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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,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
Bài 1:
a)
\(|x+\frac{4}{15}|-|-3,75|=-|-2,15|\)
\(\Leftrightarrow |x+\frac{4}{15}|-3,75=-2,15\)
\(\Leftrightarrow |x+\frac{4}{15}|=-2,15+3,75=\frac{8}{5}\)
\(\Rightarrow \left[\begin{matrix} x+\frac{4}{15}=\frac{8}{5}\\ x+\frac{4}{15}=-\frac{8}{5}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{4}{3}\\ x=\frac{-28}{15}\end{matrix}\right.\)
b )
\(|\frac{5}{3}x|=|-\frac{1}{6}|=\frac{1}{6}\)
\(\Rightarrow \left[\begin{matrix} \frac{5}{3}x=\frac{1}{6}\\ \frac{5}{3}x=-\frac{1}{6}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{1}{10}\\ x=-\frac{1}{10}\end{matrix}\right.\)
c)
\(|\frac{3}{4}x-\frac{3}{4}|-\frac{3}{4}=|-\frac{3}{4}|=\frac{3}{4}\)
\(\Leftrightarrow |\frac{3}{4}x-\frac{3}{4}|=\frac{3}{2}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}x-\frac{3}{4}=\frac{3}{2}\\ \frac{3}{4}x-\frac{3}{4}=-\frac{3}{2}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=-1\end{matrix}\right.\)
Bài 3:
a) Ta thấy:
\(|x+\frac{15}{19}|\geq 0, \forall x\Rightarrow A\ge 0-1=-1\)
Vậy GTNN của $A$ là $-1$ khi \(x+\frac{15}{19}=0\Leftrightarrow x=-\frac{15}{19}\)
b)Vì \(|x-\frac{4}{7}|\geq 0, \forall x\Rightarrow B\geq \frac{1}{2}+0=\frac{1}{2}\)
Vậy GTNN của $B$ là $\frac{1}{2}$ khi \(x-\frac{4}{7}=0\Leftrightarrow x=\frac{4}{7}\)
Lời giải:
Trị tuyệt đối của một số luôn không âm nên :
\(M=|x+\frac{15}{19}|\geq 0\)
Vậy GTNN của $M$ là $0$ khi \(|x+\frac{15}{19}|=0\Leftrightarrow x=\frac{-15}{19}\)
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\(|x-\frac{4}{7}|\geq 0\Rightarrow Q=|x-\frac{4}{7}|-\frac{1}{2}\geq 0-\frac{1}{2}=-\frac{1}{2}\)
Vậy GTNN của $Q$ là \(-\frac{1}{2}\) khi \(|x-\frac{4}{7}|=0\Leftrightarrow x=\frac{4}{7}\)
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Nếu \(x> 5\) thì \(|x-1|=x-1; |x-5|=x-5\)
\(\Rightarrow Q=x-1+x-5=2x-6> 2.5-6=4\)
Nếu \(x<1 \Rightarrow |x-1|=1-x; |x-5|=5-x\)
\(\Rightarrow Q=1-x+5-x=6-2x>6-2.1=4\)
Nếu \(1\leq x\leq 5\Rightarrow |x-1|=x-1; |x-5|=5-x\)
\(\Rightarrow Q=x-1+5-x=4\)
Vậy GTNN của $Q$ là $4$ khi \(1\leq x\leq 5\)
\(M=\left|x+\dfrac{15}{19}\right|\)
Vì giá trị tuyệt đối của mọi số luôn luôn lớn hơn hoặc bằng 0
\(\Rightarrow M=\left|x+\dfrac{15}{19}\right|=0\)
\(\Rightarrow x+\dfrac{15}{19}=0\Rightarrow x=-\dfrac{15}{19}\)
\(Q=\left|x-\dfrac{4}{7}\right|-\dfrac{1}{2}\)
\(\left|x-\dfrac{4}{7}\right|\ge0\Rightarrow x-\dfrac{4}{7}=0\)
\(\Rightarrow x=-\dfrac{4}{7}\)
\(S=\left|x-1\right|+\left|x-5\right|\)
a: \(A=\left|x-\dfrac{7}{4}\right|+\dfrac{8}{5}>=\dfrac{8}{5}\)
Dấu = xảy ra khi x=7/4
b: \(B=\left|5-x\right|+\left|x+\dfrac{3}{4}\right|>=\left|5-x+x+\dfrac{3}{4}\right|=\dfrac{23}{4}\)
Dấu = xảy ra khi (x-5)(x+3/4)<=0
=>-3/4<=x<=5
1, \(x\left(x+\dfrac{2}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)
2, a, \(\left|x+\dfrac{4}{6}\right|\ge0\)
Để \(\left|x+\dfrac{4}{6}\right|\) đạt GTNN thì \(\left|x+\dfrac{4}{6}\right|=0\)
\(\Leftrightarrow x+\dfrac{4}{6}=0\Rightarrow x=\dfrac{-2}{3}\)
Vậy, ...
