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\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\\ A_{min}=\dfrac{1}{9}\Leftrightarrow x=\dfrac{3}{5}\\ B=\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\le\dfrac{2009}{2008}\\ B_{max}=\dfrac{2009}{2008}\Leftrightarrow x=\dfrac{3}{5}\\ C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\le1\dfrac{2}{3}\\ C_{max}=1\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=-4\Leftrightarrow x=-12\)
\(1,\\ a,=\left(\dfrac{1}{4}\right)^3\cdot32=\dfrac{1}{64}\cdot32=\dfrac{1}{2}\\ b,=\left(\dfrac{1}{8}\right)^3\cdot512=\dfrac{1}{512}\cdot512=1\\ c,=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\\ d,=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=3\\ 2,\\ a,A=\left|x-\dfrac{3}{4}\right|\ge0\\ A_{min}=0\Leftrightarrow x=\dfrac{3}{4}\\ b,B=1,5+\left|2-x\right|\ge1,5\\ A_{min}=1,5\Leftrightarrow x=2\\ c,A=\left|2x-\dfrac{1}{3}\right|+107\ge107\\ A_{min}=107\Leftrightarrow2x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{6}\)
\(d,M=5\left|1-4x\right|-1\ge-1\\ M_{min}=-1\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\\ 3,\\ a,C=-\left|x-2\right|\le0\\ C_{max}=0\Leftrightarrow x=2\\ b,D=1-\left|2x-3\right|\le1\\ D_{max}=1\Leftrightarrow x=\dfrac{3}{2}\\ c,D=-\left|x+\dfrac{5}{2}\right|\le0\\ D_{max}=0\Leftrightarrow x=-\dfrac{5}{2}\)
a/ Với mọi x ta có :
\(\left|x+\dfrac{15}{19}\right|\ge0\)
\(\Leftrightarrow M\ge0\)
Dấu "=" xảy ra khi :
\(\Leftrightarrow\left|x+\dfrac{15}{19}\right|=0\Leftrightarrow x=-\dfrac{15}{19}\)
Vậy \(M_{Min}=0\Leftrightarrow x=-\dfrac{15}{19}\)
b/ Với mọi x ta có :
\(\left|x-\dfrac{4}{7}\right|\ge0\)
\(\Leftrightarrow\left|x-\dfrac{4}{7}\right|-\dfrac{1}{2}\ge-\dfrac{1}{2}\)
\(\Leftrightarrow N\ge-\dfrac{1}{2}\)
Dấu "=" xảy ra khi :
\(\Leftrightarrow\left|x-\dfrac{4}{7}\right|=0\Leftrightarrow x=\dfrac{4}{7}\)
Vậy \(N_{Min}=-\dfrac{1}{2}\Leftrightarrow x=\dfrac{4}{7}\)
\(A=0,6+\left|\dfrac{1}{2}-x\right|\\ Vì:\left|\dfrac{1}{2}-x\right|\ge\forall0x\in R\\ Nên:A=0,6+\left|\dfrac{1}{2}-x\right|\ge0,6\forall x\in R\\ Vậy:min_A=0,6\Leftrightarrow\left(\dfrac{1}{2}-x\right)=0\Leftrightarrow x=\dfrac{1}{2}\)
\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\\ Vì:\left|2x+\dfrac{2}{3}\right|\ge0\forall x\in R\\ Nên:B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\forall x\in R\\ Vậy:max_B=\dfrac{2}{3}\Leftrightarrow\left|2x+\dfrac{2}{3}\right|=0\Leftrightarrow x=-\dfrac{1}{3}\)
Bài 1:
$M=\frac{27}{x-15}-1$
Để $M$ min thì $\frac{27}{x-15}$ min.
Để $\frac{27}{x-15}$ min thì $x-15$ là số âm lớn nhất
$\Rightarrow x$ là số nguyên lớn nhất nhỏ hơn 15
$\Rightarrow x=14$
Khi đó: $M_{\min}=\frac{42-14}{14-15}=-28$
Bài 2:
\(\left(\dfrac{1}{2}\right)^x+\left(\dfrac{1}{2}\right)^{x-4}=17\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{x-4}\left[\left(\dfrac{1}{2}\right)^4+1\right]=17\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{x-4}.\dfrac{17}{16}=17\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{x-4}=16=\left(\dfrac{1}{2}\right)^{-4}\)
$\Rightarrow x-4=-4\Leftrightarrow x=0$
a,ta có:
\(\left|x+\dfrac{15}{19}\right|\ge0\\ \Rightarrow\min\limits M=0\)
khi và chỉ khi \(x+\dfrac{15}{19}=0\\ \Rightarrow x=-\dfrac{15}{19}\)
Vậy minM=0
\(C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\)
\(\Rightarrow C=-2\left|\dfrac{1}{3}x+4\right|+\dfrac{5}{3}\)
mà \(-2\left|\dfrac{1}{3}x+4\right|\le0,\forall x\)
\(\Rightarrow C=-2\left|\dfrac{1}{3}x+4\right|+\dfrac{5}{3}\le\dfrac{5}{3}\)
\(\Rightarrow GTLN\left(C\right)=\dfrac{5}{3}\left(tạix=-12\right)\)
a) Ta có : $M=|x+\dfrac{15}{19}|\geq 0$
Vậy MinM = 0 khi và chỉ khi $<=>x+\dfrac{15}{19}=0=>x=\dfrac{-15}{19}$.
b) Ta có : $|x-\dfrac{4}{7}|\geq 0$
$=>N=|x-\dfrac{4}{7}|-\dfrac{1}{2}\geq \dfrac{-1}{2}$
Vậy MinN = $\dfrac{-1}{2}$ khi và chỉ khi $<=>|x-\dfrac{4}{7}|=0=>x=\dfrac{4}{7}$
a, \(M=\left|x+\dfrac{15}{19}\right|\ge0\)
Dấu " = " khi \(\left|x+\dfrac{15}{19}\right|=0\Leftrightarrow x=\dfrac{-15}{19}\)
Vậy \(MIN_M=0\) khi \(x=\dfrac{-15}{19}\)
b, \(\left|x-\dfrac{4}{7}\right|\ge0\Leftrightarrow N=\left|x-\dfrac{4}{7}\right|-\dfrac{1}{2}\ge\dfrac{-1}{2}\)
Dấu " = " khi \(\left|x-\dfrac{4}{7}\right|=0\Leftrightarrow x=\dfrac{4}{7}\)
Vậy \(MIN_N=\dfrac{-1}{2}\) khi \(x=\dfrac{4}{7}\)