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\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).........\left(1-\frac{1}{19}\right).\left(1-\frac{1}{20}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...........\frac{18}{19}.\frac{19}{20}=\frac{1.2.3.4.5........19}{2.3.4.5.6.7......20}=\frac{1}{20}\)

1: =>x/5=6/5

=>x=6

2: =>x^2=36

=>x=6 hoặc x=-6

3: =>x-5/2=3+1/6=19/6

=>x=19/6+5/2=19/6+15/6=34/6=17/3

4: =>x=1/2:2/5=1/2*5/2=5/4

a)  \(\left|18-2x\right|=18\Rightarrow\orbr{\begin{cases}18-2x=18\\18-2x=-18\end{cases}\Rightarrow\orbr{\begin{cases}2x=0\\2x=36\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=18\end{cases}}}\)

b) \(\left|1-x\right|=1-x\Rightarrow\orbr{\begin{cases}1-x=1-x\\1-x=-\left(1-x\right)=-1+x\end{cases}}\)

  \(\Rightarrow\orbr{\begin{cases}x=\left\{\mp1;\mp2;...\right\}\\-x-x=-1-1\Rightarrow x=1\end{cases}}\)

Làm từng nãy đã , mỏi tay

a)  ta có  \(|\)18 - 2x\(|\)=18 
       =>       18-2x = 18                            hoặc  18 -2x=-18
                   -2x    =    0                                          -2x =  -36
                       x    = 0                                               x =  18
                  vậy x = 0 hoặc x=18
 
                  

a. ĐKXĐ:...

\(\Leftrightarrow2\left(\dfrac{x^2}{4}+\dfrac{9}{x^2}\right)=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)

\(\Leftrightarrow2\left(\dfrac{x^2}{4}+\dfrac{9}{x^2}-3+3\right)=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)

\(\Leftrightarrow2\left(\dfrac{x}{2}-\dfrac{3}{x}\right)^2+6=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)

Đặt \(\dfrac{x}{2}-\dfrac{3}{x}=t\Rightarrow2t^2-13t+6=0\Rightarrow\left[{}\begin{matrix}t=6\\t=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}-\dfrac{3}{x}=6\\\dfrac{x}{2}-\dfrac{3}{x}=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-12x-6=0\\x^2-x-6=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

\(\Leftrightarrow x\left(x-1\right)-\dfrac{x-1}{x^2}=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{x^2}\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-1\right)=0\)

\(\Leftrightarrow x=1\)

a) (2x + 1)(1 - 2x) + (1 - 2x)² = 18

= ( 1 - 2x) \(\left[\left(2x+1+1-2x\right)\right]\) = 18

= 2(1 - 2x)  - 18 = 0

= 2 - 4x - 18 = 0

= -16 - 4x = 0

= -4x = 16

= x = \(\dfrac{16}{-4}=-4\)

b) 2(x + 1)² -(x - 3)(x + 3) - (x - 4)² = 0

= 2 (x² + 2x + 1) - (x² - 9) - (x² - 8x + 16) = 0

= 2x² + 4x + 2 - x² + 9 - x² + 8x - 16 = 0

= 12x - 5 = 0

= 12x = 5

= x = \(\dfrac{5}{12}\)

c) (x - 5)² - x(x - 4) = 9

= x² - 10x + 25 - x² + 4x - 9 = 0

= -6x + 16 = 0

= -6x = -16

= x = \(\dfrac{-16}{-6}=\dfrac{8}{3}\)

d) (x - 5)² + (x - 4)(1 - x)

= x² - 10x + 25 + 5x - x² - 4 = 0

= -5x + 21 = 0

= -5x = -21

= x = \(\dfrac{-21}{-5}=\dfrac{21}{5}\) 

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