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)\(\left(x+0,9\right)\left(1-0,4\right)=2412\)

\(\Leftrightarrow\left(x+0,9\right)\cdot0,6=2412\)

\(\Leftrightarrow x+0,9=4020\)

\(\Leftrightarrow x=1019,1\)

\(x+\dfrac{2}{-15}=\dfrac{5}{3}\)

\(x=\dfrac{9}{5}\)

\(x+\dfrac{2}{-15}=\dfrac{-5}{3}\)

\(x=\dfrac{-5}{3}-\dfrac{-2}{15}\)

\(x=\dfrac{-23}{15}\)

Lời giải:

$x+\frac{2}{-15}=\frac{-5}{3}$

$x=\frac{-5}{3}-\frac{2}{-15}=\frac{-5}{3}+\frac{2}{15}$

$x=\frac{-23}{15}$

\(x=\dfrac{1}{8}\)

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?