Ta có: \(3x^2+3y^2+4xy+2x-2y+2=0\)

\(\Leftrightarrow x^2+2x+1+y^2-2y+1+2x^2+4xy+2y^2=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x^2+2xy+y^2\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2=0\)

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-1\right)^2\ge0\forall y\)

\(2\left(x+y\right)^2\ge0\forall x,y\)

Do đó: \(\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2\ge0\forall x,y\)

Dấu '=' xảy ra khi 

\(\left\{{}\begin{matrix}x+1=0\\y-1=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\\-1+1=0\left(đúng\right)\end{matrix}\right.\)

Thay x=-1 và y=1 vào biểu thức \(M=\left(x+y\right)^{2016}+\left(x+2\right)^{2017}+\left(y-1\right)^{2018}\), ta được: 

\(M=\left(-1+1\right)^{2016}+\left(-1+2\right)^{2017}+\left(1-1\right)^{2018}\)

\(=0^{2016}+1^{2017}+0^{2018}=1\)

Vậy: M=1

\(A=\left(x-1\right)\left(2x-1\right)\left(2x^2-3x-1\right)+2017\)

\(=\left(2x^2-3x+1\right)\left(2x^2-3x-1\right)+2017\)

\(=\left(2x^2-3x\right)^2-1+2017\)

\(=\left(2x^2-3x\right)^2+2016\ge2016\)

\(\Leftrightarrow2x^2-3x=0\Leftrightarrow x\left(2x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)

Vậy \(A_{min}=2016\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)

ai thấy mình làm đúng thì k cho mình nha!

A=\(\left(2x^2-3x+1\right)\left(2x^2-3x-1\right)+2017\)

ĐẶT \(2x^2-3x=t\)

\(\Leftrightarrow\left(t+1\right)\left(t-1\right)+2017\)

\(\Leftrightarrow t^2-1+2017\)

\(\Leftrightarrow t^2+2016\ge2016\left(do.t^2\ge0\right)\)

DẤU ''='' XẢY RA KHI VÀ CHỈ KHI \(t^2=0\Leftrightarrow2x^2-3x=0\Leftrightarrow x\left(2x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=0\end{cases}}\)

VẬY GTNN CỦA A LÀ 2016 TẠI X=0 HOẶC X=3/2

\(C=\dfrac{\left|X-2017\right|+2018}{\left|X-2017\right|+2019}=\dfrac{\left(\left|X-2017\right|+2019\right)-1}{\left|X-2017\right|+2019}=1-\dfrac{1}{\left|X-2017\right|+2019}\)

\(\text{Biểu thức C đạt giá trị nhỏ nhất khi }\left|x-2017\right|+2019\text{ có giá trị nhỏ nhất}\)

\(\text{Mà }\left|x-2017\right|\ge0\text{ nên }\left|x-2017\right|+2019\ge2019\)

\(\text{Dấu "=" xảy ra khi }x=2017\Rightarrow C=\dfrac{2018}{2019}\)

\(\text{Vậy giá trị nhỏ nhất của C là }\dfrac{2018}{2019}\text{ khi }x=2017\)

1.

Ta có: \(\left|2x-\dfrac{1}{2}\right|\ge0\)

\(\Rightarrow\left|2x-\dfrac{1}{2}\right|-2017\ge-2017\)

\(\Rightarrow A\ge-2017\)

Dấu "=" xảy ra \(\Leftrightarrow\left|2x-\dfrac{1}{2}\right|=0\)

\(\Leftrightarrow2x-\dfrac{1}{2}=0\)

\(\Leftrightarrow2x=\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

Vậy, Min_A = -2017 \(\Leftrightarrow x=\dfrac{1}{4}\)

1) Tìm giá trị nhỏ nhất của biểu thức:

A = \(\left|2x-\dfrac{1}{2}\right|\) - 2017

Ta có:

\(\left|2x-\dfrac{1}{2}\right|\) ≥ 0

=> \(\left|2x-\dfrac{1}{2}\right|\) - 2017 ≥ -2017

Dấu " = " xảy ra khi \(2x-\dfrac{1}{2}\) = 0 hay x = \(\dfrac{1}{4}\)

Vậy Min A = -2017 khi x = \(\dfrac{1}{4}\)

1/ Ta có :

\(\left|2x-\dfrac{1}{2}\right|\ge0\)

\(\Leftrightarrow\left|2x-\dfrac{1}{2}\right|-2017\ge-2017\)

\(\Leftrightarrow A\ge-2017\)

Dấu "=" xảy ra khi :

\(\left|2x-\dfrac{1}{2}\right|=0\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

Vậy ....

2/ Ta có :

\(\left|x+2017\right|\ge0\)

\(\Leftrightarrow-\left|x+2017\right|\le0\)

\(\Leftrightarrow2017-\left|x+2017\right|\le2017\)

\(\Leftrightarrow B\le2017\)

Dấu "=" xảy ra khi :

\(\left|x+2017\right|=0\Leftrightarrow x=-2017\)

Vậy..

a, \(A=\left|x-2017\right|+\left|2018-x\right|\ge\left|x-2017+2018-x\right|=1\)

Vậy \(Min=1\Leftrightarrow2017\le x\le2018\)

b, \(B=\dfrac{x^2+4+8}{x^2+4}=1+\dfrac{8}{x^2+4}\)

Thấy : \(x^2+4\ge4\)

\(\Rightarrow B=1+\dfrac{8}{x^2+4}\le3\)

Vậy \(Max=3\Leftrightarrow x=0\)

Lời giải:

Ta thấy: \(f(x)=\frac{x^3}{1-3x+3x^2}\Rightarrow f(1-x)=\frac{(1-x)^3}{1-3(1-x)+3(1-x)^2}=\frac{(1-x)^3}{3x^2-3x+1}\)

\(\Rightarrow f(x)+f(1-x)=\frac{x^3}{1-3x+3x^2}+\frac{(1-x)^3}{3x^2-3x+1}=\frac{x^3+(1-x)^3}{3x^2-3x+1}=1\)

Do đó:

\(f\left(\frac{1}{2017}\right)+f\left(\frac{2016}{2017}\right)=1\)

\(f\left(\frac{2}{2017}\right)+f\left(\frac{2015}{2017}\right)=1\)

............

\(f\left(\frac{1008}{2017}\right)+f\left(\frac{1009}{2017}\right)=1\)

Cộng theo vế:

\(\Rightarrow A=f\left(\frac{1}{2017}\right)+f\left(\frac{2}{2017}\right)+f\left(\frac{3}{2017}\right)+...f\left(\frac{2015}{2017}\right)+f\left(\frac{2016}{2017}\right)\)

\(=\underbrace{1+1+1...+1}_{1008}=1008\)