Câu I ạ

Mn ơi giúp mik câu 7 với ạ, câu 7 mik chọn I prefer watching sports to playing sports có đúng không ạ?
giups mình câu i và câu L đến hết, cảm ơn ạ
i: ĐKXĐ: \(\left[{}\begin{matrix}x\ge\sqrt{5}+1\\x\le-\sqrt{5}+1\end{matrix}\right.\)
l:ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le-5\end{matrix}\right.\)
m: ĐKXĐ: \(\left[{}\begin{matrix}x\ge4\\x\le3\end{matrix}\right.\)
n: ĐKXĐ: \(x\in R\)
bài 11 , câu i và h ạ
11:
i: \(=\dfrac{2x\left(3x-1\right)+1}{3x-1}=2x+\dfrac{1}{3x-1}\)
h: \(=\dfrac{1}{2}\cdot\dfrac{54x^3+2x^2-2x+2}{-2x+1}\)
\(=\dfrac{1}{2}\cdot\dfrac{54x^3-27x^2+29x^2-14.5x+12.5x-6.25+8.25}{-2x+1}\)
\(=\dfrac{-1}{2}\left(27x^2+14.5x+6.25\right)+\dfrac{-1}{2}\cdot\dfrac{8.25}{2x-1}\)
Giúp mk câu 4,5 bài I với ạ :< 
I.4. My family sometimes plays games together
5. Did you play badminton when you was small?
~HT~
When I was a child, I (not like) ______________ sports. chỉ em câu này với ạ
didn't like
câu này chia ở thì quá khứ đơn
Giải câu g,h,i,k,l,m thôi ạ
Câu g,h,i,k,l,m của bài 1 hay bài 2 ấy bạn?
g) \(\sqrt{7-4\sqrt{3}}+\sqrt{4-2\sqrt{3}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=\left|2-\sqrt{3}\right|+\left|\sqrt{3}-1\right|=2-\sqrt{3}+\sqrt{3}-1=1\)
h) \(\sqrt{11+6\sqrt{2}}-\sqrt{3-2\sqrt{2}}=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=3-\sqrt{2}-\sqrt{2}+1=4\)
i) \(\sqrt{15+6\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(3+\sqrt{6}\right)^2}+\sqrt{\left(3-\sqrt{6}\right)^2}=3+\sqrt{6}+3-\sqrt{6}=6\)
k) \(\sqrt{33+8\sqrt{2}}-\sqrt{33-8\sqrt{2}}=\sqrt{\left(4\sqrt{2}+1\right)^2}-\sqrt{\left(4\sqrt{2}-1\right)^2}=4\sqrt{2}+1-4\sqrt{2}+1=2\)
Làm nhanh dùm e câu g,h,i với ạ.

m: \(\Leftrightarrow\left\{{}\begin{matrix}-3x< =-4\\2x< 5\end{matrix}\right.\Leftrightarrow\dfrac{4}{3}< x< \dfrac{5}{2}\)
Câu 1 mng làm từ câu i trở xuống cũng được ạ. Em tha thiết xin mng luôn í :((
Câu 1:
a: \(\frac37+\left(-\frac52\right)+\frac35\)
\(=\frac37-\frac52+\frac35\)
\(=\frac{30}{70}-\frac{175}{70}+\frac{42}{70}=\frac{30+42-175}{70}=-\frac{103}{70}\)
b: \(0,8-\left(-\frac27\right)-\frac{7}{10}=\frac45+\frac27-\frac{7}{10}\)
\(=\frac{56}{70}+\frac{20}{70}-\frac{49}{70}=\frac{56+20-49}{70}=\frac{56-29}{70}=\frac{27}{70}\)
c: \(\frac34-\left\lbrack\left(-\frac53\right)-\left(\frac{1}{12}+\frac29\right)\right\rbrack\)
\(=\frac34+\frac53+\frac{1}{12}+\frac29\)
\(=\frac{27}{36}+\frac{60}{36}+\frac{3}{36}+\frac{8}{36}=\frac{98}{36}=\frac{49}{18}\)
d: \(19\frac13\cdot\frac45-\frac45\cdot15\frac13\)
