a Ta có 4x² - 4x + 3 = (4x² - 4x + 1) + 2 = (2x - 1)² + 2 \(\ge\)2 > 0 (đpcm)

b) Ta có y - y² - 1 

= -(y² - y + 1)

= -(y² - y + 1/4) - 3/4

= -(y - 1/2)² - 3/4 \(\le-\frac{3}{4}< 0\)(đpcm)

a) 4x² - 4x + 3 = ( 4x² - 4x + 1 ) + 2 = ( 2x - 1 )² + 2 ≥ 2 > 0 ∀ x ( đpcm )

b) y - y² - 1 = -( y² - y + 1/4 ) - 3/4 = -( y - 1/2 ) - 3/4 ≤ -3/4 < 0 ∀ x ( đpcm )

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

\(=x^2+5x-2x-10\)

\(=\left(x^2+5x\right)-\left(2x+10\right)\)

\(=x\left(x+5\right)-2\left(x+5\right)\)

\(=\left(x-2\right)\left(x+5\right)\)

Thích hđt thì chiều :))

x² + 3x - 10

= ( x² + 3x + 9/4 ) - 49/4

= ( x + 3/2 )² - ( 7/2 )²

= ( x + 3/2 - 7/2 )( x + 3/2 + 7/2 )

= ( x - 2 )( x + 5 )

1) x - 2 = ( x - 2 )²

<=> ( x - 2 ) - ( x - 2 )² = 0

<=> ( x - 2 )[ 1 - ( x - 2 ) ] = 0

<=> ( x - 2 )( 1 - x + 2 ) = 0

<=> ( x - 2 )( 3 - x ) = 0

<=> \(\orbr{\begin{cases}x-2=0\\3-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

2) ( x² + 1 )( 2x - 1 ) + 2x = 1

<=> ( x² + 1 )( 2x - 1 ) + ( 2x - 1 ) = 0

<=> ( 2x - 1 )[ ( x² + 1 ) + 1 ] = 0

<=> ( 2x - 1 )( x² + 1 + 1 ) = 0

<=> ( 2x - 1 )( x² + 2 ) = 0

<=> \(\orbr{\begin{cases}2x-1=0\\x^2+2=0\end{cases}}\Leftrightarrow x=\frac{1}{2}\)( do x² + 2 ≥ 2 > 0 ∀ x )

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

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

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

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

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

\(4x^2-4+1-4x^2+10x+8x-20=-13\) 

\(18x-3=-13\)

\(18x=-10\)

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

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