a: Đặt góc có dấu ? là x

Theo hình, ta sẽ suy ra được:

x là góc đối đỉnh với góc 50 độ

=>x=50 độ

b: Đặt góc có dấu ? là x

Theo hình, ta sẽ có được:

x là góc đối đỉnh với góc 123 độ

=>x=123 độ

good

good->better

good => better

6 đội,16 người

Theo hình có:

\(\widehat{COD}=\widehat{AOB}=147^o\) (tính chất hai góc đối đỉnh)

Theo tính chất kể bù có:

\(\widehat{AOD}+\widehat{COD}=180^o\\ \Rightarrow\widehat{AOD}=180^o-147^o=33^o\)

Theo tính chất hai góc đối đỉnh có:

\(\widehat{BOC}=\widehat{AOD}=33^o\)

o: \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)

\(=\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+\dfrac{1}{9\cdot11}+\dfrac{1}{11\cdot13}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)=\dfrac{1}{2}\cdot\dfrac{10}{39}=\dfrac{5}{39}\)

p: \(\dfrac{1}{2}+\dfrac{1}{14}+\dfrac{1}{35}+\dfrac{1}{65}+\dfrac{1}{104}+\dfrac{1}{152}\)

\(=\dfrac{2}{4}+\dfrac{2}{28}+\dfrac{2}{70}+\dfrac{2}{130}+\dfrac{2}{208}+\dfrac{2}{304}\)

\(=\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+\dfrac{2}{7\cdot10}+\dfrac{2}{10\cdot13}+\dfrac{2}{13\cdot16}+\dfrac{2}{16\cdot19}\)

\(=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{16\cdot19}\right)\)

\(=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{16}-\dfrac{1}{19}\right)\)

\(=\dfrac{2}{3}\left(1-\dfrac{1}{19}\right)=\dfrac{2}{3}\cdot\dfrac{18}{19}=\dfrac{2\cdot6}{19}=\dfrac{12}{19}\)

a: \(\left(x+10\right)\left(x-5\right)=0\)

=>\(\left[{}\begin{matrix}x+10=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=5\end{matrix}\right.\)

b: \(\left(2x+10\right)\left(4+x\right)=0\)

=>\(\left[{}\begin{matrix}2x+10=0\\4+x=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-4\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-5\end{matrix}\right.\)

c: \(\left(4x+20\right)\left(12x-24\right)=0\)

=>\(\left[{}\begin{matrix}4x+20=0\\12x-24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-20\\12x=24\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

d: \(\left(x-2024\right)\left(4x+4\right)=0\)

=>\(\left[{}\begin{matrix}x-2024=0\\4x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2024\\4x=-4\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-1\\x=2024\end{matrix}\right.\)

e: \(\left(2x-6\right)\left(7+x\right)=0\)

=>\(\left[{}\begin{matrix}2x-6=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\x=-7\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)

g: (4x+8)(6-x)=0

=>\(\left[{}\begin{matrix}4x+8=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x=6\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-2\\x=6\end{matrix}\right.\)

h: (2x+2)(4x-8)=0

=>2(x+1)*4*(x-2)=0

=>(x+1)(x-2)=0

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

i: (2x-2024)(8x-16)=0

=>\(2\left(x-1012\right)\cdot8\cdot\left(x-2\right)=0\)

=>\(\left(x-1012\right)\left(x-2\right)=0\)

=>\(\left[{}\begin{matrix}x-1012=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1012\\x=2\end{matrix}\right.\)

I believe that with the improvement of technology, in the future there will be more modern transportation for humankind. There will be flying cars, trains that goes underwater,... anything we can imagine .Flying cars will make us be able to go with high-speed and less accidents. underwater trains will help us to travel overseas. There might be some disadvantages but those can help clear traffic due to the growth of today's population. Moreover, transport will also improve its security. To sum up, future's mean of transport will be better.

nếu ok thì cho mình like nha, nếu không ok thì không cho. :)

25,6 nhé

74,24:2,9= 25,6 nha bé

25,6 NHA

HT

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