Chose the corret answer :

1. Phong is wearing a school .............

A. shoes      B. uniform      C. bag      D. hats

2. I am having a maths lesson but I forgot my ........I have some difficulty

A. calculator         B. bicycle          C. rubber          D. pencil sharpener

3. In science, we have a lot of ............ ...........

A. books - to have                      B. homework - to do

C. music - to study                     D. vocabulary - to play

1.B

2.A

3.B

hoc tot

 (x-6)^2=9

(x-6)^2=3^2

=>x-6=3

       x=3+6

      x=9

nhớ mik nha đa tạ ~~

=>(x-6)²=+-3

=>x-6=3 hoặc x-6=-3

=>x=9 hoặc x=3

 1. D

 2. C

 3. B

1. D

2. C

3. B

Chúc bn học tốt

 A. subject          B. club          C. put          D. lunch

2. A. science         B. li ke            C. music     D. ice

3. A. lunch             B. student      C. hungry   D. sun

Find the word which has a diffrent sound in the part inderlined

1. A. subject          B. club          C. put          D. lunch

2. A. science         B. l.i.k.e            C. music     D. ice

3. A. lunch             B. student      C. hungry   D. sun

Find the word which has a diffrent sound in the part inderlined

1. A. hope               B. homework          C. one             D. post

2. A. classmates     B. bag                     C. calculator   D. activity

1.C

2.C

-HT-

Fimd the word which has a diffrent sound in the part inderlined

1. A. cartoon     B. smart         C. art              D. same

2. A. like            B. history       C. exercise     D. library

\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

\(< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)

\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)\(< \)\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)\(< \)\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)\(< \)\(\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)

Vậy: \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)\(< \)\(\frac{1}{2}\)

5 x 2^2 + 9 x ( X + 3 ) = 5^2

5 x 4 + 9 nhân ( X + 3 ) =25

20 +  9 ( x + 3 ) = 25

9( x + 3) = 5

x + 3 = 5 / 9

x = -4 / 9

ta có \(\frac{1}{10}+\frac{1}{11}+..+\frac{1}{19}< \frac{1}{10}+\frac{1}{10}+..+\frac{1}{10}=1\)

\(\frac{1}{20}+\frac{1}{21}+..+\frac{1}{39}< \frac{1}{20}+\frac{1}{20}+..+\frac{1}{20}=1\)

\(\frac{1}{40}+\frac{1}{41}+..+\frac{1}{79}< \frac{1}{40}+\frac{1}{40}+..+\frac{1}{40}=1\)

Vậy \(P< \frac{1}{9}+1+1+1=\frac{28}{9}\)

ta có 

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+..+\frac{2}{30.31.32}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+..+\frac{1}{30.31}-\frac{1}{31.32}\)

\(=\frac{1}{1.2}-\frac{1}{31.32}=\frac{1}{2}-\frac{1}{992}=\frac{495}{992}\)

Vậy \(A=\frac{495}{1984}\)

\(A=\left(\frac{1}{1.2.3}\right)+\left(\frac{1}{2.3.4}\right)+\left(\frac{1}{3.4.5}\right)+...+\left(\frac{1}{30.31.32}\right)\)

Ta có :

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{30.31.32}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{30.31}\)

\(=\frac{1}{1.2}-\frac{1}{31.32}=\frac{1}{2}-\frac{1}{992}=\frac{495}{992}\)

Vậy ...........