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\(A=\frac{1}{2017}-\frac{2}{2017x}+\frac{1}{x^2}=\left(\frac{1}{2017}-\frac{1}{x}\right)^2+\frac{1}{2017}-\frac{1}{2017^2}=\left(\frac{1}{2017}-\frac{1}{x}\right)^2+\frac{2016}{2017^2}\)
\(\Rightarrow A\ge\frac{2016}{2017^2}\)Dấu "=" xảy ra khi \(\left(\frac{1}{2017}-\frac{1}{x}\right)^2=0\Rightarrow x=2017\)
Vây ......
A = \(\dfrac{x^2-2x+2020}{2021x^2}\)
= \(\dfrac{2020x^2-2.2020.x+2020^2}{2021.2020x^2}\)
\(=\dfrac{2019x^2}{2021.2020x^2}+\dfrac{x^2-2.2020.x+2020^2}{2021.2020x^2}\)
= \(\dfrac{2019}{2021.2020}+\dfrac{\left(x-2020\right)^2}{2021.2020x^2}\ge\dfrac{2019}{2021.2020}\)
Dấu "=" xảy ra <=> x - 2020 = 0
<=> x = 2020
Vậy minA = \(\dfrac{2019}{2021.2020}\)đạt được tại x = 2020
\(A=\dfrac{2x}{x\left(x+y\right)}+\dfrac{6x}{\left(x-y\right)\left(x+y\right)}-\dfrac{3}{x-y}\)
\(=\dfrac{2\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}+\dfrac{6x}{\left(x-y\right)\left(x+y\right)}-\dfrac{3\left(x+y\right)}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{2x-2y+6x-3x-3y}{\left(x-y\right)\left(x+y\right)}=\dfrac{5\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{5}{x+y}\)
\(A=x^3+2x^2+3x\\ =x\left(x^2+2x+1\right)\\ =x\left(x+1\right)^2\\ =1999.\left(1999+1\right)=1999.2000\\ =3998000\)
\(B=x^4-2017x^3+2017x^2-2017x+2018\\ =x^4-2016x^3-x^3+2016x^2+x^2-2016x-x+2016+2\\ =x^3\left(x-2016\right)-x^3\left(x-2016\right)+x\left(x-2016\right)-\left(x-2016\right)+2\\ =\left(x-2016\right)\left(x^3+x-1\right)+2=0+2=0\)
Bạn xem lại đề câu a nhé , theo mk thì phải là 2 thì tính ms nhanh đc, 3 thì cũng giải đc nhưng ko hợp lí lắm
\(a,ĐK:x\ne1;x\ne-1\\ b,C=\dfrac{x^2+x+x^2+1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{2x^2+2x+1}{2x^2-2}\\ c,C=-\dfrac{1}{2}\Leftrightarrow2-2x^2=2x^2+2x+1\\ \Leftrightarrow4x^2+2x-1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}-1}{4}\\x=\dfrac{-\sqrt{5}-1}{4}\end{matrix}\right.\\ d,C>0\Leftrightarrow2x^2-2>0\left(2x^2+2x+1>0\right)\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)>0\\ \Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)
Câu b rút gọn C sai rồi, phải là \(\dfrac{1}{2\left(x+1\right)}\) chứ.
A=1/2017-2/2017x+2018/2017x^2
1/2017x=y
A=2018y^2-2y+1/2017
A=2018(y^2-2.y./2018+1/2018^2)
-1/2018+1/(2017)
A=2018(y-1/2018)^2-1/(2018.2017)
GTNNA=1/(2017.2018)
khi y=1/2018
x=2017/2018