( x + 2 )( x² - x + 1 ) - ( x³ + 2x² )

= x( x² - x + 1 ) + 2( x² - x + 1 ) - x³ - 2x²

= x³ - x² + x + 2x² - 2x + 2 - x³ - 2x²

= -x² - x + 2 = ( 1 - x )( x + 2 )

Ta có C = x² - 4x + y² - y + 5 

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

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

=> Min C = 3/4

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2=0\\y-\frac{1}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=\frac{1}{2}\end{cases}}\)

Vậy Min C = 3/4 <=> x = 2 ; y = 1/2 

C = ( x² - 4x + 4 ) + ( y² - y + 1/4 ) + 3/4

= ( x - 2 )² + ( y - 1/2 )² + 3/4 ≥ 3/4 ∀ x.y 

Dấu "=" xảy ra <=> x = 2 ; y = 1/2 . Vậy MinC = 3/4

a) A = (2x + 6)(4x² − 12x + 36) − 8x³ + 10.

=8x³+216-8x³+10

=226

b) B = (2x − 1)(4x² + 2x + 1) − 8(x³ + 1). 

=8x³-1-8x³-8

=-9

c) C = (2 + a)(2 − a)(4 + 2a + a² )(a² − 2a + 4). 

=[(2+a)(a² − 2a + 4)] [((2 − a)(4 + 2a + a² )]

=[(a+2)(a² − 2a + 4)] [((2 − a)(4 + 2a + a² )]

=(a³+8)(8-a³)

=8a³-a⁶+64-8a³

=-a⁶+64

=64-a⁶

=(8-a³)(8+a³)

d) D = (a³ b³ − 1)(a³ b³ + 1) − a³ b³ .

=a⁶b⁶-1-a³b³

a, ( 2x - 3 )²- (2x + 1)² = -3

4x²-12x+9-4x²+4x-1=-3

-8x-1=-3

-8x=-2

x=\(\frac{1}{4}\)

b, (5x - 1) ² - (5x + 4)(5x - 4) = 7

25x²-10x+1-25x²+16=7

-10x+17=7

-10x=-10

x=1

c, ( x- 5)² + (x-3)(x+3) - 2(x + 1)²=0

x²-10x+25+x²-9-2x²-4x-2=0

-14x+14=0

-14(x-1)=0

=>x-1=0

x=1

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

\(\Leftrightarrow4x^2-12x+9-4x^2-4x-1=-3\)

\(\Leftrightarrow-16x+8=-3\)

\(\Leftrightarrow-16x=-11\)

\(\Leftrightarrow x=\frac{11}{16}\)

b)\(\left(5x-1\right)^2-\left(5x+4\right)\left(5x-4\right)=7\)

\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)

\(\Leftrightarrow-10x+17=7\)

\(\Leftrightarrow-10x=-10\)

\(\Leftrightarrow x=1\)

c)\(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+1\right)^2=0\)

\(\Leftrightarrow x^2-10x+25+x^2-9-2\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow2x^2-10x-16-2x^2-4x-2=0\)

\(\Leftrightarrow-14x-18=0\)

\(\Leftrightarrow-14x=18\)

\(\Leftrightarrow x=-\frac{9}{7}\)

#H

Ta có (x - y)² - (x + y)² 

= (x - y + x + y)(x - y - x - y) 

= 2x.(-2y)

= -4xy (đpcm) 

VT=(x-y)²-(x+y)²

=x²-2xy+y²-x²-2xy-y²

=-4xy=VP (đccm)

#H

a) B = x - x² + 2

= \(-\left(x^2-x+\frac{1}{4}-\frac{1}{4}-2\right)=-\left(x-\frac{1}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\)

=> Max B = 9/4 

Dấu "=" xảy ra <=> x - 1/2 = 0 <=> x = 1/2

Vậy Max B = 9/4 <=> x = 1/2

d) Ta có P = \(x-x^2-1=-\left(x^2-x+\frac{1}{4}-\frac{1}{4}+1\right)=-\left(x-\frac{1}{2}\right)^2-\frac{3}{4}\le-\frac{3}{4}\)

=> Max P = -3/4 

Dấu "=" xảy ra <=> x -1/2 = 0 <=> x = 1/2

Vậy Max P = -3/4 <=> x = 1/2 

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mọi người ơi giúp mình trả lồi câu hỏi này vớiiiiiiiiiiii

Ta có: 

K = x² + y² - 6x + y + 10

K = (x² - 6x + 9) + (y² + y + 1/4) + 3/4

K = (x - 3)² + (y + 1/2)² + 3/4 \(\ge\)3/4 \(\forall\)x; y (vì (x - 3)² \(\ge\)0 và (y + 1/2)² \(\ge\)0)

Dấu "=" xảy ra<=> \(\hept{\begin{cases}x-3=0\\y+\frac{1}{2}=0\end{cases}}\) <=> \(\hept{\begin{cases}x=3\\y=-\frac{1}{2}\end{cases}}\)

Vậy MinK = 3/4 <=> x = 3 và y = -1/2