I 3x-1I =I\(\dfrac{-1}{3}\)x+2I
Tính x
a, I 3x-2I<4
b, I3-2xI<x+1
c, I3x-1I>5
d, I3x+1I>I x-2I
e, I x-1I> I x+2I -3
g, Ix-1I+Ix+5I>8
h, Ix-3I +Ix+1I<8
a: \(\Leftrightarrow\left\{{}\begin{matrix}3x-2>-4\\3x-2< 4\end{matrix}\right.\Leftrightarrow-\dfrac{2}{3}< x< 2\)
c: \(\Leftrightarrow\left[{}\begin{matrix}3x-1>5\\3x-1< -5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -\dfrac{4}{3}\end{matrix}\right.\)
d: \(\Leftrightarrow\left[{}\begin{matrix}3x+1>x-2\\3x+1< -x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x>-3\\4x< 1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{3}{2}\\x< \dfrac{1}{4}\end{matrix}\right.\)
a) I x+1I +I x+2I + I x+3 I = 4x
b) I x+1I +I x+2I + I x+3 I + I x+4 I = 5x
c) I x+2I +I x+3/5I + x + 1/2 = 4x
a) |x+1|+|x+2+|x+3|=4x
<=> x+1+x+2+x+3=4x
<=> 3x+6=4x
<=> 6=4x-3x
<=> x=6
b) |x+1|+|x+2|+|x+3|+|x+4|=5x
<=> x+1+x+2+x+3+x+4=5x
<=> 4x+10=5x
<=> 10=5x-4x
<=> x=10
a, Tìm số thực thoả mãn I 3x - 1I = I 2x + 5I
b, Tìm số thực x,y,z thoả mãn (x-1)2 + I3y-1I + Iz+2I = 0
\(\left|3x-1\right|=\left|2x+5\right|\)
\(\Rightarrow\orbr{\begin{cases}3x-1=2x+5\\3x-1+2x+5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x-2x=5+1\\5x+4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=6\\x=-\frac{4}{5}\end{cases}}\)
Ta có: \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left|3y-1\right|\ge0\\\left|z+2\right|\ge0\end{cases}}\Rightarrow\left(x-1\right)^2+\left|3y-1\right|+\left|z+2\right|\ge0\)
Dấu "="\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left|3y-1\right|=0\\\left|z+2\right|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\3y-1=0\\x+2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{3}\\z=-2\end{cases}}\)
Vậy x = 1, \(y=\frac{1}{3}\),z = -2
a)\([x.\dfrac{1}{2}]^{3}=\dfrac{1}{27}\)
b)\([x+\dfrac{1}{2} ]^{2}=\dfrac{4}{5} \)
c) I 3x-4/5 I = 11/5
d) I 2x - 2I = 0
\(a,\left(x.\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x.\dfrac{1}{2}=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{3}:\dfrac{1}{2}=\dfrac{2}{3}\\ ---\\ b,\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{5}=\left(\dfrac{2}{\sqrt{5}}\right)^2=\left(-\dfrac{2}{\sqrt{5}}\right)^2 \\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{\sqrt{5}}\\x+\dfrac{1}{2}=-\dfrac{2}{\sqrt{5}}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\\x=-\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\end{matrix}\right.\\ Vậy:x=\pm\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\)
\(c,\left|3x-\dfrac{4}{5}\right|=\dfrac{11}{5}\\ \Rightarrow\left[{}\begin{matrix}3x-\dfrac{4}{5}=\dfrac{11}{5}\\3x-\dfrac{4}{5}=-\dfrac{11}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=\dfrac{11}{5}+\dfrac{4}{5}=3\\3x=-\dfrac{11}{5}+\dfrac{4}{5}=-\dfrac{7}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{3}=1\\x=-\dfrac{7}{5}:3=-\dfrac{7}{15}\end{matrix}\right.\\ ---\\ d,\left|2x-2\right|=0\\ \Leftrightarrow2x-2=0\\ \Leftrightarrow2x=2\\ \Leftrightarrow x=1\)
a: (x*1/2)^3=1/27
=>x*1/2=1/3
=>x=1/3:1/2=2/3
b: \(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{\sqrt{5}}\\x+\dfrac{1}{2}=-\dfrac{2}{\sqrt{5}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\sqrt{5}}{5}-\dfrac{1}{2}=\dfrac{4\sqrt{5}-5}{10}\\x=\dfrac{-4\sqrt{5}-5}{10}\end{matrix}\right.\)
c: =>3x-4/5=11/5 hoặc 3x-4/5=-11/5
=>3x=3 hoặc 3x=-7/5
=>x=-7/15 hoặc x=1
d: =>2x-2=0
=>2x=2
=>x=1
I x+1I +I x+2I + I x+3 I + I x+4 I = 5x
TÌM x, y, z, thuộc Q biết:
a,I x+1/2I+I y-3/4I+I z+1I=0
\(\left|x+\frac{1}{2}\right|+\left|y-\frac{3}{4}\right|+\left|z+1\right|=0\)
\(\Rightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|y-\frac{3}{4}\right|=0\\\left|z+1\right|=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=0-\frac{1}{2}\\y=0+\frac{3}{4}\\z=0-1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=\frac{3}{4}\\z=-1\end{cases}}\)
a) -Ix+1I+Ix+2I=x-1
b) Ix-3I- I 2x+1I
tim x :
a) I x+1I+I x -2I+I x+3I = 6
b) 2Ix+2I+I4-xi= 11
c) IxI - I 2x+3I = x-1
tim x :
a) I x+1I+I x -2I+I x+3I = 6
b) 2Ix+2I+I4-xi= 11
c) IxI - I 2x+3I = x-1