`20-2(x-1)^2 =2`

`=>2(x-1)^2 =18`

`=>(x-1)^2 =9`

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

`TH1:x-1=-3`

`=>x=-2`

`TH2:x-1=3`

`=>x=4`

Vậy `x in {-2;4}`

`17.64+17.36+322`

`=17(64+36)+322`

`=17.100+322`

`=1700+322`

`=2022`

`15+(x+2)^2 :3=18`

`=>(x+2)^2 :3=3`

`=>(x+2)^2 =9`

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

`TH1:x+2=-3`

`=>x=-5`

`TH2:x+2=3`

`=>x=1`

Vậy `x in {-5;1}`

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

`=[(x-1)+(x+1)]^2`

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

`=(2x)^2`

`=4x^2`

`5x^2 +y^2 -4x-2xy+2023`

`=x^2 -2xy+y^2 +4x^2 -4x+1+2022`

`=(x-y)^2 + (2x-1)^2 +2022>= 2022`

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\2x-1=0\end{matrix}\right.\)

\(\Leftrightarrow x=y=\dfrac{1}{2}\)

Vậy `M_(min) = 2022 <=> x=y= 1/2`

`x^2 -9+(x-3)^2`

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

`=(x-3)[(x+3)+(x-3)]`

`=(x-3)(x+3+x-3)`

`=2x(x-3)`

`ĐKXĐ: x >= 5`

`x-5 -sqrt(x-5)=0`

`<=>sqrt((x-5)^2) -sqrt(x-5)=0`

`<=>sqrt(x-5)(sqrt(x-5) -1)=0`

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-5}=0\\\sqrt{x-5}-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\\sqrt{x-5}=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

Vậy phương trình có tập nghiệm `S={5;6}`

`(x+2)(x+3)+(x-4)(x+5)`

`=(x^2 +2x+3x+6)+(x^2 -4x+5x-20)`

`=(x^2 +5x+6)+(x^2 +x-20)`

`=x^2 +5x+6+x^2 +x-20`

`=2x^2 +6x-14`

`M=(x-3)(x^2 +3x+9)-(x-27)-x(x-1)(x+1)`

`=(x^3 -27)-x+27-x(x^2 -1)`

`=x^3 -27-x+27-(x^3 -x)`

`=x^3 -x-x^3 +x`

`=0`

Vậy biểu thức không phụ thuộc vào biến

`A=sqrt(1-2x+x^2)+sqrt(x^2 -6x+9)`

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

`=|1-x|+|x-3|`

`>= |1-x+x-3|=|-2|=2`

Dấu "=" xảy ra `<=>(1-x)(x-3) >= 0 <=> 1 <= x <= 3`

Vậy `A_(min) =2 <=> 1 <= x <= 3`