Lời giải:
Đánh từ trang 1 đến trang 9 cần $(9-1):1+1=9$ (chữ số)

Đánh từ trang 10 đến trang 99 cần: $2\times [(99-10):1+1]=180$ (chữ số)

Đánh từ trang 100 đến trang 240 cần: $3\times [(240-100):1+1]=423$ (chữ số)

Tổng số chữ số cần sử dụng: $9+180+423=612$ (chữ số)

a) Ta có: \(Q=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)

\(=\dfrac{3x+3\sqrt{x}-3-\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

b) Thay \(x=4+2\sqrt{3}\) vào Q, ta được:

\(Q=\dfrac{\sqrt{3}+1+1}{\sqrt{3}+1-1}=\dfrac{2+\sqrt{3}}{\sqrt{3}}=\dfrac{2\sqrt{3}+3}{3}\)

c) Để Q=3 thì \(\sqrt{x}+1=3\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\sqrt{x}=-3-1\)

\(\Leftrightarrow2\sqrt{x}=4\)

hay x=4

d) Để \(Q>\dfrac{1}{2}\) thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{1}{2}>0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}+2-\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}>0\)

\(\Leftrightarrow\sqrt{x}-1>0\)

\(\Leftrightarrow x>1\)

Kết hợp ĐKXĐ, ta được: x>1

e) Để Q nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)

\(\Leftrightarrow2⋮\sqrt{x}-1\)

\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;2;3\right\}\)

hay \(x\in\left\{0;4;9\right\}\)

Cơ chế hình thành cây có kiểu gen Aaa là do rồi loạn giảm phân, diễn ra ở kì sau của giảm phân 1. 

Sơ đồ lai: 

P:       Aa          x             Aa

GP: Aa ; 0  ;           A   ;    a

F1:  Aaa             ;            a

lỗi ảnh 

???

Câu 19: B

Câu 20: A

\(\Leftrightarrow2^{x+3}=4\)

=>x+3=2

hay x=-1

\(2^{x+3}\cdot4^2=64\\ \Leftrightarrow2^{x+3}\cdot4^2=4^3\\ \Leftrightarrow2^{x+3}=4=2^2\\ \Leftrightarrow x+3=2\Leftrightarrow x=-1\)

x=−1

Bài 8: 

a: Ta có: \(\left(5x+1\right)^2=\dfrac{36}{49}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{7}\\5x=\dfrac{-13}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)

b: Ta có: \(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)

\(\Leftrightarrow x-\dfrac{2}{9}=\dfrac{4}{9}\)

hay \(x=\dfrac{2}{3}\)

\(7,\\ a,=\dfrac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot\left[-\left(3^7\right)\right]}=\dfrac{3^8}{-5}=-\dfrac{6561}{5}\\ b,=8+3-\dfrac{1}{4}\cdot4+\left(4:\dfrac{1}{2}\right)\cdot8\\ =8+3-1+64=74\\ 8,\\ a,\left(5x+1\right)^2=\dfrac{36}{49}\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\)

\(b,\Rightarrow8x-1=5\Rightarrow x=\dfrac{3}{4}\\ d,\Rightarrow\left\{{}\begin{matrix}x-3,5=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\left[\left(x-3,5\right)^2\ge0;\left(y-\dfrac{1}{10}\right)^4\ge0\right]\\ \Rightarrow\left\{{}\begin{matrix}x=3,5\\y=\dfrac{1}{10}\end{matrix}\right.\)

a: góc DAC=90-40=50 độ

b: góc ADB=90 độ

c: góc DAB=90-80=10 độ

=>góc BAE=10+50=60 độ

góc AED=180-60=120 độ

14.He asked me to help my withe the exercises.

15.He said he left early on Friday.

16.She asked me where he would meet that night.

17.Milk could be used for making butter and cheese.

18.Bottles of milk are brought to houses by the milkman.

19.A lot of beautiful toys are made from recyced plastic.

20.The concerts usually are held at the university.