\(A=x^4-3x^3+4x^2-3x+10=\left(x^4-3x^3+4x^2-3x+1\right)+9=\left(x-1\right)^2\left(x^2-x+1\right)+9\ge9\)(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

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

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

\(\Leftrightarrow2x-8=2\)

\(\Leftrightarrow2x=10\)

\(\Leftrightarrow x=5\)

https://olm.vn/hoi-dap/detail/187016317316.html

1)

a) \(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2+2\left(ab^2c+a^2bc+abc^2\right)\)\(=a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=a^2b^2+b^2c^2+c^2a^2\)(vì a+b+c=0)

b) \(a+b+c=0\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right]\)

\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)=2\left(ab+bc+ca\right)^2\left(theoa\right)\)

Ta có: \(a+b+c=3\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=9\)

\(\Rightarrow2\left(ab+bc+ca\right)=9-\left(a^2+b^2+c^2\right)=6\Rightarrow ab+bc+ca=3\)

\(\Rightarrow a^2+b^2+c^2=ab+bc+ca\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)

Mà a + b + c = 3 nên a = b = c = 1

Suy ra \(P=\left(-1\right)^{2019}+\left(-1\right)^{2020}+\left(-1\right)^{2021}=-1\)

\(50\left(y+4\right)^2-18\left(y-2\right)^2\)

\(=50\left(y^2+8y+16\right)-18\left(y^2-4y+4\right)\)

\(=50y^2+400y+800-18y^2+72y-72\)

\(=32y^2+472y+728\)