(4x+1)²-(2x-1)²+(x+2)³

=[ (4x+1)²-(2x-1)² ] + (x+2)³

=(4x+1-2x-1)(4x+1+2x-1) + x³+6x²+12x+8

=2x.6x+x³+6x²+12x+8

= 12x+x³+6x²+12x+8

= 24x +x³+6x²+8

c)Xét \(\Delta KPN\),có:

NI vừa là đường phân giác vừa là đường cao 

\(\Rightarrow\)\(\Delta KPN\)cân tại N

\(\Rightarrow\widehat{KPN}=\widehat{PKN}\)

và NI là đường trung tuyến của \(\Delta KPN\)

\(\Rightarrow I\)là trung điểm của \(KP\)

Xét \(\Delta MKP\)vuông tại M,có :

MI là đường trung tuyến 

\(\Rightarrow MI=\frac{KP}{2}=KI\)

\(\Rightarrow\Delta MKI\)cân tại I

\(\Rightarrow\widehat{MKI}=\widehat{KMI}\)

Xét \(\Delta KMI\)và \(\Delta KPN\),có:

\(\hept{\begin{cases}\widehat{PKN}:chung\\\widehat{KMI}=\widehat{KPN}\left(=\widehat{PKN}\right)\end{cases}}\)

\(\Rightarrow\Delta KMI~\Delta KPN\left(g.g\right)\)

Rút gọn nha mn

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

= [ x + 3 - ( x - 3 ) ][ ( x + 3 )² + ( x + 3 )( x - 3 ) + ( x - 3 )² ] + 4x² - 1

= ( x + 3 - x + 3 )( x² + 6x + 9 + x² - 9 + x² - 6x + 9 ) + 4x² - 1

= 6( 3x² + 9 ) + 4x² - 1

= 18x² + 54 + 4x² - 1 = 22x² + 53 

( 3x - 1 )² - ( 2x - 1 )( 2x + 1 ) = 9x² - 6x + 1 - ( 4x² - 1 )

= 9x² - 6x + 1 - 4x² + 1 = 5x² - 6x + 2

Ta có a + b + c = 0 

<=> (a + b + c)² = 0

<=> a² + b² + c² + 2(ab + bc + ca) = 0 

<=> ab + bc + ca = \(-\frac{1}{2}\)

=> \(\left(ab+bc+ca\right)^2=\frac{1}{4}\)

<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2ab^2c+2a^2bc+2abc^2=\frac{1}{4}\)

<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=\frac{1}{4}\)

<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=\frac{1}{4}\)

Lại có a² + b² + c² = 1

=> (a² + b² + c²)² = 1

<= > a⁴ + b⁴ + c⁴ + 2[(ab)² + (bc)² + (ca)²] = 1 

<=> \(a^4+b^4+c^4+2.\frac{1}{4}=1\)

<=> \(a^4+b^4+c^4=\frac{1}{2}\)

Từ a + b + c = 0 => ( a + b + c )² = 0 <=> a² + b² + c² + 2ab + 2bc + 2ca = 0

<=> ab + bc + ca = -1/2 => ( ab + bc + ca )² = 1/4

<=> a²b² + b²c² + c²a² + 2ab²c + 2bc²a + 2a²bc = 1/4

<=> a²b² + b²c² + c²a² + 2abc( a + b + c ) = 1/4

<=> a²b² + b²c² + c²a² = 1/4 ( vì a + b + c = 0 )

Từ a² + b² + c² = 1 => ( a² + b² + c² )² = 1 <=> a⁴ + b⁴ + c⁴ + 2a²b² + 2b²c² + 2c²a² = 1

<=> a⁴ + b⁴ + c⁴ + 2( a²b² + b²c² + c²a² ) = 1 

<=> a⁴ + b⁴ + c⁴ + 1/2 = 1 <=> a⁴ + b⁴ + c⁴ = 1/2

Vậy A = 1/2

\(\left(x^2+3\right)\left(3-x^2\right)\)

\(\left(x^2+3\right)\left(-x^2+3\right)\)

\(\left(-x^2+3\right).x^2+3\left(-x^2+3\right)\)

\(-x^2.x^2+3x^2+3\left(-x^2+3\right)\)

\(-x^2.x^2+3x^2-3x^2+9\)

\(-x^2.x^2+9\)

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

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

\(4x^2-10x+5\left(2x-5\right)\)

\(4x^2-10x+10x-25\)

\(4x^2-25\)

( x + y )³ - ( x - 2y )³ = [ ( x + y ) - ( x - 2y ) ][ ( x + y )² + ( x + y )( x - 2y ) + ( x - 2y )² ]

= ( x + y - x + 2y )( x² + 2xy + y² + x² - xy - 2y² + x² - 4xy + 4y² )

= 9y( x² - xy + y² )

\(\left(x+y\right)^3-\left(x-2y\right)^3\)

\(=\left(x+y-x+2y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-2y\right)+\left(x-2y\right)^2\right]\)

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

\(=3y\left(3x^2-3xy+7y^2\right)=9x^2y-9xy^2+21y^3\)

Ta có A = 2x² + 12x + 1 

= \(2\left(x^2+6x+\frac{1}{2}\right)=2\left(x^2+6x+9-\frac{17}{2}\right)=2\left(x+3\right)^2-17\ge-17\)

=> Min A = -17

Dấu "=" xảy ra <=> x + 3 = 0 

<=> x = -3

Vậy Min A = -17 <=> x = -3

b) Ta có B = x² + 3x + 2 

= \(x^2+2.\frac{3}{2}x+\frac{9}{4}-\frac{1}{4}=\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

=> Min B = -1/4

Dấu "=" xảy ra <=> x + 3/2 = 0 <=> x = -3/2

Vậy Min B = -1/4 <=> x = -3/2