Tìm x
\(|0,5.x-2|-|x+3|\)
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NQ
2
10 tháng 6 2020
a) \(\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)
0,5 x + 0,6 ( x - 2 ) = 3
0,5 x + 0.6 x - 1,2 = 3
1,1 x = 4,2
x = \(\frac{42}{11}\)
Kết luận:
b) \(\frac{1}{3}x-0,5x=0,75\)
\(\frac{1}{3}x-\frac{1}{2}x=\frac{3}{4}\)
\(-\frac{1}{6}x=\frac{3}{4}\)
\(x=-\frac{9}{2}\)
Kết luận:
c) \(\frac{3}{-2}x-0,5x=75\%\)
-1,5x - 0,5x = 0,75
-2x = 0,75
x = -0,375
Kết luận:
d) \(-\frac{2}{5}x+\frac{1}{4}=75\%-\frac{3}{4}x\)
-0,4 x + 0,25 = 0,75 - 0,75 x
-0,4 x + 0,75 x = 0,75 - 0,25
0,35 x = 0,5
x = \(\frac{10}{7}\)
Kết luận:
F
10 tháng 6 2020
\(a,\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)
\(\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)
\(\left(\frac{1}{2}+\frac{3}{5}\right)x-\frac{6}{5}=3\)
\(\frac{11}{10}x-\frac{6}{5}=3\)
\(\frac{11}{10}x=\frac{21}{5}\)
\(x=\frac{42}{11}\)
\(b,\frac{1}{3}x-\frac{1}{2}x=\frac{3}{4}\)
\(\left(\frac{1}{3}-\frac{1}{2}\right)x=\frac{3}{4}\)
\(\frac{-1}{6}x=\frac{3}{4}\)
\(x=\frac{-9}{2}\)
\(c,\frac{3}{-2}x-\frac{1}{2}x=75\%\)
\(\left(\frac{3}{-2}-\frac{1}{2}\right)x=\frac{3}{4}\)
\(-2x=\frac{3}{4}\)
\(x=\frac{-3}{8}\)
\(\frac{-2}{5}x+\frac{1}{4}=75\%-\frac{3}{4}x\)
\(\frac{-2}{5}x+\frac{3}{4}x=\frac{3}{4}-\frac{1}{4}\)
\(\left(\frac{-2}{5}+\frac{3}{4}\right)x=\frac{1}{2}\)
\(\frac{7}{20}x=\frac{1}{2}\)
\(x=\frac{10}{7}\)
HN
3
H9
HT.Phong (9A5)
CTVHS
11 tháng 7 2023
1) \(5-\left(1+\dfrac{1}{3}\right):\left(1-\dfrac{1}{3}\right)\)
\(=5-\dfrac{4}{3}:\dfrac{2}{3}\)
\(=5-\dfrac{4}{3}\cdot\dfrac{3}{2}\)
\(=5-\dfrac{4}{2}\)
\(=5-2\)
\(=3\)
b) \(\left(1+\dfrac{2}{3}-\dfrac{5}{4}\right)-\left(1-\dfrac{5}{4}\right)+2022-\dfrac{2}{3}\)
\(=1+\dfrac{2}{3}-\dfrac{5}{4}-1+\dfrac{5}{4}++2022-\dfrac{2}{3}\)
\(=\left(1-1\right)+\left(\dfrac{2}{3}-\dfrac{2}{3}\right)+\left(-\dfrac{5}{4}+\dfrac{5}{4}\right)+2022\)
\(=0+0+0+2022\)
\(=2022\)
2) \(0,7^2\cdot x=0,49^2\)
\(\Rightarrow x=\dfrac{0,49^2}{0,7^2}\)
\(\Rightarrow x=\left(\dfrac{0,49}{0,7}\right)^2\)
\(\Rightarrow x=\left(0,7\right)^2\)
\(\Rightarrow x=0,49\)
b) \(x:\left(-0,5\right)^3=\left(0,5\right)^2\)
\(\Rightarrow x=\left(0,5\right)^2\cdot\left(-0,5\right)^3\)
\(\Rightarrow x=\left(-0,5\right)^5\)
\(\Rightarrow x=-\dfrac{1}{32}\)
10 tháng 7 2023
2:
a: =>x*0,49=0,49^2
=>x=0,49
b: =>x=(0,5)^2*(-1)*(0,5)^3=-(0,5)^5
NL
0
YT
3
H9
HT.Phong (9A5)
CTVHS
9 tháng 8 2023
a) \(1=\left(2x+0,5\right)^{600}\)
\(\Rightarrow1^{600}=\left(2x+0,5\right)^{600}\)
\(\Rightarrow\left[{}\begin{matrix}2x+0,5=1\\2x+0,5=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=0,5\\2x=-1,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0,25\\x=-0,75\end{matrix}\right.\)
b) \(\left(x-0,125\right)^2=0,25\)
\(\Rightarrow\left(x-0,125\right)^2=0,5^2\)
\(\Rightarrow\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)
c) \(\left(x-3\right)^{11}=\left(x-3\right)^{41}\)
\(\Rightarrow\left(x-3\right)^{11}-\left(x-3\right)^{41}=0\)
\(\Rightarrow\left(x-3\right)^{11}\left[1-\left(x-3\right)^{30}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-3=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
9 tháng 8 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
`1 = (2x + 0,5)^600`
`=> (2x+0,5)^600 = (+-1)^600`
`=> \text {TH1: } 2x + 0,5 = 1`
`=> 2x = 1 - 0,5`
`=> 2x = 0,5`
`=> x = 0,5 \div 2`
`=> x = 0,25`
`\text {TH2: } 2x + 0,5 = -1`
`=> 2x = -1 - 0,5`
`=> 2x = -1,5`
`=> x = -1,5 \div 2`
`=> x = -0,75`
Vậy, `x \in {-0,75; 0,25}.`
`b)`
`(x - 0,125)^2 = 0,25`
`=> (x - 0,125)^2 = (+-0,5)^2`
`=> `\(\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0,5+0,125\\x=-0,5+0,125\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)
Vậy, `x \in {-0,375; 0,625}.`
`c)`
`(x - 3)^11 = (x - 3)^41`
`=> (x - 3)^11 - (x - 3)^41 = 0`
`=> (x - 3)^11 * [ 1 - (x - 3)^30] = 0`
`=>`\(\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\1-\left(x-3\right)^{30}=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Vậy, `x \in {3; 4}.`
NQ
1
3 tháng 1 2023
a,3/4 . (-5/12)+3/4.(-7/12)
` 3/4 . [ - ( 5/12 + 7/12 ) ] `
`3/4 . (-1) = -3/4 `
`2/3 . x - 0,5 = 3/4 `
` x - 0,5 = 3/4 - 2/3 `
` x-0,5 = 1/12 `
` x = 1/12 + 0,5 `
` x= 7/12 `
3 tháng 1 2023
Bài 2:
a: =>2/3x=3/4+1/2=3/4+2/4=5/4
=>x=5/4:2/3=5/4*3/2=15/8
b:=>-2x+4=3x-12
=>-5x=-16
=>x=16/5
27 tháng 3 2016
0,5.x-2/3.x=5/12
(0,5-2/3).x =5/12
(-1/6).x =5/12
x =5/12:(-1/6)
x =-5/2
Vậy x=-5/2
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