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a) \(\frac{3}{4}-\left|2x+1\right|=\frac{7}{8}\)
\(\left|2x+1\right|=\frac{3}{4}-\frac{7}{8}\)
\(\left|2x+1\right|=-\frac{1}{8}\)
\(\Rightarrow x\in\varnothing\)
b) \(2.\left|2x-3\right|=\frac{1}{2}\)
\(\left|2x-3\right|=\frac{1}{4}\)
TH1: 2x - 3 = 1/4
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TH2: 2x -3 = -1/4
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rùi bn tự lm típ nhé! câu c dựa vào phần a;b là lm đk
d)\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\left|x+\frac{4}{15}\right|-3,75=-2,15\)
\(\left|x+\frac{4}{15}\right|=1,6\)
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a)\(2\left|2x-3\right|=\frac{1}{2}\)
\(\Leftrightarrow\left|2x-3\right|=\frac{1}{4}\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=\frac{1}{4}\\2x-3=-\frac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{8}\\x=\frac{11}{8}\end{matrix}\right.\)
Vậy....
b)\(7,5-3\left|5-2x\right|=-4,5\)
\(\Leftrightarrow\left|5-2x\right|=4\)
\(\Rightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{9}{2}\end{matrix}\right.\)
VẬy...
c)\(\left|3x-4\right|+\left|5-2x\right|=0\)
Có: \(\left|3x-4\right|\ge0với\forall x\\ \left|5-2x\right|\ge0với\forall x\)
\(\Rightarrow\left[{}\begin{matrix}3x-4=0\\5-2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow x\in\varnothing\)
a. \(\dfrac{3}{4}-\left|2x+1\right|=\dfrac{7}{8}\)
=> \(\left|2x+1\right|=\dfrac{3}{4}-\dfrac{7}{8}\)
=> \(\left|2x+1\right|=\dfrac{-1}{8}\)
=> \(\left\{{}\begin{matrix}2x+1=\dfrac{-1}{8}\\2x+1=\dfrac{1}{8}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=\dfrac{-9}{16}\\x=\dfrac{-7}{16}\end{matrix}\right.\)
#Yiin
b. \(2.\left|2x-3\right|=\dfrac{1}{2}\)
=> \(\left|2x-3\right|=\dfrac{1}{4}\)
=> \(\left\{{}\begin{matrix}2x-3=\dfrac{1}{4}\\2x-3=\dfrac{-1}{4}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=\dfrac{13}{8}\\x=\dfrac{11}{8}\end{matrix}\right.\)
|2x-1|=1,5
TH(1)2x-1=1,5
2x =1,5+1
2x =2,5
x =2,5 :2
x =1,25
TH(2) 2x-1=-1,5
2x =-1,5+1
2x =-0,5
x =-0,5:2
x =-0,25
các câu khác cứ tương tự bạn nhé
b) \(7,5-\left|5-2x\right|=-4,5\)
\(\left|5-2x\right|=7,5+4,7\)
\(\left|5-2x\right|=12\)
th1 :\(5-2x=12\)
\(2x=5-12\)
\(2x=-7\)
\(x=-7:2\)
\(x=-3,5\)
th2: \(5-2x=-12\)
\(2x=5+12\)
\(2x=17\)
\(x=17:2\)
\(x=8,5\)
c) \(-3+\left|x\right|=-1\)
\(\left|x\right|=-1+3\)
\(\left|x\right|=2\)
th1: \(x=-2\)
th2 : \(x=2\)
d)\(\left|2\dfrac{1}{3}-x\right|=\dfrac{1}{6}\)
\(\left|\dfrac{7}{3}-x\right|=\dfrac{1}{6}\)
th1 :\(\dfrac{7}{3}-x=\dfrac{1}{6}\)
\(x=\dfrac{7}{3}-\dfrac{1}{2}\)
\(x=\dfrac{11}{6}\)
th2: \(\dfrac{7}{3}-x=\dfrac{-1}{6}\)
\(x=\dfrac{7}{3}+\dfrac{1}{6}\)
\(x=\dfrac{-5}{2}\)
e) \(\dfrac{5}{7}-\left|x+1\right|=\dfrac{1}{14}\)
\(\left|x+1\right|=\dfrac{5}{7}-\dfrac{1}{14}\)
\(\left|x+1\right|=\dfrac{9}{14}\)
th1 :\(x+1=\dfrac{9}{14}\)
\(x=\dfrac{9}{14}-1\)
\(x=\dfrac{-5}{14}\)
th2 : \(x+1=\dfrac{-9}{14}\)
\(x=\dfrac{-9}{14}-1\)
\(x=\dfrac{-5}{14}\)
c ) \(\left|3,4-2x\right|:\frac{1}{2}=\frac{3}{4}\)
\(\left|3,4-2x\right|=\frac{3}{8}\)
\(\Rightarrow\left[\begin{array}{nghiempt}3,4-2x=\frac{3}{8}\\3,4-2x=-\frac{3}{8}\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=\frac{121}{80}\\x=\frac{151}{80}\end{array}\right.\)
Vậy ...........
e) \(\left|x+2,8\right|=3,5\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+2,8=3,5\\x+2,8=-3,5\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0,7\\x=-6,3\end{array}\right.\)
Vậy ....................
f ) \(2.\left|2x-3\right|=\frac{1}{2}\)
\(\left|2x-3\right|=\frac{1}{4}\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-3=\frac{1}{4}\\2x-3=-\frac{1}{4}\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=\frac{13}{8}\\x=\frac{11}{8}\end{array}\right.\)
Vậy ................
g ) \(7,5-3.\left|5-2x\right|=4,5\)
\(3.\left|5-2x\right|=4,5\)
\(\left|5-2x\right|=\frac{3}{2}\)
\(\Rightarrow\left[\begin{array}{nghiempt}5-2x=\frac{3}{2}\\5-2x=-\frac{3}{2}\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=\frac{7}{4}\\x=\frac{13}{4}\end{array}\right.\)
Vậy ..............
7,5 - 3 . /5-2.x/ = -4,5
4,5 . /5-2.x/ = -4,5
/5-2.x/=-4,5 : 4,5
/5-2.x/ = -1
Vì giá trị tuyệt đối k bao giờ xuống âm => x không thỏa mãn yêu cầu đề bài
\(7,5-3\cdot\left|5-2x\right|=-4,5\)
\(\Rightarrow3\cdot\left|5-2x\right|=12\)
\(\Rightarrow\left|5-2x\right|=4\)
\(\Rightarrow\left[{}\begin{matrix}5-2x=4\Rightarrow x=0,5\\5-2x=-4\Rightarrow x=4,5\end{matrix}\right.\)
3|5-2x| = 7,5 - (-4,5) = 12
=> |5-2x| = 12 : 3 = 4
=> 5-2x=4 hoặc 5-2x=-4
=> x=1/2 hoặc x=9/2
Tk mk nha
=> 3|5-2x|=7,5-(-4,5)
=> 3|5-2x|=12
=> |5-2x| = 4
=> \(\orbr{\begin{cases}5-2x=4\\5-2x=-4\end{cases}\Rightarrow\orbr{\begin{cases}2x=1\\2x=9\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{9}{2}\end{cases}}}}\)