Ta có : 

\(B=6b-b^2-10\)

\(\Leftrightarrow B=-b^2+6b-9-1\)

\(\Leftrightarrow B=-\left(b^2-6b+9\right)-1\)

\(\Leftrightarrow B=-\left(b-3\right)^2-1< 0\)( luôn đúng với mọi b )

Vậy ..........

( x + 2 )³ - x²( x - 6 ) = 4

⇔ x³ + 6x² + 12x + 8 - x³ + 6x² - 4 = 0

⇔ 12x² + 12x + 4 = 0

⇔ 4( 3x² + 3x + 1 ) = 0

⇔ 3x² + 3x + 1 = 0

Ta có : 3x² + 3x + 1 = 3( x² + x + 1/4 ) + 1/4 = 3( x + 1/2 )² + 1/4 ≥ 1/4 > 0 ∀ x

=> Phương trình vô nghiệm

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

Mà : x + y = 1 

\(\Rightarrow M=x^2-xy+y^2=\frac{1}{2}x^2+\frac{1}{2}x^2-xy+\frac{1}{2}y^2+\frac{1}{2}y^2\)

\(=\frac{1}{2}\left(x^2+y^2\right)+\left(\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)^2\ge\frac{1}{2}\left(x^2+y^2\right)\)

Ta có : \(x+y=1\)

\(\Leftrightarrow\left(x+y\right)^2=1\)

\(\Leftrightarrow x^2+y^2+2xy=1\)

\(\Leftrightarrow2\left(x^2+y^2\right)-\left(x-y\right)^2=1\)( bước này tự tách từ trên ra nhé ) 

\(\Rightarrow2\left(x^2+y^2\right)\ge1\)

\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\Leftrightarrow x=y\)

Ta lại có : \(M\ge\frac{1}{2}\left(x^2+y^2\right)=\frac{1}{2}.\frac{1}{2}=\frac{1}{4}\)

Dấu bằng xảy ra \(\Leftrightarrow x=y=\frac{1}{4}\)

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

a) 2x² - 5x³ = 0

⇔ x²( 2 - 5x ) = 0

⇔ \(\orbr{\begin{cases}x^2=0\\2-5x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{5}\end{cases}}\)

b) ( x + 1 )( 2 - x ) - ( 3x + 5 )( x + 2 ) = -4x² + 2

⇔ -x² + x + 2 - ( 3x² + 11x + 10 ) + 4x² - 2 = 0

⇔ 3x² + x - 3x² - 11x - 10 = 0

⇔ -10x - 10 = 0

⇔ -10x = 10

⇔ x = -1

c) ( x + 3 )( x² - 3x + 9 ) - x( x - 2 )² = 27

⇔ x³ + 27 - x( x² - 4x + 4 ) - 27 = 0

⇔ x³ - x³ + 4x² - 4x = 0

⇔ 4x( x - 1 ) = 0

⇔ \(\orbr{\begin{cases}4x=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

d) ( x - 1 )( x - 5 ) + 3 = 0

⇔ x² - 6x + 5 + 3 = 0

⇔ x² - 6x + 8 = 0

⇔ x² - 2x - 4x + 8 = 0

⇔ x( x - 2 ) - 4( x - 2 ) = 0

⇔ ( x - 2 )( x - 4 ) = 0

⇔ \(\orbr{\begin{cases}x-2=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}\)

\(9x^2+5y^2-8x+3y=0\)

\(\left(9x^2-8x\right)+\left(5y^2+3y\right)=0\)

\(x\left(9x-8\right)+y\left(5y+3\right)=0\)

\(\orbr{\begin{cases}x=0\\9x-8=0\end{cases}}\)\(=>\orbr{\begin{cases}x=0\\x=\frac{8}{9}\end{cases}}\)

\(\orbr{\begin{cases}y=0\\5y+3=0\end{cases}}\)\(=>\orbr{\begin{cases}y=0\\y=\frac{-3}{5}\end{cases}}\)

Vậy \(x=\left\{0,\frac{8}{9}\right\},y=\left\{0,\frac{-3}{5}\right\}\)

đéo biết

1) \(A=-2x^2-10y^2+4xy+4x+4y+2013=-2\left(x-y-1\right)^2-8\left(y-\frac{1}{2}\right)^2+2017\le2017\forall x,y\inℝ\)Đẳng thức xảy ra khi x = ³/₂; y = ¹/₂

2) \(A=a^4-2a^3+2a^2-2a+2=\left(a^2+1\right)\left(a-1\right)^2+1\ge1\)

Đẳng thức xảy ra khi a = 1

3) \(N=\left(x-y\right)\left(x-2y\right)\left(x-3y\right)\left(x-4y\right)+y^4=\left(x^2-5xy+4y^2\right)\left(x^2-5x+6y^2\right)+y^4=\left(x^2-5xy+4y^2\right)^2+2y^2\left(x^2-5xy+4y^2\right)+y^4=\left(x^2-5xy+5y^2\right)^2\)(là số chính phương, đpcm)

4) \(a^3+b^3=3ab-1\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)-3ab+1=0\Leftrightarrow\left[\left(a+b\right)^3+1\right]-3ab\left(a+b+1\right)=0\)\(\Leftrightarrow\left(a+b+1\right)\left(a^2+2ab+b^2-a-b+1\right)-3ab\left(a+b+1\right)=0\Leftrightarrow\left(a+b+1\right)\left(a^2+b^2-ab-a-b+1\right)=0\)Vì a, b dương nên a + b + 1 > 0 suy ra \(a^2+b^2-ab-a-b+1=0\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2=0\Leftrightarrow a=b=1\)

Do đó \(a^{2018}+b^{2019}=1+1=2\)

5) \(A=n^3+\left(n+1\right)^3+\left(n+2\right)^3=3n\left(n^2+5\right)+9\left(n^2+1\right)⋮9\)(Do số chính phương chia 3 dư 1 hoặc 0)

tokuda and nguyễn yến cứt