\(2\left(x-2\right)=x^2-4x-4\)

\(\Leftrightarrow2x-4=x^2-4x-4\)

\(\Leftrightarrow x^2-6x=0\)

\(\Leftrightarrow x\left(x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}\)

vậy........

x=0

x=6 

Gợi ý thôi.

\(x^3-ax^2+bx-c=\left(x-a\right)\left(x-b\right)\left(x-c\right)\)

\(\Rightarrow x^3-ax^2+bx-c\)có ba nghiệm \(x=a,x=b,x=c\)

Theo định lí Vi-et:\(\hept{\begin{cases}a+b+c=a\\ab+bc+ca=b\\abc=c\end{cases}\Leftrightarrow}\hept{\begin{cases}b=-c\\ab+bc+ca=b\\c\left(ab-1\right)=0\end{cases}}\)

okeee cam on ban

Ta có A = (3x + 2)² + (x² + y² - 2xy) - (2x - 2y) + 2015

= (3x + 2)² + (x - y)² - 2(x - y) + 1 +  2014

= (3x + 2)² + (x - y - 1)² + 2014 \(\ge\)2014

Dấu "=" xảy ra <=> \(\hept{\begin{cases}3x+2=0\\x-y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{2}{3}\\y=x-1\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{2}{3}\\y=-\frac{5}{3}\end{cases}}\)

Vậy Min A = 2015 <=> x = -2/3 ; y = -5/3

\(A=\left(3x+2\right)^2+x^2+y^2-2xy-2x+2y+2015\)

\(=\left(3x+2\right)^2+\left(x^2-2xy+y^2\right)-\left(2x-2y\right)+1+2014\)

\(=\left(3x+2\right)^2+\left(x-y\right)^2-2\left(x-y\right)+1+2014\)

\(=\left(3x+2\right)^2+\left(x-y-1\right)^2+2014\)

Vì \(\left(3x+2\right)^2\ge0\forall x\); \(\left(x-y-1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(3x+2\right)^2+\left(x-y-1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(3x+2\right)^2+\left(x-y-1\right)^2+2014\ge2014\forall x,y\)

hay \(A\ge2014\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}3x+2=0\\x-y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3x=-2\\y=x-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-2}{3}\\y=\frac{-5}{3}\end{cases}}\)

Vậy \(minA=2014\)\(\Leftrightarrow x=-\frac{2}{3}\)và \(y=-\frac{5}{3}\)

can gap nha

Ta có (x + 5)³ - x² + 25 = 0

=> (x + 5)³ - (x² - 25) = 0

=> (x + 5)³ - (x + 5)(x - 5) = 0

=> (x + 5)[(x + 5)² - x + 5] = 0

=> (x + 5)(x² + 9x + 30) = 0

=> x + 5 = 0 (Vì \(x^2+9x+30=\left(x^2+9x+\frac{81}{4}\right)+\frac{39}{4}=\left(x+\frac{9}{2}\right)^2+\frac{39}{4}\ge\frac{39}{4}>0\))

=> x = -5

Vậy x = -5

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

\(=x^3+2+3x^3-6\)

\(=4x^3-4\)

\(=4.\left(x^3-1\right)\)

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

Ta có A = -x² + 4x - 6 - y² - 2y 

= -(x² - 4x + 4) - (y² + 2y + 1) - 1

= -(x - 2)² - (y + 1)² - 1 \(\le-1< 0\)

=> A < 0 với mọi x ; y

A = -x² + 4x - 6 - y² - 2y 

= -( x² - 4x + 4 ) - ( y² + 2y + 1 ) - 1

= -( x - 2 )² - ( y - 1 )² - 1 ≤ -1 < 0 ∀ x, y

=> đpcm

a) \(4x.\left(2-x\right)+\left(2x+1\right)^2=2\)

    \(8x-4x^2+4x^2+4x+1=2\)

                                         \(12x+1=2\)

                                                  \(12x=2-1\)

                                                  \(12x=1\)

                                                        \(x=\frac{1}{12}\)

b) \(\left(x+3\right)^2-5.\left(x+3\right)=0\)

      \(\left(x+3\right).\left(x+3-5\right)=0\)

                \(\left(x+3\right).\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)