + y² ] }

= -2x{ [ a² + x² + y² - 2ax - 2ay + 2xy ] + [ a² - x² + y² - 2ay ] + [ a² + x² + y² + 2ax - 2ay - 2xy ] }

= -2x{ a² + x² + y² - 2ax - 2ay + 2xy + a² - x² + y² - 2ay + a² + x² + y² + 2ax - 2ay - 2xy }

= -2x{ 3a² + x² + 3y² - 6ay } < trời ơi dài > ;-;

\(A=x^4-3x^3+4x^2-3x+10=\left(x^4-3x^3+4x^2-3x+1\right)+9=\left(x-1\right)^2\left(x^2-x+1\right)+9\ge9\)(do \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\x^2-x+1>0\forall x\end{cases}}\))

Đẳng thức xảy ra khi x = 1

Với giả thiết x² - 4x + 1 = 0 thì\(B=x^5-3x^4-3x^3+6x^2-20x+2025=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2020=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2020=\left(x^3+x^2+5\right)\left(x^2-4x+1\right)+2020=2020\)

Thank you nhiều nha . Chúc bạn học tốt. I love you <3

a) Ta có x³ + y³ = 2

<=> x³ + 3x²y + 3xy² + y³ - 3x²y - 3xy² = 2

<=> ( x³ + 3x²y + 3xy² + y³ ) - ( 3x²y + 3xy² ) = 2

<=> ( x + y )³ - 3xy( x + y ) = 2

<=> 1³ - 3xy = 2

<=> 3xy = -1

<=> xy = -1/3

Lại có x + y = 1

<=> ( x + y )⁵ = 1

<=> x⁵ + 5x⁴y + 10x³y² + 10x²y³ + 5xy⁴ + y⁵ = 1 ( HĐT bậc 5 này bạn lên mạng tra nhé :)) )

<=> x⁵ + y⁵ = 1 - ( 5x⁴y + 10x³y² + 10x²y³ + 5xy⁴ )

<=> x⁵ + y⁵ = 1 - [ ( 5x⁴y + 5xy⁴ ) + ( 10x³y² + 10x²y³ ) ]

<=> x⁵ + y⁵ = 1 - [ 5xy( x³ + y³ ) + 10x²y²( x + y ) ]

<=> x⁵ + y⁵ = 1 - [ 5xy( x³ + y³ ) + 10(xy)²( x + y ) ]

<=> x⁵ + y⁵ = 1 - [ 5.(-1/3).2 + 10.(-1/3)².1 ]

<=> x⁵ + y⁵ = 1 - [ -10/3 + 10/9 ]

<=> x⁵ + y⁵ = 1 - (-20/9) = 29/9

b) x + y = 8

<=> ( x + y )² = 64

<=> x² + 2xy + y² = 64

<=> 40 + 2xy = 64

<=> 2xy = 24

<=> xy = 12

Ta có : x³ + y³ = x³ + 3x²y + 3xy² + y³ - 3x²y - 3xy² 

                       = ( x³ + 3x²y + 3xy² + y³ ) - ( 3x²y + 3xy² ) 

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

                       = 8³ - 3.12.8

                       = 512 - 288 = 224

cam on

Cấy ni chỉ tìm được Max thôi

5x - 2x² + 1

= -2( x² - 5/2x + 25/16 ) + 33/8

= -2( x - 5/4 )² + 33/8 ≤ 33/8 ∀ x 

Dấu "=" xảy ra khi x = 5/4

Vậy GTLN của biểu thức = 33/8 <=> x = 5/4

5x - 2x² + 1

= - 2x² + 5x -\(\frac{25}{8}+\frac{33}{8}\)

= - 2 ( x -\(\frac{5}{4}\))² +\(\frac{33}{8}\le\frac{33}{8}\)

Dấu "=" xảy ra <=> - 2 ( x - 5/4 )² = 0 <=> x - 5/4 = 0 <=> x = 5/4

Vậy maxB = 33/8 <=> x = 5/4

8( x - 2 ) - 2( 3x - 4 ) = 2

<=> 8x - 16 - 6x + 8 = 2

<=> 2x - 8 = 2

<=> 2x = 10

<=> x = 5

\(8.\left(x-2\right)-2.\left(3x-4\right)=2\)

\(\Leftrightarrow8x-16-6x+8=2\)

\(\Leftrightarrow2x-8=2\)

\(\Leftrightarrow2x=10\)

\(\Leftrightarrow x=5\)