Bài 1

a) \(\dfrac{3}{8}+\dfrac{5}{8}=\dfrac{8}{8}=1\)             

b) \(\dfrac{3}{4}+\dfrac{-7}{16}=\dfrac{12}{16}+\dfrac{-7}{16}=\dfrac{5}{16}\)

c) \(2\dfrac{17}{20}-\dfrac{1}{2}+3\dfrac{3}{20}=\dfrac{57}{20}-\dfrac{1}{2}+\dfrac{63}{20}\)\(=\dfrac{47}{20}+\dfrac{63}{20}=\dfrac{110}{20}=\dfrac{11}{2}\)

d) \(\dfrac{2}{3}-2\dfrac{1}{8}+\dfrac{7}{24}=\dfrac{2}{3}-\dfrac{17}{8}+\dfrac{7}{24}=\dfrac{16}{24}-\dfrac{51}{24}+\dfrac{7}{24}=\dfrac{16-51+7}{24}=\dfrac{-28}{24}=\dfrac{-7}{6}\)

Bài 2 :

a) \(x-\dfrac{7}{4}=3\)

\(x=3+\dfrac{7}{4}\)

\(x=\dfrac{19}{4}\)

b) \(x-\dfrac{1}{2}=\dfrac{4}{16}\cdot\dfrac{8}{3}\)

   \(x-\dfrac{1}{2}=\dfrac{2}{3}\)

  \(x=\dfrac{2}{3}+\dfrac{1}{2}\)

 \(x=\dfrac{5}{6}\)

c) \(\dfrac{15}{11}\div x=\dfrac{45}{22}\)

             \(x=\dfrac{15}{11}\div\dfrac{45}{22}\)

            \(x=\dfrac{2}{3}\)

d) \(\dfrac{8}{3}-2x=\dfrac{8}{5}-1\)

    \(\dfrac{8}{3}-2x=\dfrac{3}{5}\)

           \(2x=\dfrac{8}{3}-\dfrac{3}{5}\)

           \(2x=\dfrac{31}{15}\)

            \(x=\dfrac{31}{15}\div2\)

           \(x=\dfrac{31}{30}\)

1. How much is this watch?

2. How much are these beautiful scarves?

3. Can you tell me how to get to Dong Nai post office?

4. Can you tell me how to get to the station?

5. Peter works hard

6. There aren't any bottles on the shelf

7. We don't have time to prepare the speech

giải cái gì bạn

c c a b b a d b a c

Bài 1: 

a) Ta có: \(A=x^2-2x+7\)

\(=x^2-2x+1+6\)

\(=\left(x-1\right)^2+6\ge6\forall x\)

Dấu '=' xảy ra khi x=1

b) Ta có: \(B=5x^2-20x\)

\(=5\left(x^2-4x+4-4\right)\)

\(=5\left(x-2\right)^2-20\ge-20\forall x\)

Dấu '=' xảy ra khi x=2

d) Ta có: \(D=4x^2+4x+11\)

\(=4x^2+4x+1+10\)

\(=\left(2x+1\right)^2+10\ge10\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)

Bài 2: 

a) Ta có: \(A=-x^2+10x-2\)

\(=-\left(x^2-10x+2\right)\)

\(=-\left(x^2-10x+25-23\right)\)

\(=-\left(x-5\right)^2+23\le23\forall x\)

Dấu '=' xảy ra khi x=5

b) Ta có: \(B=-2x^2+2x+3\)

\(=-2\left(x^2-x-\dfrac{3}{2}\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}-\dfrac{7}{4}\right)\)

\(=-2\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{2}\le\dfrac{7}{2}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

1 that => which

2 he => who

3 hardly => hard

4 quick => quickly

5 turning off  => turning on

6 to => and

7 For => therefore

8 therefore => so

9 to have => having

10 to swim => swimming

11 will go => go

12 don't => won't

13 was => is

14 for => since

15 didn't => haven't

\(v_o=36km/h=10m/s\\ v=54km/h=15m/s\\ s=625m\)

a) Gia tốc của xe là:

\(v^2-v_o^2=2as\rightarrow a=\dfrac{v^2-v_o^2}{2s}=\dfrac{15^2-10^2}{2.625}=0,1\left(m/s\right)\)

b) Thời gian tăng tốc:

\(a=\dfrac{v-v_o}{t}\rightarrow t=\dfrac{v-v_o}{a}=\dfrac{15-10}{0,1}=50\left(s\right)\)

Câu 1: 

A: Sai

B: Sai

C: Đúng

D: Sai

Câu 2: A

\(a,A\left(x\right)=x^2+3x^4-4x+7+x^4\)

\(=x^2+4x^4-4x+7\)

Sắp xếp : \(4x^4+x^2-4x+7\)

\(B\left(x\right)=x^4-2x^2+\left(1-5x^4+4x-4\right)\)

\(=x^4-2x^2+1-5x^4+4x-4\)

\(=-4x^4-2x^2+4x-3\) ( Đã sắp xếp )

\(b,\) \(C\left(x\right)-A\left(x\right)=B\left(x\right)\)

\(\Rightarrow C\left(x\right)=A\left(x\right)+B\left(x\right)\)

\(=4x^4+x^2-4x+7-4x^4-2x^2+4x-3\)

\(=-x^2+4\)

Đặt \(C\left(x\right)=0\Rightarrow-x^2+4=0\Rightarrow x^2=-4\left(ktm\right)\)

Vậy C(x) vô nghiệm