ta có x/6=7/9

=> x=6.7/9=14/3

thay x vào ta được 14/3=z/2

=> z=2.14/3=28/3

mà x+y+z=27

=> y=27-x-z=13

\(\frac{2}{5.8}+\frac{2}{8.11}+...+\frac{2}{230.233}\)

\(=\frac{2}{5}-\frac{2}{8}+\frac{2}{8}-\frac{2}{11}+...+\frac{2}{230}-\frac{2}{233}\)

\(=\frac{2}{5}-\frac{2}{233}\)

\(=\frac{466}{1165}-\frac{10}{1165}=\frac{456}{1165}\)

Biểu thức này ko tồn tại cả min lẫn max

\(\dfrac{1}{M}=\dfrac{\sqrt{x}-1}{2\sqrt{x}+4}=\dfrac{-\dfrac{1}{4}\left(2\sqrt{x}+4\right)+\dfrac{\sqrt{x}}{2}}{2\sqrt{x}+4}=-\dfrac{1}{4}+\dfrac{\sqrt{x}}{4\left(\sqrt{x}+2\right)}\)

Do \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}+2>0\end{matrix}\right.\) \(\Rightarrow\dfrac{\sqrt{x}}{4\left(\sqrt{x}+2\right)}\ge0\)

\(\Rightarrow\dfrac{1}{M}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(x=0\)

1 to finish

2 adding

3 to have

4 to do

5 to finish

6 working

7 writing

8 to complete

9 to get

10 to show

11 putting

12 to be

13 to email

14 to remember 

\(50\%+\dfrac{1}{2}+\dfrac{1}{7}\times\dfrac{2}{5}+\dfrac{1}{7}\times\dfrac{3}{5}\)

\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{2}{35}+\dfrac{3}{35}\)

\(=1+\dfrac{5}{35}\)

\(=1+\dfrac{1}{7}\)

\(=\dfrac{8}{7}\)

\(50\%+\dfrac{1}{2}+\dfrac{1}{7}x\dfrac{2}{5}+\dfrac{1}{7}x\dfrac{3}{5}\)

\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{7}x\left(\dfrac{2}{5}+\dfrac{3}{5}\right)\)

\(=1+\dfrac{1}{7}x1\)

\(=\dfrac{7}{7}+\dfrac{1}{7}=\dfrac{8}{7}\)

50% + 1/2 + 1/7 × 2/5 + 1/7 × 3/5

= 1/2 + 1/2 + 1/7 × (2/5 + 3/5)

= 1 + 1/7 × 1

= 1 + 1/7

= 7/7 + 1/7

= 8/7

8/7

B

B

C

B

B

A

A

A

D

8B

9B

10C

11B

12B

13B

14D

15C

16D