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)

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)

\(x=\dfrac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\right)=\sqrt{6}\)

\(y=\sqrt{\left(\sqrt{6}-1\right)^2}=\sqrt{6}-1\)

\(\Rightarrow x-y=1\Rightarrow P=1\)

\(B=x-2020-\sqrt{x-2020}+\dfrac{1}{4}+\dfrac{8079}{4}\)

\(B=\left(\sqrt{x-2020}-\dfrac{1}{2}\right)^2+\dfrac{8079}{4}\ge\dfrac{8079}{4}\)

\(B_{min}=\dfrac{8079}{4}\) khi \(x=\dfrac{8081}{4}\)

a . ta có : \(1\le1+\sqrt{2-x}\Rightarrow GTNN=1\)

\(-2\le\sqrt{x-3}-2\Rightarrow GTNN=-2\)

b. \(0\le\sqrt{4-x^2}\le2\)

\(\sqrt{2x^2-x+3}=\sqrt{2\left(x^2-\frac{x}{2}+\frac{1}{16}\right)+\frac{23}{8}}=\sqrt{2\left(x-\frac{1}{4}\right)^2+\frac{23}{8}}\ge\frac{\sqrt{46}}{4}\)

vậy \(GTNN=\frac{\sqrt{46}}{4}\)

ta có : \(0\le-x^2+2x+5=-\left(x-1\right)^2+6\le6\)

\(\Rightarrow1-\sqrt{6}\le1-\sqrt{-x^2+2x+5}\le1\)Vậy \(\hept{\begin{cases}GTNN=1-\sqrt{6}\\GTLN=1\end{cases}}\)

Sửa đề: \(M=2019\sqrt{x-2}+2020\sqrt{10-y}\)

+Có: \(\sqrt{x-2}\ge với\forall x\\ \sqrt{10-y}\ge0với\forall x\\ \Rightarrow2019\sqrt{x-2}+2020\sqrt{10-y}\ge0\\ \Leftrightarrow M\ge0\)

+Dấu ''='' xảy ra khi

\(\sqrt{x-2}=0\\ \Leftrightarrow x=2\)

\(\sqrt{10-y}=0\\ \Leftrightarrow y=10\)

+Vậy \(M_{min}=0\) khi \(x=2,y=10\)

1. B = | x - 2018 | + | x - 2019 | + | x - 2020 |

= ( | x - 2018 | + | x - 2020 | ) + | x - 2019 | 

= ( | x - 2018 | + | 2020 - x | ) + | x - 2019 |

Vì \(\hept{\begin{cases}\left|x-2018\right|+\left|2020-x\right|\ge\left|x-2018+2020-x\right|=2\\\left|x-2019\right|\ge0\end{cases}}\)=> B ≥ 2 ∀ x

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(x-2018\right)\left(2020-x\right)\ge0\\x-2019=0\end{cases}}\Rightarrow x=2019\)

Vậy MinB = 2 <=> x = 2019

2. ĐKXĐ : x ≥ 0

Ta có : \(\sqrt{x}+3\ge3\forall x\ge0\)

=> \(\frac{2019}{\sqrt{x}+3}\le673\forall x\ge0\). Dấu "=" xảy ra <=> x = 0 (tm)

Vậy MaxC = 673 <=> x = 0

Bài 1 : 

\(B=\left|x-2018\right|+\left|x-2019\right|+\left|x-2020\right|\)

Ta có : \(\left|x-2018\right|\ge0\forall x;\left|x-2019\right|\ge0\forall x;\left|x-2020\right|\ge0\forall x\)

\(\left|x-2018\right|+\left|x-2019\right|+\left|x-2020\right|\ge0\)

Dấu ''='' xảy ra khi \(x=2018;x=2019;x=2020\)

Vậy GTNN B là 0 khi x = 2018 ; x = 2019 ; x = 2020 

Mọi người giải giúp em nhé

Tính hợp lí

(2018/2017-2019/2018+2020/2019)×(1/2-

1/3-1/6)×(1/2+1/3+1/4+...+1/2020)

Em cảm ơn

Tìm Max trước thôi nhé, Min nghĩ sau:V

a) đk: \(1\le x\le4\)

Ta có: \(A=\sqrt{x-1}+\sqrt{4-x}\)

=> \(A^2=\left(\sqrt{x-1}+\sqrt{4-x}\right)\le\left(1^2+1^2\right)\left(x-1+4-x\right)=2.3=6\)

=> \(A\le\sqrt{6}\) ( BĐT Bunhiacopxki)

Dấu "=" xảy ra khi: \(x-1=4-x\Rightarrow x=\frac{5}{2}\)

Vậy Max(A) = \(\sqrt{6}\) khi x = 5/2

b) đk: \(-1\le x\le6\)

Tương tự sử dụng BĐT Bunhiacopxki:

\(B\le\sqrt{\left(1^2+1^2\right)\left(x+1+6-x\right)}=\sqrt{2.7}=\sqrt{14}\)

Dấu "=" xảy ra khi: \(x+1=6-x\Rightarrow x=\frac{5}{2}\)

Vậy Max(B) = \(\sqrt{14}\) khi \(x=\frac{5}{2}\)

Min:

Áp dụng BĐT \(\sqrt{A}+\sqrt{B}\ge\sqrt{A+B}\) . Dấu "=" xảy ra khi \(AB=0\):

\(A=\sqrt{x-1}+\sqrt{4-x}\ge\sqrt{x-1+4-x}=\sqrt{3}\)

Dấu "=" xảy ra khi \(\left(x-1\right)\left(4-x\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=4\end{cases}}\)

\(B=\sqrt{x+1}+\sqrt{6-x}\ge\sqrt{x+1+6-x}=\sqrt{7}\)

Dấu "=" xảy ra khi \(\left(x+1\right)\left(6-x\right)=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=6\end{cases}}\)