1.C

2.A

3.B

4. kh có đáp án nên mình sẽ điền là but/however

5.A

1C

2.A

3.B

4.C

5.C

6.A

7.D

8/A

9.B

10.A

chuc ban hoc tot ^^ 

\(\sqrt{36}+\sqrt{49}-\sqrt{64}=6+7-8=5\)

\(\sqrt{36}+2.\sqrt{49}-\sqrt{25}=6+14-5=15\)

\(\sqrt{100}+\sqrt{144}-4\sqrt{49}=10+12-28=-6\)

\(\sqrt{169}-\sqrt{81}-2\sqrt{64}=13-9-8=-4\)

\(8\sqrt{16}-7\sqrt{9}-2\sqrt{196}=32-21-28=-17\)

\(\sqrt{36}.\sqrt{25}+\sqrt{225}-\sqrt{256}=6\cdot5+15-16=29\)

\(\sqrt{324}-\left(\sqrt{36}-\sqrt{289}\right)=18-\left(6-17\right)=29\)

\(\sqrt{361}-\left(\sqrt{256}+\sqrt{81}\right)=19-\left(16+9\right)=-6\)

\(\sqrt{36}.\sqrt{49}-\sqrt{64}=6\cdot7-8=34\)

\(\sqrt{4}+\sqrt{9}-\sqrt{64}+\sqrt{100}=2+3-8+10=7\)

\(\sqrt{25}+\sqrt{36}.\sqrt{49}-\sqrt{64}:\sqrt{16}=5+6.7-8:4=45\)

\(\left(\sqrt{36}+\sqrt{49}-\sqrt{121}\right):2=\left(6+7-11\right):2=1\)

Bài 2:

\(\sqrt{40^2-24^2}=\left(40-24\right)\left(40+24\right)=16.64=1024\)

\(\sqrt{25^2-24^2}=\left(25-24\right)\left(25+24\right)=49\)

\(\sqrt{52^2-48^2}=\left(52-48\right)\left(52+48\right)=400\)

Bài 3:

\(2\sqrt{x}=18\\ \Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\)

\(-2\sqrt{x}=-14\Leftrightarrow\sqrt{x}=7\Leftrightarrow x=49\)

\(2\sqrt{x}+12=14\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)

\(2\sqrt{x}-3=17\\ \Leftrightarrow\sqrt{x}=10\Leftrightarrow x=100\)

\(20-3\sqrt{x}=11\Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\)

\(36-12\sqrt{x}=12\\ \Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)

1.for

2.on

3.from 7am to 5pm

4.at

5.in

từ B kẻ 1 đường thẳng xuống Cz cắt tại D ta có
\(\Delta BCD=\widehat{BCD}+\widehat{CDB}+\widehat{DBC=180^0}\\ \Leftrightarrow\widehat{CDB}+\widehat{DBC}=140^0\left(1\right)\)

mà : \(\widehat{CBD}+\widehat{DBy}=140^0\left(2\right)\)

\(\Rightarrow\widehat{DBC}=\widehat{DBy}\)

mà hai góc ở vị trí so le trong \(\Rightarrow CD//By\) hay \(Cz//By\left(3\right)\)

Cách làm tương tự ta có \(By//Ax\left(4\right)\)

Từ (3)(4) =>Ax//By//Cz

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2009.2010}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2009}-\dfrac{1}{2010}\)

\(=1-\dfrac{1}{2010}=\dfrac{2009}{2010}\)

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)

\(3A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(=1-\dfrac{1}{103}=\dfrac{102}{103}\)

\(\Rightarrow A=\dfrac{102}{309}=\dfrac{34}{103}\)

\(-\dfrac{1}{2011.2010}-\dfrac{1}{2010.2009}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{2009.2010}+\dfrac{1}{2010.2011}\right)\)

\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2009}-\dfrac{1}{2010}+\dfrac{1}{2010}-\dfrac{1}{2011}\right)\)

\(=-\left(1-\dfrac{1}{2011}\right)=-\dfrac{2010}{2011}\)

\(\dfrac{-1}{3}+\dfrac{-1}{15}+\dfrac{-1}{35}+\dfrac{-1}{63}+...+\dfrac{-1}{9999}\)

\(=-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{9999}\right)\)

câu này b thông cảm mình kh tách được 9999 thành tích của 2 số lẻ liên tiếp nhau 

thật ra mấy cái này có nhiều trên google lắm bạn, thay vì hỏi bạn có thể search ở trên đó luôn

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

ĐKXĐ: \(x>0;x\ne1\)

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

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

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

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