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Bài 1:\(\left|2x-1\right|=2x-1\) khi \(x>0\)
b)\(\left|0,5-3x\right|=3x-0.5\) khi x= 4
c)\(\left|5x+1\right|-10x=0,5\) khi x= 0,1
Bài 2:Min A=0
Min B=-2
Bài 1:
a, \(\left|2x-1\right|=2x-1\)
+) Xét \(x\ge\dfrac{1}{2}\) ta có:
\(2x-1=2x-1\)
\(\Rightarrow x\) tùy ý với \(x\ge\dfrac{1}{2}\)
+) Xét \(x< \dfrac{1}{2}\) ta có:
\(1-2x=2x-1\)
\(\Rightarrow4x=2\)
\(\Rightarrow x=\dfrac{1}{2}\) ( không t/m )
Vậy...
b, \(\left|0,5-3x\right|=3x-0,5\)
+) Xét \(x\ge\dfrac{1}{6}\) ta có:
\(0,5-3x=3x-0,5\)
\(\Rightarrow6x=1\)
\(\Rightarrow x=\dfrac{1}{6}\) ( t/m )
+) Xét \(x< \dfrac{1}{6}\) ta có:
\(3x-0,5=3x-0,5\)
\(\Rightarrow x\) tùy ý với \(x< \dfrac{1}{6}\)
Vậy \(x\le\dfrac{1}{6}\)
c, \(\left|5x+1\right|-10x=0,5\)
+) Xét \(x\ge\dfrac{-1}{5}\) ta có:
\(5x+1-10x=0,5\)
\(\Rightarrow-5x=-0,5\)
\(\Rightarrow x=\dfrac{1}{10}\) ( t/m )
+) Xét \(x< \dfrac{-1}{5}\) ta có:
\(-5x-1-10x=0,5\)
\(\Rightarrow-15x=1,5\)
\(\Rightarrow x=\dfrac{-1}{10}\) ( không t/m )
Vậy \(x=\dfrac{1}{10}\)
Bài 2:
a, Ta có: \(-\left|x-3,5\right|\le0\)
\(\Rightarrow A=0,5-\left|x-3,5\right|\le3,5\)
Dấu " = " xảy ra khi \(-\left|x-3,5\right|=0\Rightarrow x=3,5\)
Vậy \(MIN_A=0,5\) khi x = 3,5
b, Ta có: \(-\left|1,4-x\right|\le0\)
\(\Rightarrow B=-\left|1,4-x\right|-2\le-2\)
Dấu " = " xảy ra khi \(-\left|1,4-x\right|=0\Rightarrow x=1,4\)
Vậy \(MIN_B=-2\) khi \(x=1,4\)
Ta có : (2x + 1)4 = (2x + 1)6
=> (2x + 1)4 - (2x + 1)6 = 0
<=> (2x + 1)4[1 - (2x + 1)2] = 0
\(\Leftrightarrow\orbr{\begin{cases}\left(2x+1\right)^4=0\\1-\left(2x+1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\\left(2x+1\right)^2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=-1\\\left(2x+1\right)=1;-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\2x=0;-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=0;-1\end{cases}}\)
Vậy x thuộc \(-\frac{1}{2};0;-1\)
a)
TH1: \(x< \dfrac{-2}{3}\)
<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=2-0,5x\\\left|x+\dfrac{2}{3}\right|=-x-\dfrac{2}{3}\end{matrix}\right.\)
PT <=> \(2-0,5x+x+\dfrac{2}{3}=0< =>x=\dfrac{-16}{3}\left(c\right)\)
TH2: \(\dfrac{-2}{3}\le x< 4\)
<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=2-0,5x\\\left|x+\dfrac{2}{3}\right|=x+\dfrac{2}{3}\end{matrix}\right.\)
PT <=> \(2-0,5x-x-\dfrac{2}{3}=0< =>x=\dfrac{8}{9}\left(c\right)\)