b, \(\left|x-\dfrac{1}{3}\right|\ge0\)
Để \(\left|x-\dfrac{1}{3}\right|\) đạt GTLN thì \(\left|x-\dfrac{1}{3}\right|=0\)
\(\Leftrightarrow x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)
Vậy, ...
1)
a)
\(x\cdot\left(x+\dfrac{2}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)
2)
a)
\(\left|x+\dfrac{4}{6}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x+\dfrac{4}{6}=0\Leftrightarrow x=\dfrac{-4}{6}\Leftrightarrow x=\dfrac{-2}{3}\)
Vậy \(Min_{\left|x+\dfrac{4}{6}\right|}=0\text{ khi }x=\dfrac{-2}{3}\)
b)
\(\left|x-\dfrac{1}{3}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy \(Min_{\left|x-\dfrac{1}{3}\right|}=0\text{ khi }x=\dfrac{1}{3}\)
\(a,C=\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\)
Ta có \(\left|\dfrac{1}{3}x+4\right|\ge0\)
\(\Rightarrow\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\ge1\dfrac{2}{3}\)
Dấu "=" xảy ra khi \(\left|\dfrac{1}{3}x+4\right|=0\)
\(\Leftrightarrow\dfrac{1}{3}x+4=0\)
\(\Leftrightarrow\dfrac{1}{3}x=0-4=-4\)
\(\Leftrightarrow x=-4:\dfrac{1}{3}\)
\(\Leftrightarrow x=-12\)
Vậy \(\min\limits_C=1\dfrac{2}{3}\Leftrightarrow x=-12\)
\(b,D=\left|x-6\right|+\left|x+\dfrac{5}{4}\right|\)
Ta có : \(\left\{{}\begin{matrix}\left|x-6\right|\ge-x+6\\\left|x+\dfrac{5}{4}\right|\ge x+\dfrac{5}{4}\end{matrix}\right.\)
\(\Rightarrow\left|x-6\right|+\left|x+\dfrac{5}{4}\right|\ge-x+6+x+\dfrac{5}{4}=\dfrac{29}{4}\)
Dấu "=" xảy ra khi
\(\left\{{}\begin{matrix}-x+6\ge0\\x+\dfrac{5}{4}\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le6\\x\ge-\dfrac{5}{4}\end{matrix}\right.\)
Vậy \(\min\limits_D=\dfrac{29}{4}\Leftrightarrow-\dfrac{5}{4}\le x\le6\)
b) \(D=\left|x-6\right|+\left|x+\dfrac{5}{4}\right|\)
\(D=\left|6-x\right|+\left|x+\dfrac{5}{4}\right|\ge\left|6-x+x+\dfrac{5}{4}\right|=\dfrac{29}{4}\)
Dấu = xảy ra khi \(\left(6-x\right)\left(x+\dfrac{5}{4}\right)\ge0\Leftrightarrow-\dfrac{5}{4}\le x\le6\)
vậy \(D_{min}=\dfrac{29}{4}\) khi \(-\dfrac{5}{4}\le x\le6\)
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 \(x=\dfrac{4}{7}\)