1: Đặt A=1+11+...+11...1(10 chữ số 1)

=>\(9A=9+99+...+99...9\)

=>\(9A=\left(10-1\right)+\left(10^2-1\right)+...+\left(10^{10}-1\right)\)

=>\(9A=\left(10+10^2+...+10^{10}\right)-10\)

=>\(9A=\dfrac{10\left(1-10^{10}\right)}{1-10}-10=\dfrac{10-10^{11}}{1-10}-10=\dfrac{10-10^{11}}{-9}-10\)

=>\(9A=\dfrac{10^{11}-10}{9}-10=\dfrac{10^{11}-10-90}{9}=\dfrac{10^{11}-100}{9}\)

=>\(A=\dfrac{10^{11}-100}{81}\)

\(6+66+...+66...6\)(10 chữ số 6)

=6(1+11+...+11...1(10 chữ số 1))

=6A

\(=\dfrac{2}{27}\left(10^{11}-100\right)\)

2: \(1+2+3+...+x=465\)

=>\(\dfrac{x\left(x+1\right)}{2}=465\)

=>x(x+1)=930

=>\(x^2+x-930=0\)

=>(x+31)(x-30)=0

=>\(\left[{}\begin{matrix}x=-31\left(loại\right)\\x=30\left(nhận\right)\end{matrix}\right.\)

1 has broken her leg.

2 Her English has improved.

3 He has grown a beard.

4 It has arrived.

5 It has gone up.

1 She has broken her leg

2 Her English has improved

3 He has grown a beard

4 The letter has arrived

5 The price has gone up

\(n^2=36=\left(\pm6\right)^2\)

\(\Rightarrow n=\pm6\)

Mà: n là STN

\(\Rightarrow n=6\)

Vậy: n=6

\(n^2=36\\ n^2=6^2\\ n=\left\{6;-6\right\}\)

vậy n = 6 hoặcd -6

\(\left(4,5-2x\right)\cdot\left(-1\dfrac{4}{7}\right)=\dfrac{11}{14}\\ \left(4,5-2x\right)\cdot\left(-\dfrac{11}{7}\right)=\dfrac{11}{14}\\ 4,5-2x=-\dfrac{1}{2}\\ 2x=5\\ x=2,5\)

`(4,5 - 2x).`(-1 `4/7`) `= 11/4`

`(4,5 - 2x). (- 3/7).      = 11/4`

`(4,5 - 2x)                    = 11/4 ÷ (-3/7)`

`(4,5 - 2x)                   = -77/12`

`9/2- 2x                      = -77/12`

          `2x                    = 9/2 -- 77/12`

          ` 2x                  = 9/2 + 77/12`

         ` 2x                  = 131/12`

            `x                  = 131/12 ÷ 2`

           `x                  = 131/24`

Vậy `x = 131/24`

We did not arrive early enough to see that film.

We didn't arrive early enough to see that film

\(\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{6}{9}+\dfrac{2}{9}=\dfrac{8}{9}\)

2/3 + 2/9 = 6/9 + 2/9 = 8/9

Bài 1

1 two pieces of furniture

2 Two plates of pasta

3 four sheets of paper

4 Three glasses of lemonade

5 Two cups of yoghurt

6 Three bottles of oil

7 Two tubes of toothpaste

8 Five steaks of salmon

IV

30 every

31 everyone

32 every

33 nothing

34 somewhere

\(B=\left(-\dfrac{1}{2}\right)+\left(-\dfrac{1}{6}\right)+\left(-\dfrac{1}{12}\right)+\left(-\dfrac{1}{20}\right)+\left(-\dfrac{1}{30}\right)+\left(-\dfrac{1}{42}\right)+\left(-\dfrac{1}{56}\right)+\left(-\dfrac{1}{72}\right)+\left(-\dfrac{1}{90}\right)\)

\(=-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\)

\(=-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\)

\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=-\left(1-\dfrac{1}{10}\right)=-\dfrac{9}{10}\)