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) \(\dfrac{8}{3}-2x=\dfrac{8}{5}-1\)

    \(\dfrac{8}{3}-2x=\dfrac{3}{5}\)

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

           \(2x=\dfrac{31}{15}\)

            \(x=\dfrac{31}{15}\div2\)

           \(x=\dfrac{31}{30}\)

42

bố 42 con 12

42 tuổi

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