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3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
Ta có bất đẳng thức giá trị tuyệt đối:
\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
Dấu \(=\)khi \(AB\ge0\).
d) \(\left|x+1\right|+\left|x+2\right|+\left|2x-3\right|\)
\(\ge\left|x+1+x+2\right|+\left|2x-3\right|\)
\(=\left|2x+3\right|+\left|3-2x\right|\)
\(\ge\left|2x+3+3-2x\right|=6\)
Dấu \(=\)khi \(\hept{\begin{cases}\left(x+1\right)\left(x+2\right)\ge0\\\left(2x+3\right)\left(3-2x\right)\ge0\end{cases}}\Leftrightarrow-1\le x\le\frac{3}{2}\).
e) \(\left|x+1\right|+\left|x+2\right|+\left|x-3\right|+\left|x-5\right|\)
\(=\left(\left|x+1\right|+\left|3-x\right|\right)+\left(\left|x+2\right|+\left|5-x\right|\right)\)
\(\ge\left|x+1+3-x\right|+\left|x+2+5-x\right|\)
\(=4+7=11\)
Dấu \(=\)khi \(\hept{\begin{cases}\left(x+1\right)\left(3-x\right)\ge0\\\left(x+2\right)\left(5-x\right)\ge0\end{cases}}\Leftrightarrow-1\le x\le3\).
Do đó phương trình đã cho vô nghiệm.
a) \(\left|2,5-x\right|-1,3=0\)
th1: \(2,5-x\ge0\Leftrightarrow x\le2,5\)
\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow2,5-x-1,3=0\Leftrightarrow x=1,2\left(tmđk\right)\)
th2: \(2,5-x< 0\Leftrightarrow x>2,5\)
\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow x-2,5-1,3=0\Leftrightarrow x=3,8\left(tmđk\right)\)
vậy \(x=1,2;x=3,8\)
b) \(1,6.\left|x-0,2\right|=0\Leftrightarrow\left|x-0,2\right|=0\Leftrightarrow x-0,2=0\Leftrightarrow x=0,2\) vậy \(x=0,2\)
c) \(\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\)
th1: \(\dfrac{1}{3}-x\ge0\Leftrightarrow x\le\dfrac{1}{3}\)
\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow\dfrac{1}{3}-x-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{-2}{21}\left(tmđk\right)\)
th2: \(\dfrac{1}{3}-x< 0\Leftrightarrow x>\dfrac{1}{3}\)
\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow x-\dfrac{1}{3}-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{16}{21}\left(tmđk\right)\)
vậy \(x=\dfrac{-2}{21};x=\dfrac{16}{21}\)
d) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
th1: \(x+\dfrac{4}{15}\ge0\Leftrightarrow x\ge\dfrac{-4}{15}\)
\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow x+\dfrac{4}{15}-3,75=-2,15\)
\(\Leftrightarrow x=\dfrac{4}{3}\left(tmđk\right)\)
th2: \(x+\dfrac{4}{15}< 0\Leftrightarrow x< \dfrac{-4}{15}\)
\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow-x-\dfrac{4}{15}-3,75=-2,15\)
\(\Leftrightarrow x=\dfrac{-28}{15}\left(tmđk\right)\)
vậy \(x=\dfrac{4}{3};x=\dfrac{-28}{15}\)
e) ta có : \(\left|x-1,5\right|\ge0\forall x\) và \(\left|2,5-x\right|\ge0\forall x\)
\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|=0\Leftrightarrow\left\{{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\) 2 giá trị này khác nhau \(\Rightarrow\) phương trình vô nghiệm
Ta có : \(\left|x+\frac{13}{14}\right|=-\left|x-\frac{3}{7}\right|\)
\(\Rightarrow\left|x+\frac{13}{14}\right|+\left|x-\frac{3}{7}\right|=0\)
Mà : \(\left|x+\frac{13}{14}\right|\ge0\forall x\)
\(\left|x-\frac{3}{7}\right|\ge0\forall x\)
Nên : \(\orbr{\begin{cases}\left|x+\frac{13}{14}\right|=0\\\left|x-\frac{3}{7}\right|=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{13}{14}=0\\x-\frac{3}{7}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{13}{14}\\x=\frac{3}{7}\end{cases}}\)
3.
