Tìm giá trị nhỏ nhất của:P=/x-2016/+/x-2017/.

Áp dụng BĐT /a+b/. ≤/a/+/b/. ⇒ P=/x-2016/+/x-2017/= /x-2016/+/2017-x/ lớn hơn hoặc bằng /x-2016+2017-x/=1.

Vậy GTNN của P là 1 <=> 0. ≤(x-2016)(2017-x) <=> 2016. ≤x. ≤2017. 

2/\(ĐKXĐ:x\ne-1\)

\(Q=\frac{2x^2+2}{\left(x+1\right)^2}=\frac{2\left(x+1\right)^2-4\left(x+1\right)+4}{\left(x+1\right)^2}\)

  \(=2-\frac{4}{x+1}+\frac{4}{\left(x+1\right)^2}\)

Đặt \(\frac{2}{x+1}=t\)

\(\Rightarrow Q=t^2-2t+2=\left(t-1\right)^2+1\ge1\forall t\)

\(\Rightarrow minQ=1\Leftrightarrow t=1\)

                           \(\Leftrightarrow\frac{2}{x+1}=1\)

                         \(\Leftrightarrow x=1\left(tmđkxđ\right)\)             

Ta có: \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}=\frac{2^2}{2}=2\)

=> \(A\le\frac{2019}{2.2+2016}=\frac{2019}{2020}\)

Dấu "=" xảy ra <=> a = b = 1

Câu 2)

*) Cách khác : ( Dùng delta hoặc mò để xét hiệu tìm ra GTNN và GTLN )

ĐKXĐ : \(x\ne-1\)

Ta có : \(Q-1=\frac{2x^2+2}{\left(x+1\right)^2}-1\)

\(=\frac{2x^2+2-\left(x+1\right)^2}{\left(x+1\right)^2}\)

\(=\frac{2x^2+2-x^2-2x-1}{\left(x+1\right)^2}\)

\(=\frac{\left(x-1\right)^2}{\left(x+1\right)^2}\)

Ta thấy : \(\left(x-1\right)^2\ge0\forall x,\left(x+1\right)^2\ge0\forall x\ne-1\)

\(\Rightarrow\frac{\left(x-1\right)^2}{\left(x+1\right)^2}\ge0\forall x\ne-1\)

hay : \(Q-1\ge0\Rightarrow Q\ge1\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x=1\) ( thỏa mãn ĐKXĐ )

Vậy : min \(Q=1\) tại \(x=1\)

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 

Ta có: \(\frac{x-2019}{2018}+\frac{x-2018}{2017}=\frac{x-2017}{2016}+\frac{x-2016}{2015}\)

\(\Leftrightarrow\left(\frac{x-2019}{2018}+1\right)+\left(\frac{x-2018}{2017}+1\right)=\left(\frac{x-2017}{2016}+1\right)+\left(\frac{x-2016}{2015}+1\right)\)

\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}=\frac{x-1}{2016}+\frac{x-1}{2015}\)

\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}-\frac{x-1}{2016}-\frac{x-1}{2015}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)

\(\Leftrightarrow x-1=0\)( vì \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\))

\(\Leftrightarrow x=1\)

Vạy x=1

À mình ghi nhầm!! •_•"' Sửa lại như vầy nha

c) C=(26-x)/(x-20) đạt GTNN

TXĐ: \(D=\left(-1;1\right)\)

\(B=\frac{2018x+2019\sqrt{1-x^2}+2020}{\sqrt{1-x^2}}\)

\(=\frac{2018x+2020}{\sqrt{1-x^2}}+2019\)

Đặt  \(A=\frac{2018x+2020}{\sqrt{1-x^2}}>0\)vì \(-1< x< 1\)

=> \(\sqrt{1-x^2}.A=2018x+2020\)

=> \(\left(1-x^2\right)A^2=2018^2x^2+2.2018.2020x+2020^2\)

<=> \(\left(2018^2+A^2\right)x^2+2.2018.2020x+2020^2-A^2=0\)

pt trên có nghiệm <=> \(\Delta\ge0\)<=> \(\left(2018.2020\right)^2-\left(2018^2+A^2\right).\left(2020^2-A^2\right)\ge0\)

<=> \(A^4-\left(2020^2-2018^2\right)A^2\ge0\)

<=> \(A^2-8076\ge0\)

<=> \(A\ge\sqrt{8076}\)

"=" xảy ra <=> \(x=-\frac{1009}{1010}\left(tm\right)\)

Vậy GTNN của B = \(\sqrt{8076}+2019\) đạt tại  \(x=-\frac{1009}{1010}\)

Lời giải:

Áp dụng BĐT Bunhiacopxky:

$(x^2+y^2+z^2)(1+1+1)\geq (x+y+z)^2$

$\Leftrightarrow 3B\geq (x+y+z)^2$

$\Leftrightarrow B\geq \frac{(x+y+z)^2}{3}=\frac{2019^2}{3}=1358787$

Vậy $B_{\min}=1358787$. Giá trị này đạt tại $x=y=z=673$