TH3: \(x\ge4\)
<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=0,5x-2\\\left|x+\dfrac{2}{3}\right|=x+\dfrac{2}{3}\end{matrix}\right.\)
PT <=> \(0,5x-2-x-\dfrac{2}{3}=0< =>x=\dfrac{-16}{3}\left(l\right)\)
KL: x \(\left\{\dfrac{-16}{3};\dfrac{8}{9}\right\}\)
b) TH1: \(x\ge-1< =>\left|x+1\right|=x+1\)
PT <=> 2x - x -1 = \(\dfrac{-1}{2}\)
<=> x = \(\dfrac{1}{2}\) (c)
TH2: x < -1 <=> \(\left|x+1\right|=-x-1\)
PT <=> 2x + x + 1 = \(\dfrac{-1}{2}\)
<=> x = \(\dfrac{-1}{2}\) (l)
KL: x \(\in\left\{\dfrac{1}{2}\right\}\)
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) \(\left(x-3\right)^{11}=\left(x-3\right)^{41}\)
\(\Rightarrow\left(x-3\right)^{11}-\left(x-3\right)^{41}=0\)
\(\Rightarrow\left(x-3\right)^{11}\left[1-\left(x-3\right)^{30}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-3=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
`@` `\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}.`
a) 9x-1/4=3/2
=>9x=3/2+1/4
=>9x=7/4
=>x=7/4:9
=>x=7/36
Vậy x=7/36
b)(4x+2):2,5=3,2:0,5
=>(4x+2):2,5=6,4
=>4x+2=6,4.2,5
=>4x+2=16
=>4x=16-2
=>4x=14
=>x=14:4
=>x=7/2
Vậy x=7/2
c) 5,4/x-2=6/7
=>5,4/x=6/7+2
=>5,4/x=20/7
=>x=5,4 :20/7
=>x=1,89
Vậy x= 1,89
d) 0,5:2=3:(2x+7)
=>3:(2x+7)=0,25
=>2x+7=3:0,25
=>2x+7=12
=>2x=12-7
=>2x=5
=>x=5/2
Vậy x=5/2
a) 9x-1/4=3/2
=>9x=3/2+1/4
=>9x=7/4
=>x=7/4:9
=>x=7/36
Vậy x=7/36
b)(4x+2):2,5=3,2:0,5
=>(4x+2):2,5=6,4
=>4x+2=6,4.2,5
=>4x+2=16
=>4x=16-2
=>4x=14
=>x=14:4
=>x=7/2
Vậy x=7/2
c) 5,4/x-2=6/7
=>5,4/x=6/7+2
=>5,4/x=20/7
=>x=5,4 :20/7
=>x=1,89
Vậy x= 1,89
d) 0,5:2=3:(2x+7)
=>3:(2x+7)=0,25
=>2x+7=3:0,25
=>2x+7=12
=>2x=12-7
=>2x=5
=>x=5/2
Vậy x=5/2
(5 - \(x\))(9\(x^2\) - 4) =0
\(\left[{}\begin{matrix}5-x=0\\9x^2-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\9x^2=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x^2=\dfrac{4}{9}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x=-\dfrac{2}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x\) \(\in\) { - \(\dfrac{2}{3}\); \(\dfrac{2}{3}\); \(5\)}
72\(x\) + 72\(x\) + 3 = 344
72\(x\) \(\times\) ( 1 + 73) = 344
72\(x\) \(\times\) (1 + 343) = 344
72\(x\) \(\times\) 344 = 344
72\(x\) = 344 : 344
72\(x\) = 1
72\(x\) = 70
\(2x\) = 0
\(x\) = 0
Kết luận: \(x\) = 0
trả lời giúp em
1/3.|2x+1|=1.5
|2x+1|=4.5
-->2x+1=4.5 hoặc 2x+1=-4.5
-->x=1.75 hoặc x=-2.75