a) \(\left|x\right|=2,1\Leftrightarrow\left[\begin{array}{nghiempt}x=2,1\\x=-2,1\end{array}\right.\)
b)\(\left|x\right|=\frac{17}{9}\Leftrightarrow x=-\frac{17}{9}\) (vì x/<0)
c) \(\left|x\right|=1\frac{2}{5}\Leftrightarrow\left|x\right|=\frac{7}{5}\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{7}{5}\\x=-\frac{7}{5}\end{array}\right.\)
d) \(\left|x\right|=0,35\Leftrightarrow x=0,35\) (Vì x>0)
2.
a) \(\left|x-1,7\right|=2,3\)
\(\Leftrightarrow\begin{cases}x\ge1,7\\x-1,7=2,3\end{cases}\) hoặc \(\begin{cases}x< 1,7\\1,7-x=2,3\end{cases}\)
\(\Leftrightarrow\begin{cases}x\ge1,7\\x=4\left(tm\right)\end{cases}\) hoặc \(\begin{cases}x< 1,7\\x=-0,6\left(tm\right)\end{cases}\)
Vậy x={4;-0,6}
b) đề thiếu
a) |- 2,5| = 2,5 b) |- 2,5| = - 2,5 c) |- 2,5| = - |- 2,5|
>.<
\(\left(-\frac{28}{19}\right)\times\left(-\frac{38}{14}\right)=\frac{14\times2\times19\times2}{19\times14}=4\)
>.<
\(\left(-\frac{21}{16}\right)\times\left(-\frac{24}{7}\right)=\frac{7\times3\times8\times3}{8\times2\times7}=\frac{9}{2}\)
>.<
\(\left(-\frac{12}{17}\right)\times\left(-\frac{34}{9}\right)=\frac{3\times4\times17\times2}{17\times3\times3}=\frac{8}{3}\)
>.<
\(\left|x\right|=2,1\)
\(x=\pm2,1\)
>.<
\(\left|x\right|=\frac{17}{9}\)
\(x=\pm\frac{17}{9}\)
x < 0
\(x=-\frac{17}{9}\)
>.<
\(\left|x\right|=1^2_5\)
\(x=\pm\frac{7}{5}\)
>.<
\(\left|x\right|=0,35\)
\(x=\pm0,35\)
x > 0
x = 0,35
>.<
\(\left|x-1,7\right|=2,3\)
\(x-1,7=\pm2,3\)
Th1:
x - 1,7 = 2,3
x = 2,3 + 1,7
x = 4
Th2:
x - 1,7 = - 2,3
x = - 2,3 + 1,7
x = - 0,6
>.<
\(\left|x+\frac{3}{4}\right|-\frac{1}{3}=0\)
\(\left|x+\frac{3}{4}\right|=\frac{1}{3}\)
\(x+\frac{3}{4}=\pm\frac{1}{3}\)
Th1:
x + 3/4 = 1/3
x = 1/3 - 3/4
x = \(-\frac{5}{12}\)
Th2:
\(x+\frac{3}{4}=-\frac{1}{3}\)
\(x=-\frac{1}{3}-\frac{3}{4}\)
\(x=-\frac{13}{12}\)
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
đề bài là j vậy bn???
Ta có: \(\left|x+2,5\right|+\left|x+6,5\right|+\left|x+9,5\right|=7\)
\(\Rightarrow\left(\left|x+2,5\right|+\left|x+9,5\right|\right)+\left|x+6,5\right|=7\)
Ta có: \(\left|x+2,5\right|+\left|x+9,5\right|=\left|x+2,5\right|+\left|-x-9,5\right|\ge\left|x+2,5-x-9,5\right|=\left|-7\right|=7\) ( * )
Dấu " = " xảy ra
\(\Leftrightarrow\left(x+2,5\right).\left(-x-9,5\right)\ge0\)
\(\Leftrightarrow-9,5\le x\le-2,5\)
Ta có: \(\left|x+6,5\right|\ge0\) ( ** )
Dấu " = " xảy ra
\(\Leftrightarrow\left|x+6,5\right|=0\)
\(\Leftrightarrow x+6,5=0\)
\(\Leftrightarrow x=-6,5\)
Từ ( * ) ; ( ** )
\(\Rightarrow\left|x+2,5\right|+\left|x+6,5\right|+\left|x+9,5\right|\ge7+0\)
\(\Rightarrow\left|x+2,5\right|+\left|x+6,5\right|+\left|x+9,5\right|\ge7\)
\(\Rightarrow GTNN\) của \(\left|x+2,5\right|+\left|x+6,5\right|+\left|x+9,5\right|=7\)
Dấu " = " xảy ra
\(\Leftrightarrow\hept{\begin{cases}-9,5\le x\le-2,5\\x=-6,5\end{cases}}\)
\(\Leftrightarrow x=-6,5\)