166+167. Tom was a more marvelous performer than Minh

à Tom …performed more marvelously than Minh……………..

à Minh didn’t……perform as marvelously as Tom…….

168. It is relaxing to listen to music in the evening.

à Listening …to music in the evening is relaxing…...

169+170. I played soccer more skillfully than my brother

à I …was a more skillful soccer player than my brother……….

à My brother didn’t……play soccer as skillfully as me…………………

171. Working for this company is tiring and boring.

à It …tiring and boring to work for this company………

172+173. I was a more excellent cook than my mother.

à I ……cooked more excellently than my mother……….

à My mother didn’t……cook as excellently as me………..

174. I am keen on learning French

à I ……am interested in learning French……….

175. When is your birthday?

à When were………you born?…

Số lớn nhất là 172, số nhỏ nhất là 159

R = 172 - 159 = 13

165.a           166.b       167.c            168.a                  169.c

165:A

166:B

167:C

168:A

169:C

166. “Are Tam and Hoa late for class?” Tuan asked Lan
………………………tuan asked lan if tam and hoa were late for class………………
167. She said to me “Can you speak Chinese?”
…………………… she asked if/whether I could speak Chinese…………………

168. “Will she be here for five days?” Tam asked Thu
…………………Tam asked Thu if she would be there for 5 days……………………
169. “Were you reading this book at 8 o’clock last Sunday?” She asked Ba
………………………she asked ba if he had been reading that book at 8 o’clock the previous Sunday………………

Câu 3:

Theo đề, ta có;

\(\left\{{}\begin{matrix}4a+2b+c=3\\-\dfrac{b}{2a}=1\\-\dfrac{b^2-4ac}{4a}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-2a\\b^2-4ac=-8a\\4a+2b+c=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=-2a\\4a^2-4ac=-8a\\4a+2b+c=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-2a\\4a\left(a-c\right)=4a\cdot\left(-2\right)\\4a+2b+c=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=-2a\\c=a+2\\4a-4a+a+2=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-2\\c=3\end{matrix}\right.\)

Câu 3:

Theo đề, ta có;

\(\left\{{}\begin{matrix}4a+2b+c=3\\-\dfrac{b}{2a}=1\\-\dfrac{b^2-4ac}{4a}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-2a\\b^2-4ac=-8a\\4a+2b+c=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=-2a\\4a^2-4ac=-8a\\4a+2b+c=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-2a\\4a\left(a-c\right)=4a\cdot\left(-2\right)\\4a+2b+c=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=-2a\\c=a+2\\4a-4a+a+2=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-2\\c=3\end{matrix}\right.\)

Đáp án là C