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\(B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6+3\)
vì \(B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6\le0,\forall x\inℝ\)
\(\Rightarrow B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6+3\le3\)
Dấu "=" xảy ra khi và chỉ khi
\(\dfrac{4}{9}x-\dfrac{2}{15}=0\Rightarrow\dfrac{4}{9}x=\dfrac{2}{15}\Rightarrow x=\dfrac{9}{15}\)
Vậy \(GTLN\left(B\right)=3\left(tạix=\dfrac{9}{15}\right)\)
\(A=\left(2x+\dfrac{1}{3}\right)^4-1\)
vì \(\left(2x+\dfrac{1}{3}\right)^4\ge0,\forall x\inℝ\)
\(\Rightarrow A=\left(2x+\dfrac{1}{3}\right)^4-1\ge-1\)
Dấu "=" xảy ra khi và chỉ khi
\(2x+\dfrac{1}{3}=0\Rightarrow2x=-\dfrac{1}{3}\Rightarrow x=-\dfrac{1}{6}\)
\(\Rightarrow GTNN\left(A\right)=-1\left(tạix=-\dfrac{1}{6}\right)\)
a: =>|x-2009|=2009-x
=>x-2009<=0
=>x<=2009
b: =>2x-1=0 và y-2/5=0 và x+y-z=0
=>x=1/2 và y=2/5 và z=x+y=1/2+2/5=5/10+4/10=9/10
bai 1
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right).....\left(\dfrac{1}{10}-1\right)\)
\(A=\left(\dfrac{1-2}{2}\right)\left(\dfrac{1-3}{3}\right).....\left(\dfrac{1-9}{10}\right)\)
\(A=-\left(\dfrac{1.2.3.....8.9}{2.3....9.10}\right)=-\dfrac{1}{10}>-\dfrac{1}{9}\)
\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\\ A_{min}=\dfrac{1}{9}\Leftrightarrow x=\dfrac{3}{5}\\ B=\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\le\dfrac{2009}{2008}\\ B_{max}=\dfrac{2009}{2008}\Leftrightarrow x=\dfrac{3}{5}\\ C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\le1\dfrac{2}{3}\\ C_{max}=1\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=-4\Leftrightarrow x=-12\)
\(a,B=4,2+\left|x+1,5\right|\ge4,2\\ B_{min}=4,2\Leftrightarrow x+1,5=0\Leftrightarrow x=-1,5\\ b,C=\dfrac{4}{5}-\left|2x+1\right|\le\dfrac{4}{5}\\ C_{max}=\dfrac{4}{5}\Leftrightarrow2x+1=0\Leftrightarrow x=-\dfrac{1}{2}\)
a, Do |x +1,5| ≥ 0 ⇒ 4,2 + |x + 1,5| ≥ 4,2
Dấu "=" xảy ra ⇔ x + 1,5 = 0 ⇔ x = - 1,5
Vậy Bmin= 4,2 ⇔ x= -1,5
b, Do |2x + 1| ≥ 0 ⇒ \(\dfrac{4}{5}-\left|2x+1\right|\le\dfrac{4}{5}\)
Dấu "=" xảy ra ⇔ 2x + 1 = 0 ⇔ 2x = -1 ⇔ \(x=-\dfrac{1}{2}\)
Vậy Cmax = \(\dfrac{4}{5}\Leftrightarrow x=-\dfrac{1}{2}\)
\(A=\left|2021-x\right|+\dfrac{1}{2}\left|4040-2x\right|\)
\(A=\left|2021-x\right|+\left|2020-x\right|\)
\(A=\left|2021-x\right|+\left|x-2020\right|\ge\left|2021-x+x-2020\right|=1\)
\(A_{min}=1\) khi \(2020\le x\le2021\)
a: \(\Leftrightarrow-\dfrac{3}{2x-3}=\dfrac{2}{5}-\dfrac{3}{2}-3=\dfrac{-41}{10}\)
=>41(2x-3)=30
=>82x-123=30
=>82x=153
hay x=153/82
b: \(\Leftrightarrow\left(x-1\right)\left(7-2x\right)=0\)
=>x=1 hoặc x=7/2
c: \(\Leftrightarrow\left(\dfrac{x+1}{2018}+1\right)+\left(\dfrac{x+2}{2017}+1\right)+\left(\dfrac{x+3}{2016}+1\right)=\left(\dfrac{x+10}{2009}+1\right)+\left(\dfrac{x+11}{2008}+1\right)+\left(\dfrac{x+12}{2007}+1\right)\)
=>x+2019=0
hay x=-2019
a: \(=\dfrac{2008}{2007}-2009\cdot2-\dfrac{2009}{2007}+2009\cdot2\)
=-1/2007
b: \(=\dfrac{5^5\cdot5^3\cdot2^6-5^4\cdot5^3\cdot2^6+5^7\cdot2^{10}}{5^6\cdot2^{10}}\)
\(=\dfrac{5^8\cdot2^6-5^7\cdot2^6+5^7\cdot2^{10}}{5^6\cdot2^{10}}\)
\(=\dfrac{5^7\cdot2^6\left(5-1+2^4\right)}{5^6\cdot2^{10}}=\dfrac{5}{16}\cdot\dfrac{20}{1}=\dfrac{100}{16}=\dfrac{25}{4}\)
a: \(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)
nên \(\left\{{}\begin{matrix}2x-1=0\\y-\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=x+y=\dfrac{9}{10}\end{matrix}\right.\)
b: Bạn xem lại đề, nghiệm rất xấu
\(A=\dfrac{1}{2}+\left|2x-1\right|\ge\dfrac{1}{2}\forall x\)
\(minA=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2}\)
\(B=\dfrac{\left|x\right|+2007}{2008}\ge\dfrac{0+2007}{2008}=\dfrac{2007}{2008}\)
\(minB=\dfrac{2007}{2008}\Leftrightarrow x=0\)