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a)\(1,5-2\left|x\right|=-0,5\)
\(\Leftrightarrow2\left|x\right|=1,5+0,5\)
\(\Leftrightarrow\left|x\right|=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
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}.`
1)\(x+0,5+x+1,5+x+2,5=33\)
\(\Leftrightarrow3x=33-0,5-1,5-2,5=28,5\)
\(\Leftrightarrow x=9,5\)
2)\(\left(x+0,9\right)\left(1-0,4\right)=2412\)
\(\Leftrightarrow\left(x+0,9\right)\cdot0,6=2412\)
\(\Leftrightarrow x+0,9=4020\)
\(\Leftrightarrow x=1019,1\)
Lời giải:
$x+\frac{2}{-15}=\frac{-5}{3}$
$x=\frac{-5}{3}-\frac{2}{-15}=\frac{-5}{3}+\frac{2}{15}$
$x=\frac{-23}{15}$
a) \(\dfrac{3,5}{15}=\dfrac{-2}{x}\)
\(\Rightarrow x=\dfrac{15.-2}{3,5}\)
\(\Rightarrow x=-8,57\)
b) \(2\left(3x-2\right)-3\left(x-2\right)-=-1\)
\(\Rightarrow6x-4-3x+6=-1\)
\(\Rightarrow6x-3x=-1+4-6\)
\(\Rightarrow3x=-3\)
\(\Rightarrow x=-\dfrac{3}{3}=-1\)
a) \(\dfrac{x.2}{-15}=\dfrac{-5}{3}\)
\(\dfrac{x.2}{-15}=\dfrac{25}{-15}\)
x.2=25
x=12,5
b) \(\dfrac{x-1}{-12}=\dfrac{-3}{x-1}\)
(x-1)2=-3.(-12)
(x-1)2=36
⇒(x-1)2\(\Rightarrow\left[{}\begin{matrix}x-1=6\\x-1=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)
2,8:0,5=|x-1|:1,5
|x-1|:1,5=5,6
|x-1|=8,1
\(\Rightarrow\orbr{\begin{cases}x-1=8,1\\x-1=-8,1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=9,1\\x=-7,1\end{cases}}\)
Vậy x=9,1 hoặc x=-7,1
\(2,8:0,5=\left|x-1\right|:1,5\)
\(\Rightarrow5,6=\left|x-1\right|:1,5\)
\(\Rightarrow\left|x-1\right|=5,6.1,5\)
\(\Rightarrow\left|x-1\right|=8,4\)
\(\Rightarrow\orbr{\begin{cases}x-1=8,4\\x-1=-8,4\end{cases}\Rightarrow\orbr{\begin{cases}x=8,4+1\\x=-8,4+1\end{cases}\Rightarrow}\orbr{\begin{cases}x=9,4\\x=-7,4\end{cases}}}\)
Vậy \(x\in\left\{9,4;-7,4\right\}\)
Chúc em học tốt nhé!
\(ĐK:x\ge0\\ PT\Leftrightarrow\sqrt{x}=\dfrac{6}{2}=3\Leftrightarrow x=9\left(tm\right)\)
a. \(2\sqrt{x}+1=7\)
\(2\sqrt{x}=7-1\)
\(2\sqrt{x}=6\)
\(\sqrt{x}=6:2\)
\(\sqrt{x}=3\)
\(\Rightarrow\) \(x=3^2\)
\(x=9\)
\(\left(\frac{1}{4}x-1,5\right)+\left(\frac{5}{6}-3\right)-\left(\frac{5}{8}x-0,5\right)=45\)
\(\frac{1}{4}x-1,5-\frac{13}{6}-\frac{5}{8}x+0,5=45\)
\(-\frac{3}{8}x-\frac{19}{6}=45\)
\(-\frac{3}{8}x=\frac{289}{6}\)
\(\Rightarrow x=-\frac{1156}{9}\)
cảm ơn bạn anh nha mà cho em hỏi tại sao x lại ghép đc với nhau z?
\(1,5-2\cdot|x|=-0,5\)
=> \(2\cdot|x|=1,5+0,5\)
=> \(2\cdot|x|=2\)
=> \(|x|=1\)
=> \(x=1\) hoặc \(x=-1\)
ik
1,5-2./x/=-0,5
2./x/=1,5-(-0,5)
2./x/=2
/x/=2:2
/x/=1
-->x=+-1