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ý đúng rồi

Bạn tự viết lại đề bài nha

\(\frac{1}{2}\)x\(\frac{2}{3}\)x\(\frac{3}{4}\)x\(\frac{4}{5}\)x...x\(\frac{18}{19}\)x\(\frac{19}{20}\)

=\(\frac{1x2x3x4x...x18x19}{2x3x4x5x...x19x20}\)

=\(\frac{1}{20}\)

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

= \(\frac{1}{2}\)

tk cho mk nha, mik đg âm điểm huhu

Đáp số là \(\frac{1}{20}\)

a)

`127+246+273+354`

`=(127+273)+(246+354)`

`=400+600`

`=1000`

b)

`1,58+3,04+6,96+3,42`

`=(1,58+3,42)+(3,04+6,96)`

`=5+10`

`=15`

c)

`1/2+1/3+1/5+1/6`

`=(1/2+1/3+1/6)+1/5`

`=(3/6+2/6+1/6)+1/5`

`=6/6+1/5`

`=1+1/5`

`=5/5+1/5`

`=6/5`

a, 127 + 246 + 273 + 354

= ( 127 + 273) + ( 246 + 354)

= 400 + 600

= 1000

b, 1,58 + 3,04 + 6,96 + 3,42

= ( 1,58 + 3,42) + ( 3,04 + 6,96)

= 5 + 10

= 15

c, \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{5}\) + \(\dfrac{1}{6}\)

=  ( \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\)) + 0,2

= 1 + 0,2

= 1,2 

a)

$127+246+273+354$

$=(127+273)+(246+354)$

$=400+600$

$=1000$

b)

$1,58+3,04+6,96+3,42$

$=(1,58+3,42)+(3,04+6,96)$

$=5+10$

$=15$

c)

$1/2+1/3+1/5+1/6$

$=(1/2+1/3+1/6)+1/5$

$=(3/6+2/6+1/6)+1/5$

$=6/6+1/5$

$=1+1/5$

$=5/5+1/5$

$=6/5$

xin lỗi em mới học lớp 6 

\(\left(x+2\right)^2=\left(2x-1\right)^2\\ \Leftrightarrow\left(x+2\right)^2-\left(2x-1\right)^2=0\\\Leftrightarrow\left[x+2-\left(2x-1\right)\right]\left[x+2+2x-1\right]=0\\ \Leftrightarrow\left(x+2-2x+1\right)\left(x+2+2x-1\right)=0\\ \Leftrightarrow\left(-x+3\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-x+3=0\\3x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=-3\\3x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{3}\end{matrix}\right.\)

\(\left(x+2\right)^2=\left(2x-1\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=2x-1\\x+2=-\left(2x-1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2x=-1-2\\x+2=-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=-3\\x+2x=1-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\3x=-1\end{matrix}\right.\)

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

(x+2)2=(2x−1)2⇔(x+2)2−(2x−1)2=0⇔[x+2−(2x−1)][x+2+2x−1]=0⇔(x+2−2x+1)(x+2+2x−1)=0⇔(−x+3)(3x+1)=0⇔[−x+3=03x+1=0​⇔[−x=−33x=−1​
 

=) x = 3 hoặc -1/3.

Chúc bạn hok tốt

1/1x2 + 1/2x3 +1/3x4 + ......+1/98x99+1/99x100

=1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +......+ 1/98 - 1/99 + 1/99 + 1/100

=(1-1/100)+(1/2 - 1/2 ) + ( 1/3 - 1/3 ) + ...... + (1/98 - 1/98 ) + ( 1/99 - 1/99 )

= 100/100 - 1/100 + 0 + 0 +.....+ 0 + 0

=99/100

vậy GTBT = 99/100

bn vào câu hỏi tương tự là có

= 1/1 + 1/100

=1011/100

Tích nha

TH1:\(\hept{\begin{cases}x+\frac{1}{2}>0\\x-\frac{1}{3}>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-\frac{1}{2}\\x>\frac{1}{3}\end{cases}\Leftrightarrow}x>\frac{1}{3}}\)

TH2:\(\hept{\begin{cases}x+\frac{1}{2}< 0\\x-\frac{1}{3}< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -\frac{1}{2}\\x< \frac{1}{3}\end{cases}\Leftrightarrow}x< -\frac{1}{2}}\)

vậy để biểu thức \(\left(x+\frac{1}{2}\right)\left(x-\frac{1}{3}\right)>0\)thì x > 1/3 hoặc x < (-1/2)