IX
2. j
3. i
4. f
5. c
6. a
7. h
8. e
9. g
10. d

XI
2. part => parts
3. a => an
4. a => an
5. a => the
6. are => will be (không chắc lắm)
7. taking => take
8. are => is

C. 
Bài 1
1. C
2. B
3. C
4. B

(Nên double-check trước khi chép)
 

\(x\left(x^2-1\right)=6\)

\(\Leftrightarrow x^3-x-6=0\)

\(\Leftrightarrow x^3-2x^2+2x^2-4x+3x-6=0\)

\(\Leftrightarrow x^2\left(x-2\right)+2x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^2+2x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x^2+2x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\\left(x+1\right)^2=-2\left(KĐS\right)\end{cases}}\)

Vậy x = 2 là ngiệm của pt trên.

a: \(\left(\frac{1}{24}-\frac{3}{16}\right):\left(-\frac38+\frac12\right)\)

\(=\left(\frac{2}{48}-\frac{9}{48}\right):\left(-\frac38+\frac48\right)\)

\(=\frac{-7}{48}:\frac18=-\frac{7}{48}\cdot8=-\frac76\)

b: \(-\frac57:\left(5-4\frac47\right)+\left(15\frac13\cdot\frac{-1}{23}\right)\)

\(=-\frac57:\left(5-4-\frac47\right)+\frac{46}{3}\cdot\frac{-1}{23}=-\frac57:\left(1-\frac47\right)+\frac{-2}{3}\)

\(=-\frac57:\frac37+\frac{-2}{3}=-\frac53-\frac23=-\frac73\)

c: \(\left(-2\frac23\cdot1\frac45+2\frac56\right):\left(\frac16-\frac34\right)\)

\(=\left(-\frac83\cdot\frac95+\frac{17}{6}\right):\left(\frac{2}{12}-\frac{9}{12}\right)\)

\(=\left(-\frac{24}{5}+\frac{17}{6}\right):\frac{-7}{12}=\left(-\frac{144}{30}+\frac{85}{30}\right):\frac{-7}{12}\)

\(=-\frac{59}{30}\cdot\frac{12}{-7}=\frac{59\cdot2}{7\cdot5}=\frac{118}{35}\)

d: \(\left(\frac{1}{3^2}-\frac{1}{7^2}\right)\left(\frac{1}{4^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{49^2}-\frac{1}{7^2}\right)\)

\(=\left(\frac{1}{3^2}-\frac{1}{7^2}\right)\left(\frac{1}{4^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{7^2}-\frac{1}{7^2}\right)\cdot\left(\frac{1}{8^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{49^2}-\frac{1}{7^2}\right)\)

\(=\left(\frac{1}{3^2}-\frac{1}{7^2}\right)\left(\frac{1}{4^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{49}-\frac{1}{49}\right)\cdot\left(\frac{1}{8^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{49^2}-\frac{1}{7^2}\right)\)

\(=\left(\frac{1}{3^2}-\frac{1}{7^2}\right)\left(\frac{1}{4^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot0\cdot\left(\frac{1}{8^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{49^2}-\frac{1}{7^2}\right)\)

=0

e: \(\left(1-\frac23\right)\left(1-\frac25\right)\left(1-\frac27\right)\cdot\ldots\cdot\left(1-\frac{2}{2023}\right)\)

\(=\frac13\cdot\frac35\cdot\frac57\cdot\ldots\cdot\frac{2021}{2023}=\frac{1}{2023}\)

f: \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\cdot\ldots\cdot\left(\frac{1}{100^2}-1\right)\)

\(=\left(\frac12-1\right)\left(\frac13-1\right)\cdot\ldots\cdot\left(\frac{1}{100}-1\right)\left(\frac12+1\right)\left(\frac13+1\right)\cdot\ldots\cdot\left(\frac{1}{100}+1\right)\)

\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\ldots\cdot\frac{-99}{100}\cdot\frac32\cdot\frac43\cdot\ldots\cdot\frac{101}{100}=\frac{-1}{100}\cdot\frac{101}{2}=-\frac{101}{200}\)

g:

Ta có công thức sau:

\(1-\frac{2}{n\left(n+1\right)}\)

\(=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}\)

\(=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

\(\left(1-\frac13\right)\left(1-\frac16\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)\cdot\ldots\cdot\left(1-\frac{1}{780}\right)\)

\(=\left(1-\frac26\right)\left(1-\frac{2}{12}\right)\left(1-\frac{2}{20}\right)\left(1-\frac{2}{30}\right)\cdot\ldots\cdot\left(1-\frac{2}{1560}\right)\)

\(=\left(1-\frac{2}{2\cdot3}\right)\left(1-\frac{2}{3\cdot4}\right)\cdot\ldots\cdot\left(1-\frac{2}{39\cdot40}\right)\)

\(=\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\cdot\ldots\cdot\frac{\left(39+2\right)\left(39-1\right)}{39\left(39+1\right)}\)

\(=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\ldots\cdot\frac{41\cdot38}{39\cdot40}=\frac{4\cdot5\cdot\ldots\cdot41}{3\cdot4\cdot..\cdot40}\cdot\frac{1\cdot2\cdot\ldots\cdot38}{2\cdot3\cdot\ldots\cdot39}\)

\(=\frac{41}{3}\cdot\frac{1}{39}=\frac{41}{117}\)

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