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\(\left|2x+1\right|+\left|x+y-\frac{1}{2}\right|\le0\)
Nhận thấy: \(\left|2x+1\right|\ge0\); \(\left|x+y-\frac{1}{2}\right|\ge0\)
=> \(\left|2x+1\right|+\left|x+y-\frac{1}{2}\right|\ge0\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x+1=0\\x+y-\frac{1}{2}=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=-\frac{1}{2}\\y=1\end{cases}}\)
đến đây bạn thay x,y tìm đc vào A để tính nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề : a) Tìm GTNN A
a) \(A=\left|x-5\right|+3\)có : \(\left|x-5\right|\ge0\Rightarrow\left|x-5\right|+3\ge0\)
\(\Leftrightarrow A\ge3\)dấu "=" xảy ra khi : \(\left|x-5\right|=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)
Vậy GTNN A = 3 khi x = 5.
b) \(C=-\left|x+1\right|+5\)có : \(-\left|x+1\right|\le0\Rightarrow-\left|x+1\right|+5\le5\)
\(\Leftrightarrow C\le5\)dấu "=" xảy ra khi : \(-\left|x+1\right|=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy GTLN C = 5 khi x = -1.
\(D=5-\left|2x+3\right|\)có : \(-\left|2x+3\right|\le0\Rightarrow5-\left|2x+3\right|\le5\)
\(\Leftrightarrow D\le5\)dấu "=" xảy ra khi : \(-\left|2x+3\right|=0\Leftrightarrow2x+3=0\Leftrightarrow x=-\frac{3}{2}\)
Vậy GTLN D = 5 khi x = -3/2.
c) \(\left|x-3\right|+\left|y+1\right|=0\)có \(\left|x-3\right|\ge0;\left|y+1\right|\ge0\Rightarrow\left|x-3\right|+\left|y+1\right|\ge0\)
\(\Rightarrow\hept{\begin{cases}\left|x-3\right|=0\\\left|y+1\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-1\end{cases}}.\)
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a; \(x\) - \(\dfrac{3}{5}\) = 1 - \(\dfrac{4}{5}\) + \(\dfrac{1}{6}\)
\(x\) - \(\dfrac{3}{5}\) = \(\dfrac{30}{30}\) - \(\dfrac{24}{30}\) + \(\dfrac{5}{30}\)
\(x\) - \(\dfrac{3}{5}\) = \(\dfrac{6}{30}\) + \(\dfrac{5}{30}\)
\(x\) - \(\dfrac{3}{5}\) = \(\dfrac{11}{30}\)
\(x\) = \(\dfrac{11}{30}\) + \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{11}{30}\) + \(\dfrac{18}{30}\)
\(x\) = \(\dfrac{29}{30}\)
Vậy \(x\) = \(\dfrac{29}{30}\)
b; (- \(\dfrac{10}{4}\)) + \(\dfrac{1}{4}\) = \(\dfrac{3}{4}\) thế \(x\) của em đâu nhỉ???
c; - \(\dfrac{3}{2}\) + (\(x\) - \(\dfrac{1}{2}\)) = \(\dfrac{1}{2}\)
\(x\) - \(\dfrac{1}{2}\) = \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\)
\(x\) - \(\dfrac{1}{2}\) = 2
\(x\) = 2 + \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{4}{2}\) + \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{5}{2}\)
Vậy \(x=\dfrac{5}{2}\)
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a) \(\left|x\right|=2,1\)
x= +- 2,1
b) \(\left|x\right|=\frac{3}{4}\left(x< 0\right)\)
x= -3/4
c) \(\left|x\right|=-1\frac{2}{5}\)
\(x\in\varphi\)
d) \(\left|x\right|=0,35\left(x>0\right)\)
\(x=0,35\)
a) |x| = 2,1 <=> \(\orbr{\begin{cases}x=2,1\\x=-2,1\end{cases}}\)
b) |x| = 3/4 <=> x = - 3/4 ( do x < 0 )
c) ko tim dc x vi |x| >= 0 voi moi x
d) |x| = 0,35 <=> x = 0,35 ( do x>0 )
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left|x+\frac{2}{3}\right|\)
Ta có: \(\left|x+\frac{2}{3}\right|\ge0\forall x\)
\(A=0\Leftrightarrow\left|x+\frac{2}{3}\right|=0\Leftrightarrow x=-\frac{2}{3}\)
Vậy \(A_{min}=0\Leftrightarrow x=-\frac{2}{3}\)
\(B=\left|x\right|+\frac{1}{2}\)
Ta có: \(\left|x\right|\ge0\forall x\)
\(\Rightarrow\left|x\right|+\frac{1}{2}\ge\frac{1}{2}\forall x\)
\(B=\frac{1}{2}\Leftrightarrow\left|x\right|=0\Leftrightarrow x=0\)
Vậy \(B_{min}=\frac{1}{2}\Leftrightarrow x=0\)
Câu c,d tương tự
P/S tất cả những bài trên chỉ tìm được min, ko tìm được max.
\(x+1,25-1\frac{1}{4}+\frac{4}{5}=0\)
\(x+1,25-1,25+0,8=0\)
\(x+0,8=0\)
\(x=-0,8=-\frac{4}{5}\)
\(x+1.25-1\frac{1}{4}+\frac{4}{5}=0\)
x+1.25-1.25+4.5=0
x+1.25-1.25=0-4.5
x+1.25-1.25=-4.5
x+1.25=-4.5+1.25
x+1.25=-3.25
x=-3.25-1.25
x=-4.5