Tìm x biết
a, | 1/2 × x - 3 | = 1/2
b, | 3× x + 1/5 | - 1/3 = 2/3
c, |3× x - 1/5 | + | 15× x - 1 |= 6/2
d, | 1/2 × x - 3 | < 1/2
e, | 3× x - 4/5 | > 1/3
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a)4/5+x=2/3
x=2/3-4/5
x=-2/15
b)-5/6-x=2/3
x=-5/6-2/3
x=-3/2
c)1/2x+3/4=-3/10
1/2x=-3/10-3/4
1/2x=-21/20
x=-21/20:1/2
x=-21/10
d)x/3-1/2=1/5
x/3=1/5+1/2
x/3=7/10
10x/30=21/30
10x=21
x=21:10
x=21/10
Nguyễn Trà My
Phần a)
\(3\times\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
\(32-3x+13=76-x\)
\(116-3x=76-x\)
\(116-76=3x-x\)
\(46=2x\)
\(x=46\div2\)
\(x=13\)
a) \(\dfrac{1}{2}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{1}{2}=\dfrac{5}{6}-\dfrac{3}{6}=\dfrac{2}{6}=\dfrac{1}{3}\)
b) \(x+\dfrac{1}{4}=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2}\)
c) \(x-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Rightarrow x=\dfrac{3}{10}+\dfrac{1}{5}=\dfrac{3}{10}+\dfrac{2}{10}=\dfrac{5}{10}=\dfrac{1}{2}\)
d) \(\dfrac{5}{6}-x=\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{1}{3}=\dfrac{5}{6}-\dfrac{2}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)
e) \(\dfrac{3}{10}+x=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}-\dfrac{3}{10}=\dfrac{5}{10}-\dfrac{3}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)
g) \(x+\dfrac{1}{4}=\dfrac{3}{8}\)
\(\Rightarrow x=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{3}{8}-\dfrac{2}{8}=\dfrac{1}{8}\)
a) 12+x=5612+x=56
⇒x=56−12=56−36=26=13⇒x=56−12=56−36=26=13
b) x+14=34x+14=34
⇒x=34−14=24=12⇒x=34−14=24=12
c) x−15=310x−15=310
⇒x=310+15=310+210=510=12⇒x=310+15=310+210=510=12
d) 56−x=1356−x=13
⇒x=56−13=56−26=36=12⇒x=56−13=56−26=36=12
e) 310+x=12310+x=12
⇒x=12−310=510−310=210=15⇒x=12−310=510−310=210=15
g) x+14=38x+14=38
⇒x=38−14=38−28=18⇒x=38−14=38−28=18
Đọc tiếp
a: x*3/4=1/5
=>x=1/5:3/4=1/5*4/3=4/15
b: x*3/7=2/5
=>x=2/5:3/7=2/5*7/3=14/15
c: 1/3+2/9=2/12x
=>1/6x=3/9+2/9=5/9
=>x=5/9*6=30/9=10/3
d: 4/15*x-2/3=1/5
=>4/15*x=2/3+1/5=10/15+3/15=13/15
=>4x=13
=>x=13/4
e: x:1/7=2/3
=>x=2/3*1/7=2/21
f: 1/9:x=7/3
=>x=1/9:7/3=1/9*3/7=3/63=1/21
j: 1/4+5/12=8/3:x
=>8/3:x=3/12+5/12=8/12=2/3
=>x=4
h: =>7/4:x=1/5+1/2=7/10
=>x=7/4:7/10=10/4=5/2
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d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
a: \(\left|\dfrac{1}{2}x-3\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=\dfrac{1}{2}\\\dfrac{1}{2}x-3=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{7}{2}\\\dfrac{1}{2}x=\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=5\end{matrix}\right.\)
b: \(\left|3x+\dfrac{1}{5}\right|-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\Leftrightarrow\left|3x+\dfrac{1}{5}\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{1}{5}=1\\3x+\dfrac{1}{5}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{4}{5}\\3x=-\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{15}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d: \(\left|\dfrac{1}{2}x-3\right|< \dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-3>-\dfrac{1}{2}\\\dfrac{1}{2}x-3< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x>\dfrac{5}{2}\\\dfrac{1}{2}x< \dfrac{7}{2}\end{matrix}\right.\Leftrightarrow5< x< 7\)
e: \(\left|3x-\dfrac{4}{5}\right|>\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{4}{5}>\dfrac{1}{3}\\3x-\dfrac{4}{5}< -\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x>\dfrac{17}{15}\\3x< \dfrac{7}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{17}{45}\\x< \dfrac{7}{45}\end{matrix}\right.\)