Lời giải:
Áp dụng BĐT dạng \(|a|+|b|\geq |a+b|\) ta có:
\(|x+2017|+|x-2|=|x+2017|+|2-x|\geq |x+2017+2-x|=2019\)
\(\Rightarrow A=\frac{1}{|x+2017|+|x-2|}\leq \frac{1}{2019}\)
Vậy \(A_{\max}=\frac{1}{2019}\). Dấu "=" xảy ra khi \((x+2017)(2-x)\geq 0\Leftrightarrow -2017\leq x\leq 2\)
Câu 1 : Tìm GTLN
a) \(A=\dfrac{2003}{\left(x-2\right)^2+\left(x-y\right)^6+3}\)
b) \(B=3-\left(2x+\dfrac{1}{3}\right)^6\)
c) \(C=\dfrac{x^{2016}+2017}{x^{2016}+2015}\)
Tìm GTLN của R=\(\frac{2018}{\left(x-2\right)^2+\left(x-y\right)^4+8}\)
tìm x, biết
a) \(\left|x+2\right|+\left|x-3\right|=7\)
b) \(\left|x+2\right|-6x=1\)
c) \(\left|2x+1\right|+\left|x+8\right|=4x\)
d) \(x+\left|x+2017\right|=-2017\)
Tìm x,biết
a, \(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
Với x ∉ -2,-5,-10,-17
b,\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
Với x∉1,3,8,20
c,\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
5. Tìm x biết:
a, \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+...+\left|x+10\right|=11x+1\)
b, \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
Tìm giá trị lớn nhất hoặc nhỏ nhất của biểu thức sau
a, A = \(2x^2-2\)
b, B = \(\left|x+\dfrac{1}{3}\right|-\dfrac{1}{6}\)
c, C = \(\dfrac{\left|x\right|+2017}{2018}\)
d, D = \(3-\left(x+1\right)^2\)
e, E = \(-\left|0,1+x\right|-1,9\)
f, F = \(\dfrac{1}{\left|x\right|+2017}\)
a)\(\left(x-\frac{2}{3}\right)\). \(\left(x+\frac{1}{4}\right)\)= 0
b) \(\left(x-\frac{2}{3}\right)\). \(\left(2x-\frac{3}{4}\right)\) = \(\left(3x+\frac{1}{2}\right)\). \(\left(x+\frac{2}{3}\right)\)
Tìm x
a, \(\frac{1}{3}x-\frac{2}{5}.\left(x+1\right)=0\)
b, \(\frac{x+2}{0,5}=\frac{2x+1}{2}\)
c, \(|x+\frac{1}{5}|-4=2\)
d, \(\left(\frac{1}{4}x-1,5\right)+\left(\frac{5}{6}x-3\right)-\left(\frac{5}{8}x-0,5\right)=-4,5\)
bài 2:tìm x giá trị biểu thức
a)2-\(|^{ }_{ }\)3x-\(\frac{1}{4}|\)=\(\left|-\frac{5}{4}\right|\) f)-2,5:\(\left|\frac{3}{4}x+\frac{1}{2}\right|=3\)
b)5-3.\(\left|5-2x\right|\)=-4,5 g)\(\left|2x-5\right|=\left|x=1\right|\)
c)\(\frac{3}{2}+\frac{3}{4}.\left|x-\frac{3}{4}\right|=\frac{7}{4}\)
d)\(\frac{11}{4}+\frac{3}{2}\left|4x-\frac{1}{5}\right|=\frac{7}{2}\)
e)\(\frac{21}{5}+3:\left|\frac{x}{4}-\frac{2}{3}\right|=6\) h)\(\left|\right|2x+1|-3|=\left|-5\right|\)
