Question

[Đừng gọi tôi là Jung Hae Ri]

Tìm GTNN và GTLN của các biểu thức:

\(a,P=\sqrt{x}+\sqrt{2-x}\)

\(b,Q=\sqrt{x-2019}+\sqrt{2020-x}\)

Answer 1

[Lê Thị Thục Hiền]

a, P>0

Có \(P^2=x+2\sqrt{x\left(2-x\right)}+2-x=2+2\sqrt{2x-x^2}=\sqrt{1-\left(x^2-2x+1\right)}+2=2+\sqrt{1-\left(x-1\right)^2}\)

Luôn có: \(1-\left(x-1\right)^2\le1\)=> \(0\le\sqrt{1-\left(x-1\right)^2}\le1\)<=> \(0\le2\sqrt{1-\left(x-1\right)^2}\le4\)

<=> \(2\le2+2\sqrt{1-\left(x-1\right)^2}\le2+2\)

<=> \(2\le P^2\le4\)

<=> \(\sqrt{2}\le P\le2\)(do P>0)

minP xảy ra <=> \(\sqrt{1-\left(x-1\right)^2}=0\)

<=> \(\left(x-1\right)^2=1\) <=> \(\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)(t/m)

maxP xảy ra<=> \(\sqrt{1-\left(x-1\right)^2}=1\)

<=> \(\left(x-1\right)^2=0\) <=> x=1(t/m)

Answer 2

[Lê Thị Thục Hiền]

b, Q>0 (đk :\(2019\le x\le2020\))

Có \(Q^2=x-2019+2\sqrt{\left(x-2019\right)\left(2020-x\right)}+2020-x=1+2\sqrt{\left(x-2019\right)\left(2020-x\right)}\)

Luôn có: \(0\le2\sqrt{\left(x-2019\right)\left(2020-x\right)}\le\left(x-2019\right)+\left(2020-x\right)\)

<=> \(1\le1+2\sqrt{\left(x-2019\right)\left(2020-x\right)}\le1+1\)

<=> \(1\le Q^2\le2\)

<=> \(1\le Q\le\sqrt{2}\)( do Q>0)

minQ=1 <=> \(\sqrt{\left(x-2019\right)\left(2020-x\right)}=0\)

<=> \(\left(x-2019\right)\left(2020-x\right)=0\)

<=> x=2019(tm) hoặc x=2020(t/m)

maxQ=\(\sqrt{2}\) <=> \(x-2019=2020-x\) <=> \(x=\frac{4039}{2}\) (tm)