tìm GTNN M = |5x -2010 | + |5x-2010|
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Ta có:
\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}\)
\(=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}\)
\(=1-\frac{1}{5x+6}\)
\(=\frac{5x+5}{5x+6}=\frac{2010}{2011}\)
\(\Rightarrow5x+5=2010\)
\(\Rightarrow5x=2010-5=2005\)
\(\Rightarrow x=2005:5=401\)
Vậy x=401
135 - 5x = |-10| . \(\left(-1\right)^{2010}\)
135 - 5x = 10
5x = 135 - 10
5x = 125
x = 125 : 5
x = 25
135 - 5x = |-10|.(-1)2010
135 - 5x = 10.1
135 - 5x = 10
5x = 135 - 10
5x = 125
x = 125 : 5
x = 25
=> x = 25
\(1-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\frac{1}{5x+6}=1-\frac{2010}{2011}\)
\(\frac{1}{5x+6}=\frac{1}{2011}\)
\(5x+6=2011\)
\(5x=2011-6\)
\(5x=2005\)
\(x=2005:5\)
\(x=401\)
\(1-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\frac{1}{5x+6}=1-\frac{2010}{2011}\)
\(\frac{1}{5x+6}=\frac{1}{2011}\)
=> 5x + 6 = 2011
5x = 2011 - 6
5x = 2005
x = 2005 : 5
x = 401
\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}=\frac{2010}{2011}\)
\(\Rightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\Rightarrow1-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\Rightarrow\frac{1}{5x+6}=1-\frac{2010}{2011}\)
\(\Rightarrow\frac{1}{5x+6}=\frac{1}{2011}\)
\(\Rightarrow5x+6=2011\)
\(\Rightarrow5x=2011-6\)
\(\Rightarrow5x=2005\)
\(\Rightarrow x=401\)
f(x)=0 =>5x=0
hay x=0
f(x)=1 =>5x=1
=>x=1/5
f(x)=-5
=>5x=-5
=>x=-1
f(x)=2010
=>5x=2010
hay x=402
\(E=4x^2+4xy+y^2+x^2-2x+1+y^2+4y+4+2005\)
\(=\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+2005\ge2005\)
\(E_{min}=2005\) khi \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
1-1/6+1/6-1/11+....+1/(5x+1)-1/(5x+2)=2010/2011 <=>1-1/(5x+2)=2010/2011 <=>1/2011=1/(5x+2) <=>x=401
Ta có M = |5x - 2010| + |5x - 2010| = 2.|5x - 2010|
Nhận thấy 2|5x - 2010| \(\ge0\)
=> Min M = 0
Dấu "=" xảy ra <=> 5x - 2010 = 0
<=> 5x = 2010
<=> x = 402
Vậy Min M = 0 <=> x = 402
bạn ơi, ở vế sau là 2010 hay là 2020 vậy?