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3.
a) \(\left(x-1\right)^3=125\)
=> \(\left(x-1\right)^3=5^3\)
=> \(x-1=5\)
=> \(x=5+1\)
=> \(x=6\)
Vậy \(x=6.\)
b) \(2^{x+2}-2^x=96\)
=> \(2^x.\left(2^2-1\right)=96\)
=> \(2^x.3=96\)
=> \(2^x=96:3\)
=> \(2^x=32\)
=> \(2^x=2^5\)
=> \(x=5\)
Vậy \(x=5.\)
c) \(\left(2x+1\right)^3=343\)
=> \(\left(2x+1\right)^3=7^3\)
=> \(2x+1=7\)
=> \(2x=7-1\)
=> \(2x=6\)
=> \(x=6:2\)
=> \(x=3\)
Vậy \(x=3.\)
Chúc bạn học tốt!
a, \(A=\dfrac{2}{3}+\left|5-x\right|\)
Vì \(\left|5-x\right|\ge0\forall x\Rightarrow\dfrac{2}{3}+\left|5-x\right|\ge\dfrac{2}{3}\forall x\)
\(\Rightarrow A_{Min}=\dfrac{2}{3}\Leftrightarrow\dfrac{2}{3}+\left|5-x\right|=\dfrac{2}{3}\)
\(\Leftrightarrow\left|5-x\right|=0\Leftrightarrow5-x=0\Leftrightarrow x=5\)
Vậy, A đạt GTNN là \(\dfrac{2}{3}\Leftrightarrow x=5\)
b, \(B=5\left(x-2\right)^2+11\)
Vì \(5\left(x-2\right)^2\ge0\forall x\Rightarrow5\left(x-2\right)^2+11\ge11\forall x\)
\(\Rightarrow B_{Min}=11\Leftrightarrow5\left(x-2\right)^2+11=11\)
\(\Leftrightarrow5\left(x-2\right)^2=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy, B đạt GTNN là \(11\Leftrightarrow x=2\)
c, \(C=0,5-\left|x-4\right|=-\left|x-4\right|+0,5\)
Vì \(\left|x-4\right|\ge0\forall x\Rightarrow-\left|x-4\right|\le0\forall x\Rightarrow-\left|x-4\right|+5\le5\forall x\)
\(\Rightarrow C_{Max}=5\Leftrightarrow-\left|x-4\right|+5=5\)
\(\Leftrightarrow-\left|x-4\right|=0\Leftrightarrow\left|x-4\right|=0\Leftrightarrow x-4=0\Leftrightarrow x=4\)
Vậy, C đạt GTLN là \(5\Leftrightarrow x=4\)
d, \(D=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|=-\left|2x+\dfrac{2}{3}\right|+\dfrac{2}{3}\)
Vì \(\left|2x+\dfrac{2}{3}\right|\ge0\forall x\Rightarrow-\left|2x+\dfrac{2}{3}\right|\le0\forall x\Rightarrow-\left|2x+\dfrac{2}{3}\right|+\dfrac{2}{3}\le\dfrac{2}{3}\forall x\)
\(\Rightarrow D_{Max}=\dfrac{2}{3}\Leftrightarrow-\left|2x+\dfrac{2}{3}\right|+\dfrac{2}{3}=0\)
\(\Leftrightarrow-\left|2x+\dfrac{2}{3}\right|=\dfrac{-2}{3}\Leftrightarrow2x+\dfrac{2}{3}=\dfrac{2}{3}\Leftrightarrow2x=0\Leftrightarrow x=0\)
Vậy, D đạt GTLN là \(\dfrac{2}{3}\Leftrightarrow x=0\)
-- còn lại làm tương tự =))
Đính chính lại
\(...2^{1+2+...+x}< 2^{11}\Rightarrow2^{\dfrac{x\left(x+1\right)}{2}}< 2^{11}\Rightarrow\dfrac{x\left(x+1\right)}{2}< 11\)
\(\Rightarrow x\left(x+1\right)< 22\)
Vì \(4.5=20< 22;5.6=30>22\)
\(\Rightarrow x=4\left(x\in N\right)\) lớn nhất thỏa mãn (1)
\(2.2^2.2^3....2^x< 2^{11}\left(1\right)\)
\(\Rightarrow2^{1+2+...+x}< 2^{11}\)
\(\Rightarrow2^{x\left(x+1\right)}< 2^{11}\)
\(\Rightarrow x\left(x+1\right)< 11\)
vì \(2.\left(2+1\right)=6< 11;3.\left(3+1\right)=12>11\)
\(\Rightarrow x=2\left(x\in N\right)\) lớn nhất thỏa mãn (1)