\(x^2+x-1=0\)

=>\(x^2+x+\dfrac{1}{4}-\dfrac{5}{4}=0\)

=>\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)

=>\(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{\sqrt{5}}{2}\\x+\dfrac{1}{2}=-\dfrac{\sqrt{5}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}}{2}-\dfrac{1}{2}\\x=-\dfrac{\sqrt{5}}{2}-\dfrac{1}{2}\end{matrix}\right.\)

cứu tuii ii mn :<

   \(x^3\) + 3\(x^2\)y + 3\(xy^2\) + y³ - \(x-y\)

= (\(x^3\) + 3\(x^2\)y + 3\(xy^2\) + y³) - (\(x+y\))

= (\(x+y\))³ - (\(x+y\))

= (\(x+y\))[(\(x+y\))² - 1]

= (\(x+y\))[\(x+y-1\)][\(x+y+1\)]

`3 - (x + 1)^2`

`= 3 - (x^2 + 2x + 1)`

`= 3 - x^2 - 2x - 1`

`= 2 - x^2 - 2x`

  3 - (\(x+1\))^\(2\)

= 3 - (\(x^2\) + 2\(x\) + 1)

= 3 - \(x^2\) - 2\(x-1\)

= - \(x^2\) - 2\(x\) + (3 -1)

= - \(x^2\) - 2\(x\) + 2 

\(x^2\) x \(x\) = \(x^{2+1}\) = \(x^3\) 

cảm ơn ạ 

\(x^2\) - \(x\) + 3.(\(x-1\)) = 0

\(x\left(x-1\right)\) + 3(\(x-1\)) = 0

(\(x-1\))\(\left(x+3\right)\) = 0

\(\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-3; 1}

C = 1 - \(\dfrac{\left(x+3\right)\left(x+5\right)}{2}\)

C = 1 - \(\dfrac{x^2+5x+3x+15}{2}\)

C = 1 - \(\dfrac{x^2+\left(5x+3x\right)+15}{2}\)

C = 1 - \(\dfrac{x^2+8x+16-1}{2}\)

C = 1 - \(\dfrac{\left(x^2+2.x.4+4^2\right)}{2}\) + \(\dfrac{1}{2}\)

C = (1 + \(\dfrac{1}{2}\)) -  \(\dfrac{\left(x+4\right)^2}{2}\)

C =  \(\dfrac{3}{2}\)- \(\dfrac{\left(x+4\right)^2}{2}\)

Vì (\(x+4\))² ≥ 0 \(\forall\) \(x\) ⇒ - \(\dfrac{1}{2}\)(\(x+4\))² ≤ 0 ∀ \(x\)

    ⇒ \(\dfrac{3}{2}\) - \(\dfrac{\left(x+4\right)^2}{2}\) ≤ \(\dfrac{3}{2}\) dấu bằng xảy ra khi \(x+4\) = 0 ⇒ \(x=-4\)

Vậy giá trị lớn nhất của biểu thức C là \(\dfrac{3}{2}\) xảy ra khi \(x=-4\) 

    \(x^3\)  - \(x-y\) + y³

= (\(x^3\) + y³) - (\(x+y\))

= (\(x+y\)).(\(x^2\) - \(xy\) + y²) - (\(x+y\))

= (\(x+y\)).(\(x^2\) - \(xy+y^2\) - 1)

\(x^3-x-y+y^3\)

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

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

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

Ta có:\(A=\dfrac{x-1}{x+2}=\dfrac{x+2-3}{x+2}=\dfrac{x+2}{x+2}-\dfrac{3}{x+2}=1-\dfrac{3}{x+2}\)

Để A là giá trị nguyên => \(x+2\inƯ\left(3\right)=\left\{-1,-3,1,3\right\}\)

Ta có bảng giá trị: 

Vậy để A nguyên thì \(x\in\left\{-5,-3,-1,1\right\}\)

Bổ sung điều kiện \(x\) ≠ - 2

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

=>\(x^2-x-8x+8=0\)

=>x(x-1)-8(x-1)=0

=>(x-1)(x-8)=0

=>\(\left[{}\begin{matrix}x-1=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)

x^2-9x+8=0

(x-8)(x-1)=0

x=8 hoặc x=1.

Ta có: \(x^4-4x^3+5x^2-6x+9=0\)

=>\(x^4-4x^3+4x^2+x^2-6x+9=0\)

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

=>\(\left\{{}\begin{matrix}x^2-2x=0\\x-3=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\in\left\{0;2\right\}\\x=3\end{matrix}\right.\)

=>\(x\in\varnothing\)