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My answer is : C ) 3 

\(\widehat{xOy}=50^o;\widehat{xOz}=100^o;\widehat{xOt}=130^o\)

Solution:

\(\widehat{BOI}=\dfrac{1}{4}=\widehat{AOB}=\dfrac{1}{4}\times60^o=15^o.\)

Because OI rays are between 2 OA rays, the OB should \(\widehat{AOI}+\widehat{BOI}=\widehat{AOB}\)

Inferred \(\widehat{AOI}+15^o=60^o\) or  \(\widehat{AOI}=45^o\)To

Two corners \(xOy\) and \(yOy'\)  compensated \(\widehat{xOy}+\widehat{yOy'}=180^o\).

Imagine: \(\widehat{yOy'}=180^o-\widehat{xOy}=180^o-120^o=60^o\).

So \(\widehat{yOy'}=60^o\)

The area of the shaded region in the picture is:

                   24x48=1152 ( m2 )

The area of the shaded region is:

 \(48\times24=1152\left(m^2\right)\)

               Answer: \(1152m^2\)

mk ko nhanh như bn khác nhưng hãy k mk! *_<

Diện tích của khu vực bóng mờ là:

        24x48=1152 (m2)

                  Đáp số: 1152 m2.

We have:\(1-\dfrac{1}{1+2+......+n}=1-\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=1-\dfrac{2}{n\left(n+1\right)}=\dfrac{n\left(n+1\right)-2}{n\left(n+1\right)}=\dfrac{n^2+n-2}{n\left(n+1\right)}\)

\(=\dfrac{n^2-n+2n-2}{n\left(n+1\right)}=\dfrac{n\left(n-1\right)+2\left(n-1\right)}{n\left(n+1\right)}=\dfrac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

Instead in the expression we have:

A=\(\dfrac{4.1}{2.3}+\dfrac{5.2}{3.4}+.......+\dfrac{1988.1985}{1986.1987}\)

\(=\dfrac{1.2....1985}{3.4....1987}.\dfrac{4.5.....1988}{2.3.....1986}\)

\(=\dfrac{1.2}{1986.1987}.\dfrac{1987.1988}{2.3}=\dfrac{1988}{3.1986}=\dfrac{1988}{5958}=\dfrac{994}{2979}\)

P/s:A is the expression