$x^2-2x+5\\=x^2-2x+1+4\\=(x-1)^2+4$

Ta có: $(x-1)^2\ge 0$

$\Rightarrow (x-1)^2+4\ge 4$

$\Rightarrow P\ge 4$

$\Rightarrow$ Dấu "=" xảy ra khi $x-1=0$ hay $x=1$

Vậy P đạt GTNN là 4 tại x=1 

$\dfrac{2}{3}=\dfrac{6}{9}\\\dfrac{2}{6}=\dfrac{3}{9}\\\dfrac{3}{2}=\dfrac{9}{6}\\\dfrac{6}{2}=\dfrac{9}{3}$

$\dfrac{x}{3}=-\dfrac{16}{48}\\\Leftrightarrow x=\dfrac{-16.3}{48}=-1$

$\dfrac{y}{-9}=-\dfrac{16}{48}\\\Leftrightarrow y=\dfrac{-16.(-9)}{48}=3$

$\dfrac{-8}{z}=\dfrac{-16}{48}\\\Rightarrow x=\dfrac{-8.48}{-16}=24$

Vậy $x=-1,y=3,z=24$

Căn nhé

`x^2+3x-10=0

`<=>x^2+5x-2x-10=0`

`<=>(x^2+5x)-(2x+10)=0`

`<=>x(x+5)-2(x+5)=0`

`<=>(x-2)(x+5)=0`

$\Leftrightarrow\left[\begin{array}{1}x-2=0\\x+5=0\end{array}\right.\\\Leftrightarrow\left[\begin{array}{1}x=2\\x=-5\end{array}\right.$

Vậy $x\in\{2;-5\}$

$(x-3)2x-2x(x+5)=7\\\Leftrightarrow 2x(x-3-x-5)=7\\\Leftrightarrow 2x.(-8)=7\\\Leftrightarrow -16x=7\\\Leftrightarrow x=-\dfrac{7}{16}$

Vậy $x=-\dfrac{7}{16}$

$a^4-8\\=(a^2-\sqrt 8)(a^2+\sqrt 8)$