Question

Tìm x, biết:
a) (x+2)² + (x-3)² = 2x (x+7)

b) (x+1)³ + (x-1)³ = 2x³

c) x³ - 3x² + 3x - 126 = 0

d) x² + y² - 2x + 4y + 5 = 0

e) 2x² + 8x +y² - 2y + 9 = 0

Answer 1

a)

pt <=>     \(x^2+4x+4+x^2-6x+9=2x^2+14x\)

<=>     \(2x^2-2x+13=2x^2+14x\)

<=>     \(16x=13\)

<=>     \(x=\frac{13}{16}\)

b)

pt <=>     \(x^3+3x^2+3x+1+x^3-3x^2+3x-1=2x^3\)

<=>   \(2x^3+6x=2x^3\)

<=>   \(6x=0\)

<=>   \(x=0\)

c)

pt <=>    \(\left(x^3-3x^2+3x-1\right)-125=0\)

<=>   \(\left(x-1\right)^3=125\)

<=>   \(x-1=5\)

<=>   \(x=6\)

d)

pt <=>   \(\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)

<=>   \(\left(x-1\right)^2+\left(y+2\right)^2=0\)     (1)

CÓ:   \(\left(x-1\right)^2;\left(y+2\right)^2\ge0\forall x;y\)

=>   \(\left(x-1\right)^2+\left(y+2\right)^2\ge0\)       (2)

TỪ (1) VÀ (2) =>    DÁU "=" XẢY RA <=>   \(\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{cases}}\)

<=>     \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

e)

pt <=>   \(2x^2+8x+8+y^2-2y+1=0\)

<=>   \(2\left(x+2\right)^2+\left(y-1\right)^2=0\)

TA LUÔN CÓ:   \(2\left(x+2\right)^2+\left(y-1\right)^2\ge0\forall x;y\)

=> DẤU "=" XẢY RA <=>   \(\hept{\begin{cases}2\left(x+2\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\) 

<=>     \(\hept{\begin{cases}x=-2\\y=1\end{cases}}\)

Answer 2

a) ( x + 2 )² + ( x - 3 )² = 2x( x + 7 )

<=> x² + 4x + 4 + x² - 6x + 9 = 2x² + 14x

<=> x² + 4x + x² - 6x - 2x² - 14x = -4 - 9

<=> -16x = -13

<=> x = 13/16

b) ( x + 1 )³ + ( x - 1 )³ = 2x³

<=> x³ + 3x² + 3x + 1 + x³ - 3x² + 3x - 1 = 2x³

<=> x³ + 3x² + 3x + x³ - 3x² + 3x - 2x³ = -1 + 1

<=> 6x = 0

<=> x = 0

c) x³ - 3x² + 3x - 126 = 0

<=> ( x³ - 3x² + 3x - 1 ) - 125 = 0

<=> ( x - 1 )³ = 125

<=> ( x - 1 )³ = 5³

<=> x - 1 = 5

<=> x = 6

d) x² + y² - 2x + 4y + 5 = 0

<=> ( x² - 2x + 1 ) + ( y² + 4y + 4 ) = 0

<=> ( x - 1 )² + ( y + 2 )² = 0 (*)

\(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)

Đẳng thức xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

e) 2x² + 8x + y² - 2y + 9 = 0

<=> 2( x² + 4x + 4 ) + ( y² - 2y + 1 ) = 0

<=> 2( x + 2 )² + ( y - 1 )² = 0 (*)

\(\hept{\begin{cases}2\left(x+2\right)^2\ge0\forall x\\\left(y-1\right)^2\ge0\forall y\end{cases}}\Rightarrow2\left(x+2\right)^2+\left(y-1\right)^2\ge0\forall x,y\)

Đẳng thức xảy ra ( tức xảy ra (*) ) <=> \(\hept{\begin{cases}x+2=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=1\end{cases}}\)

Answer 3

a. \(\left(x+2\right)^2+\left(x-3\right)^2=2x\left(x+7\right)\)

\(\Leftrightarrow x^2+4x+4+x^2-6x+9=2x^2+14x\)

\(\Leftrightarrow2x^2-2x+13=2x^2+14x\)

\(\Leftrightarrow16x=13\)

\(\Leftrightarrow x=\frac{13}{16}\)

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

\(\Leftrightarrow x^3+3x^2+3x+1+x^3-3x^2+3x-1=2x^3\)

\(\Leftrightarrow2x^3+6x=2x^3\)

\(\Leftrightarrow6x=0\)

\(\Leftrightarrow x=0\)

c. \(x^3-3x^2+3x-126=0\)

\(\Leftrightarrow\left(x-6\right)\left(x^3+3x+21\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x^3+3x+21=0\left(ktm\right)\end{cases}\Leftrightarrow}x=6\)

d. \(x^2+y^2-2x+4y+5=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left(y+2\right)^2\ge0\end{cases}}=0\Leftrightarrow\orbr{\begin{cases}x-1=0\\y+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\y=-2\end{cases}}}\)

e. \(2x^2+8x+y^2-2y+9=0\)

\(\Leftrightarrow2\left(x+2\right)^2+\left(y-1\right)^2=0\)

\(\hept{\begin{cases}2\left(x+2\right)^2\ge0\\\left(y-1\right)^2\ge0\end{cases}=0\Leftrightarrow\orbr{\begin{cases}x+2=0\\y-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\y=1\end{cases}}}\)