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\(M=1-\dfrac{2}{a}+\dfrac{2008}{a^2}=2008\left(\dfrac{1}{a^2}-2.\dfrac{1}{a}.\dfrac{1}{2008}+\dfrac{1}{2008^2}\right)+\dfrac{2007}{2008}\)
\(M=2008\left(\dfrac{1}{a}-\dfrac{1}{2008}\right)^2+\dfrac{2007}{2008}\ge\dfrac{2007}{2008}\)
\(\Rightarrow M_{min}=\dfrac{2007}{2008}\) khi \(\dfrac{1}{a}-\dfrac{1}{2008}=0\Rightarrow a=2008\)
\(M-\frac{2020}{2011}=\frac{a^2-2a+2011}{a^2}-\frac{2010}{2011}\)
\(=\frac{2011a^2-2.2011a+2011^2-2010a^2}{2011a^2}\)
\(=\frac{a^2-2.2011a+2011^2}{2011a^2}=\frac{\left(a-2011\right)^2}{2011a^2}\ge0\)
\(\Rightarrow M\ge\frac{2010}{2011}\)
Vậy giá trị nhỏ nhất của \(M=\frac{2010}{2011}\) khi \(a-2011=0\Rightarrow a=2011\)
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
Giải
a, 2A+3B=0 <=> \(\dfrac{10}{2m+1}+\dfrac{12}{2m-1}=0\)
<=>10(2m-1)+ 12(2m+1) =0
<=> 44m +2 =0
<=> m=-1/22
b, AB= A+B <=> \(\dfrac{20}{\left(2m-1\right)\left(2m+1\right)}=\dfrac{5}{2m+1}+\dfrac{4}{2m-1}\)
<=> 20 = 5(2m -1) + 4(2m+1)
<=> 20 = 18m - 1
<=> m=7/6
`a)D` xác định `<=>a-1 ne 0<=>a ne 1`
`b)` Với `a ne 1` có:
`D=([a-1]/[a^2+a+1]-[1-3a+a^2]/[(a-1)(a^2+a+1)]-1/[a-1]).[1-a]/[a^2+1]`
`D=[(a-1)^2-1+3a-a^2-a^2-a-1]/[(a-1)(a^2+a+1)].[-(a-1)]/[a^2+1]`
`D=[a^2-2a+1-1+3a-a^2-a^2-a-1]/[(-a^2-1)(a^2+a+1)]`
`D=[-a^2-1]/[(-a^2-1)(a^2+a+1)]=1/[a^2+a+1]`
`c)` Với `a ne 1` có:
`1/D=1/[1/[a^2+a+1]]=a^2+a+1=(a+1/2)^2+3/4`
Vì `(a+1/2)^2 >= 0 AA a ne 1`
`=>(a+1/2)^2+3/4 >= 3/4 AA a ne 1`
Hay `1/D >= 3/4 AA a ne 1=>1/D _[mi n]=3/4`
Dấu "`=`" xảy ra `<=>a=-1/2` (t/m).
\(M=\frac{a^2-2a+2008}{a^2}\)
\(M=\frac{a^2}{a^2}-\frac{2a}{a^2}+\frac{2008}{a^2}\)
\(M=1-\frac{2}{a}+\frac{2008}{a^2}\)
\(M=1-2\cdot\frac{1}{a}+2008\cdot\left(\frac{1}{a}\right)^2\)
Đặt \(\frac{1}{a}=x\)
Ta có :
\(M=1-2x+2008x^2\)
\(M=2008\left(x^2-x\cdot\frac{1}{1004}+\frac{1}{2008}\right)\)
\(M=2008\left(x^2-2\cdot x\cdot\frac{1}{2008}+\frac{1}{2008^2}+\frac{2007}{2008^2}\right)\)
\(M=2008\left[\left(x-\frac{1}{2008}\right)^2+\frac{2007}{2008^2}\right]\)
\(M=2018\left(x-\frac{1}{2008}\right)^2+\frac{2007}{2008}\ge\frac{2007}{2008}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2008}\)