\(=\frac45\left(19+\frac13-15-\frac13\right)\)
\(=\frac45\cdot4=\frac{16}{5}\)
e: \(\frac38\cdot\left(-\frac{3}{11}\right)+\frac38\cdot\left(-\frac{8}{11}\right)+2\frac57\)
\(=\frac38\left(-\frac{3}{11}-\frac{8}{11}\right)+2+\frac57\)
\(=-\frac38+2+\frac57=\frac{13}{8}+\frac57=\frac{91}{56}+\frac{40}{56}=\frac{131}{56}\)
f: \(\left(-\frac{5}{11}:\frac{13}{8}-\frac{5}{11}:\frac{13}{3}\right)-\left|-\frac{1}{33}\right|\)
\(=-\frac{5}{11}\cdot\frac{8}{13}-\frac{5}{11}\cdot\frac{3}{13}-\frac{1}{33}\)
\(=-\frac{5}{11}\cdot\frac{11}{13}-\frac{1}{33}=-\frac{5}{13}-\frac{1}{33}=\frac{-165-13}{13\cdot33}=\frac{-178}{429}\)
g: \(\frac{4}{23}+\frac{5}{21}+\frac12-\frac{4}{23}+\frac{16}{21}\)
\(=\left(\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+\frac12\)
\(=1+\frac12=\frac32\)
h: \(\frac{13}{25}+\frac{6}{41}+\frac{-38}{25}+\frac{35}{41}-0,5\)
\(=\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-0,5\)
\(=-\frac{25}{25}+\frac{41}{41}-0,5=-0,5\)
i: \(\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-1\frac{15}{17}+\frac23\)
\(=\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac13+\frac23\right)-1-\frac{15}{17}\)
\(=1+1-1-\frac{15}{17}=1-\frac{15}{17}=\frac{2}{17}\)
j: \(\left(\frac{2022}{2023}\right)^0-\left(-3\right)_{}^3+\frac12:\left(-\frac12\right)^3\)
\(=1-\left(-27\right)+\frac12:\frac{-1}{8}\)
=28-4
=24
k: \(\left(-2\right)^3+\frac12:\frac{-1}{4}-\left|15\right|\)
\(=-8+\frac12\cdot\left(-4\right)-15\)
=-8-2-15
=-25
l: \(3:\left(-\frac32\right)^2+\frac19\cdot\sqrt{36}\)
\(=3:\frac94+\frac19\cdot6=3\cdot\frac49+\frac69=\frac{12}{9}+\frac69=\frac{18}{9}\)
=2
m: \(\sqrt{81}\cdot\frac13-\left(-3\right)^3=9\cdot\frac13-\left(-27\right)\)
=3+27
=30
n: \(\sqrt{\frac{16}{25}}\cdot\sqrt{\frac{121}{64}}-1\frac{3}{10}\)
\(=\frac45\cdot\frac{11}{8}-\frac{13}{10}\)
\(=\frac{11}{10}-\frac{13}{10}=-\frac{2}{10}=-\frac15\)
o: \(\frac{3^2}{2^3}\cdot\sqrt{144}-50\%\)
\(=\frac98\cdot12-\frac12=9\cdot\frac32-\frac12=\frac{27}{2}-\frac12=\frac{26}{2}\)
=13
p: \(\frac{6^2+3\cdot6^2+3^2}{-13}\)
\(=\frac{36+3\cdot108+9}{-13}=\frac{45+324}{-13}=\frac{369}{-13}=-\frac{369}{13}\)
q: \(\frac{5^4\cdot20^4}{25^5\cdot4^5}=\frac{\left(5\cdot20\right)^4}{\left(25\cdot4\right)^5}=\frac{100^4}{100^5}=\frac{1}{100}\)
r: \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
\(=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=\frac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot5}=\frac23\cdot\frac65=\frac{12}{15}=\frac45\)