x=1.2

x=2

X ko có giá trị để 1:2=x(thuộc N*)

Ta có : \(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)

=> (x - 1)(x + 3) = (x - 2)(x + 2) 

=> x² + 2x - 3 = x² - 4 

=> 2x = - 1

=> x = -0,5 

\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\left(x\ne-2;x\ne-3\right)\)

<=>(x-1)(x+3)=(x-2)(x+2)

<=>x²+2x-3=x²-4

<=> x²+2x-3-x²+4=0

<=> 2x+1=0

<=> x=\(\frac{-1}{2}\)(tm)

Ta có \(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)

\(\Rightarrow\left(x-1\right)\cdot\left(x+3\right)=\left(x+2\right)\cdot\left(x-2\right)\)

\(\Rightarrow x^2-x+3x-3=x^2+2x-2x-4\)

\(\Rightarrow x^2+2x-3=x^2-4\)

\(\Rightarrow x^2-x^2+2x=-4+3\)

\(\Rightarrow2x=-1\)

Vậy \(x=\frac{-1}{2}\)

giúp mình với

\(x^n-x=x\left(x^{n-1}-1\right)\text{ Ta có:}0< x< 1\Rightarrow0< x^{n-1}< 1\Rightarrow x^{n-1}-1< 0\)

\(\Rightarrow x\left(x^{n-1}-1\right)< 0\Rightarrow x^n< x\text{ Ta có điều phải chứng minh}\)

ummmm ok

      \(5x\left(1-2x\right)-3x\left(x+18\right)=0\)

\(\Leftrightarrow5x-10x^2-3x^2-54x=0\)

\(\Leftrightarrow-13x^2-49x=0\)

\(\Leftrightarrow-x\left(13x-49\right)=0\)

\(\Rightarrow\hept{\begin{cases}-x=0\Leftrightarrow x=0\\13x-49=0\Leftrightarrow x=\frac{49}{13}\end{cases}}\)

Vậy \(x=0\)hoặc \(x=\frac{49}{13}\)

5 (1-2x) - 3x(x+18) = 0

5x - 10x² - 3x ²- 54x = 0

- 13x² - 49x = 0

(-13x - 49)x = 0

TH1

:x = 0

TH2:

-13x - 49 = 0

-13x         = 49

     x         = -49/13

vậy x = 0 và x = -49/13