Đáp án+Giải thích các bước giải:
`\sqrt{x-2}+\sqrt{y+2021}+\sqrt{z-2022}=1/2(x+y+z)`
Điều kiện:`{(x>=2),(y>=-2021),(z>=2022):}`
`pt<=>2\sqrt{x-2}+2\sqrt{y+2021}+2\sqrt{z-2022}=x+y+z`
`<=>x+y+z-2\sqrt{x-2}-2\sqrt{y+2021}-2\sqrt{z-2022}=0`
`<=>x-2-2\sqrt{x-2}+1+y+2021-2\sqrt{y+2021}+1+z-2022-2\sqrt{z-2022}+1=0`
`<=>(\sqrt{x-2}-1)^2+(\sqrt{y+2021}-1)^2+(\sqrt{z-2022}-1)^2=0`
Vì `(\sqrt{x-2}-1)^2+(\sqrt{y+2021}-1)^2+(\sqrt{z-2022}-1)^2>=0`
Dấu "=" xảy ra khi `\sqrt{x-2}=\sqrt{y+2021}=\sqrt{z-2022}=1`
`<=>{(x=3),(y=-2020),(z=2023):}(TMĐK)`
Vậy `x=3,y=-2020,z=2023.`
`2)a=\sqrt{17}-1`
`<=>a+1=\sqrt{17}`
`<=>(a+1)^2=17`
`<=>a^2+2a+1=17`
`<=>a^2+2a-16=0`
`A=(a^5+2a^4-17a^3-a^2+18a-17)^2022`
`A=(a^5+2a^4-16a^3-a^3-2a^2+16a+a^2+2a-16-1)^2022`
`A=[a^3(a^2+2a-16)-a(a^2+2a-16)+a^2+2a-16-1]^2022`
`A=(0+0+0-1)^2022`
`A=1`
