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

\(=-\dfrac{1}{3}+\dfrac{\left(2x+1\right)^2}{3\left(x+\dfrac{1}{2}\right)^2+\dfrac{9}{4}}\ge-\dfrac{1}{3}\)

\(B_{min}=-\dfrac{1}{3}\) khi \(x=-\dfrac{1}{2}\)

x^2-2x+2016=(x-1)^2+2015>=2015

=> min của x^2-2x+2016=2015 khi x =1

-x^2+2x+2016=-(x-1)^2+2017=<2017

=> max -x^2+2x+2016 =2017 khi x=1

x^2-2x+1+2014>=2014 min B=2014 khi x=1

min của B = 2016

               = 0^2-2x0+2016

               =  0-0+2016

                khi  x = 0 (vì min: nhỏ nhất)

ủng hộ nhé

x khac 0

Bx^2=x^2-2x+2016

(1-B)x^2-2x+2016=0

\(\Rightarrow\Delta=1-4.\left(1-B\right).2016\ge0\Rightarrow1-4.2016+4.2016B\ge0\)

\(B\ge\frac{4.2016-1}{4.2016}=1-\frac{1}{4.2016}\)

GTNN(B)=1-1/(4.2016)

bắt hết các loại gió mùa

Ta có:

\(B=\frac{x^2-2x+2016}{x^2}\Rightarrow2016B=\frac{2015x^2+\left(x^2-2.2016x+2016^2\right)}{x^2}=2015+\frac{\left(x-2016\right)^2}{x^2}\ge2015\)

Dấu "=" xảy ra khi \(\frac{\left(x-2016\right)^2}{x^2}=0\Rightarrow x=2016\)

\(\Rightarrow2016B_{min}=2015\Rightarrow B_{min}=\frac{2015}{2016}\) khi \(x=2016\)

Giúp e với

1.

\(G=\dfrac{2}{x^2+8}\le\dfrac{2}{8}=\dfrac{1}{4}\)

\(G_{max}=\dfrac{1}{4}\) khi \(x=0\)

\(H=\dfrac{-3}{x^2-5x+1}\) biểu thức này ko có min max

2.

\(D=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{3}{2}\)

\(D_{min}=\dfrac{3}{2}\) khi \(x=4\)

\(E=\dfrac{4x^4-x^2-1}{\left(x^2+1\right)^2}=\dfrac{-\left(x^4+2x^2+1\right)+5x^4+x^2}{\left(x^2+1\right)^2}=-1+\dfrac{5x^4+x^2}{\left(x^2+1\right)^2}\ge-1\)

\(E_{min}=-1\) khi \(x=0\)

\(G=\dfrac{3\left(x^2-4x+5\right)-5}{x^2-4x+5}=3-\dfrac{5}{\left(x-2\right)^2+1}\ge3-\dfrac{5}{1}=-2\)

\(G_{min}=-2\) khi \(x=2\)

Giusp e ạ !