Để A có GTNN thì\(-x^2+2x-4\) có GTLN

Mà \(-x^2+2x-4=-\left(x^2-2x+4\right)\)

Để \(-x^2+2x-4\)có GTLN thì\(x^2-2x+4\)có GTNN

Mà \(x^2-2x+4=x^2-2x+1+3=\left(x-1\right)^2+3\)

    \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+3\ge3\)

Dấu ''='' xảy ra khi \(x-1=0\Leftrightarrow x=1\)

Vậy biểu thức A có GTNN là 3 khi và chỉ khi \(x=1\)

Vì \(x^2+x+1=x^2+2\cdot\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

\(\Rightarrow\hept{\begin{cases}x=0\\x+1=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}\)

Vậy ....

a)

(x-2y)² >= 0 V x,y

(y-2018)>=0  V y

=> P=(ghi lại đề) >= 0

vậy GTNN của p bằng 0

dấu "=" xảy ra (=) \(\hept{\begin{cases}x-2y=0\\y-2018=0\end{cases}}\left(=\right)\hept{\begin{cases}x=2y\\y=2018\end{cases}}\left(=\right)\hept{\begin{cases}y=2018\\x=4036\end{cases}}\)

b) (x+y-3)⁴ >= 0 V x,y

(x-2y)² >= V x,y

=> Q=(ghi lại đề) >= 2018

vậy GTNN của Q bằng 2018

dấu "=" xảy ra (=) \(\hept{\begin{cases}x+y-3=0\\x-2y=0\end{cases}}\left(=\right)\hept{\begin{cases}x=2y\\3y=3\end{cases}}\left(=\right)\hept{\begin{cases}y=1\\x=2\end{cases}}\)

c) 

(2x + 1/6)⁴>= 0 V x

=> N=(ghi lại đề) >= -2

vậy GTNN của N bằng -2

dấu "=" xảy ra (=) 2x+1/6=0

(=) 2x=-16

(=) x=-1/12

#Học-tốt

Menhera-Kun chữ V rồi gạch đó là j

\(ĐKXĐ:x\ne-1;x\ne2\)

\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)

\(\Rightarrow\frac{x-2+5x+5}{\left(x+1\right)\left(2-x\right)}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)

\(\Rightarrow x-2+5x+5=15\)

\(\Rightarrow6x+3=15\Leftrightarrow6x=12\Leftrightarrow x=2\)

Vậy x = 2

x=2 ko thỏa mãn nên loại

pt vô nghiêmj

a) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Rightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Mà \(\left(\frac{1}{9}< \frac{1}{8}< \frac{1}{7}< \frac{1}{6}\right)\)nên \(\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)< 0\)

\(\Rightarrow x+10=0\Rightarrow x=-10\)

Vậy x = -10

b) \(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)

\(\Rightarrow\frac{x}{2012}-1+\frac{x+1}{2013}-1+\frac{x+2}{2014}-1\)

\(+\frac{x+3}{2015}-1+\frac{x+4}{2016}-1=0\)

\(\Rightarrow\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}\)\(+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\)

\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)

Mà \(\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)nên x - 2012 = 0

Vậy x = 2012

a, (x+1)/9 +1 + (x+2)/8  =  (x+3)/7 + 1 + (x+4)/6 + 1

<=> (x+10)/9 +(x+10)/8 = (x+10)/7 + (x+10)/6

<=> (x+10). (1/9 +1/8 - 1/7 -1/6) =0

vì 1/9 +1/8 -1/7 - 1/6 khác 0

=> x+10=0

=> x=-10

\(\frac{3x+2}{x-1}+\frac{2x-4}{x+2}=5\)

<=> \(\frac{\left(3x+2\right)\left(x+2\right)+\left(x-1\right)\left(2x-4\right)}{\left(x-1\right)\left(x+2\right)}=5\)

<=> \(3x^2+8x+4+2x^2-6x+4=5\left(x^2+x-2\right)\)

<=> \(5x^2+2x+8=5x^2+5x-10\)

<=> \(5x^2-5x^2+2x-5x=-10-8\)

<=> \(-3x=-18\)

<=> \(x=6\)

Vậy S = {6}

\(\frac{3x+2}{x-1}+\frac{2x-4}{x+2}=5\)

\(\Leftrightarrow\frac{\left(3x+2\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}+\frac{\left(2x-4\right)\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=5\)

\(\Leftrightarrow\frac{3x^2+6x+2x+4+2x^2-2x-4x+4}{\left(x+2\right)\left(x-1\right)}=5\)

\(\Leftrightarrow\frac{5x^2+2x+8}{\left(x+2\right)\left(x-1\right)}=5\)

\(\Leftrightarrow5x^2+2x+8=5\left(x^2-x+2x-2\right)\)

<=> 5x²+2x+8=5(x²+x-2)

<=>5x²+2x+8=5x²+5x-10

<=> 5x²-5x²+2x-5x=-10-8

<=> -3x=-18

<=> x=6