\(\left(x+3\right)^2+\left(0,5y-1\right)^2=0\)

Do \(\left(x+3\right)^2\ge0;\left(0,5y-1\right)^2\ge0\)

\(\Rightarrow\left(x+3\right)^2+\left(0,5y-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(x+3\right)^2=0\\\left(0,5y-1\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x+3=0\\0,5y-1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-3\\y=2\end{cases}}\)

...

Vì \(\hept{\begin{cases}\left(x+3\right)^2\ge0\forall x\\\left(0.5y-1\right)^2\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x+3\right)^2+\left(0.5y-1\right)^2\ge0\forall x,y\)

Mà \(\left(x+3\right)^2+\left(0.5y-1\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}x+3=0\\0.5y-1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-3\\y=2\end{cases}}\)

Vậy ...

\(=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{20}\cdot\dfrac{20\cdot21}{2}\)

\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{21}{2}\)

=(2+3+4+...+21)/2

=(20*23/2):2=230:2=115

Bài 9:

a: \(2^{195}=8^{65}\)

\(3^{130}=9^{65}\)

mà 8<9

nên \(2^{195}< 3^{130}\)

mình gửi lại đề bài !

Where is he?

Why is he sad?

What is that?

When is the exam?

Who is she?

Why is he happy?

Where are they?

Who is crying?

When is your birthday?

Who makes dinner?

Why are you running?

Where is Jack?

trên xuống dưới và trái qua phải
where        why
what          when
who           why
what          when
where        who
where        what
when         who
why           where

56083   | 123

  688     | 455

    733   | 

    118

=> \(56083:123=455\left(dư.118\right)\)

cứu mình với mình sắp đến hạn nộp rồi :((((

Bài 6:

Vì \(ƯCLN\left(72,96\right)=24\) nên có nhiều nhất 24 hs nhận thưởng