a)

\(175\cdot19+38\cdot175+43\cdot175\\ =175\cdot19+175\cdot38+175\cdot43\\ =175\cdot\left(19+38+43\right)\\ =175\cdot100\\ =17500\)

b)

\(125\cdot75+125\cdot13-80\cdot125\\ =125\cdot75+125\cdot13-125\cdot80\\ =125\cdot\left(75+13-80\right)\\ =125\cdot10\\ =125\cdot8\\ =1000\)

a, 175. 19 + 38. 175 + 43. 175

= 175. 19 + 175. 38 + 175. 43

= 175.(19 + 38 + 43)

= 175. 100

= 17500 

2/

Xét phân số \(\dfrac{2n-3}{n+1}=\dfrac{2n+2-5}{n+1}=\dfrac{2n+2}{n+1}-\dfrac{5}{n+1}=\dfrac{2\left(n+1\right)}{n+1}-\dfrac{5}{n+1}=2-\dfrac{5}{n+1}\)

\(n\in Z\Rightarrow2n-3\inƯ\left(5\right)=\left\{-1;-5;1;5\right\}\)

Ta có bảng:

Vậy \(n\in\left\{-1;1;2;4\right\}\)

1/

(x + 1) + (x + 3) + (x + 5) + ... + (x + 999) = 500

<=> (x + x + x + ... + x) + (1 + 3 + 5 + ... + 999) = 500

Xét tổng A = 1 + 3 + 5 + ... + 999

Số số hạng của A là: (999 - 1) : 2 + 1 = 500 

Tổng A là: (999 + 1) x 500 : 2 = 250 000

Do A có 500 số hạng nên có 500 ẩn x.

Vậy ta có: 500x + 250 000 = 500

=> 500x = -249 500

=> x = 499

Vậy x = 499

67:

a: \(\frac{-x}{2}+\frac{2x}{3}+\frac{x+1}{4}+\frac{2x+1}{6}=\frac83\)

=>\(\frac{-6x}{12}+\frac{8x}{12}+\frac{3\left(x+1\right)}{12}+\frac{2\left(2x+1\right)}{12}=\frac{32}{12}\)

=>-6x+8x+3(x+1)+2(2x+1)=32

=>2x+3x+3+4x+2=32

=>9x=32-5=27

=>x=3

b: \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)

=>\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{12}{26}=\frac{6}{13}\)

=>\(\frac{6}{2x+1}=\frac{6}{13}\)

=>2x+1=13

=>2x=12

=>x=6

Bài 68:

a: \(\frac{1}{51}<\frac{1}{50};\frac{1}{52}<\frac{1}{50};...;\frac{1}{100}<\frac{1}{50}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}<\frac{1}{50}+\frac{1}{50}+\cdots+\frac{1}{50}=\frac{50}{50}=1\) (1)

Ta có: \(\frac{1}{51}>\frac{1}{100};\frac{1}{52}>\frac{1}{100};\ldots;\frac{1}{100}=\frac{1}{100}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+..+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\cdots+\frac{1}{100}\)

=>\(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}>\frac{50}{100}=\frac12\) (2)

Từ (1),(2) suy ra \(\frac12<\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}<1\)

b: Ta có: \(\frac{1}{21}<\frac{1}{20};\frac{1}{22}<\frac{1}{20};\ldots;\frac{1}{30}<\frac{1}{20}\)

Do đó: \(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{30}<\frac{1}{20}+\frac{1}{20}+\cdots+\frac{1}{20}=\frac{10}{20}=\frac12\) (3)

Ta có: \(\frac{1}{31}<\frac{1}{30};\frac{1}{32}<\frac{1}{30};\ldots;\frac{1}{40}<\frac{1}{30}\)

Do đó: \(\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{40}<\frac{1}{30}+\frac{1}{30}+\cdots+\frac{1}{30}=\frac{10}{30}=\frac13\) (4)

Từ (3),(4) suy ra \(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{40}<\frac12+\frac13=\frac56\left(5\right)\)

Ta có: \(\frac{1}{21}>\frac{1}{30};\frac{1}{22}>\frac{1}{30};\ldots;\frac{1}{30}=\frac{1}{30}\)

Do đó: \(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+\cdots+\frac{1}{30}=\frac{10}{30}=\frac13\) (6)

Ta có: \(\frac{1}{31}>\frac{1}{40};\frac{1}{32}>\frac{1}{40};\ldots;\frac{1}{40}=\frac{1}{40}\)

Do đó: \(\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+\cdots+\frac{1}{40}=\frac{10}{40}=\frac14\) (7)

Từ (6),(7) suy ra \(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}>\frac13+\frac14=\frac{7}{12}\) (8)

Từ (5),(8) suy ra \(\frac{7}{12}<\frac{1}{21}+\ldots+\frac{1}{40}<\frac56\)

a) Có 18 học sinh đi đến trường bằng xe đạp.

b) Lớp 6A có 45 học sinh.

c) Tỉ số phần trăm học sinh đi bộ đến trường là:

          (9 : 45) . 100 = 20%

Bài 2: 

a; \(x\) - \(\dfrac{1}{2}\) =  \(\dfrac{3}{10}\).\(\dfrac{5}{6}\)

    \(x\) - \(\dfrac{1}{2}\) = \(\dfrac{1}{4}\)

   \(x\)        = \(\dfrac{1}{4}\) + \(\dfrac{1}{2}\)

   \(x\)        = \(\dfrac{3}{4}\)

Vậy \(x\) = \(\dfrac{3}{4}\)

b; \(\dfrac{x}{5}\) = \(\dfrac{-3}{14}\) \(\times\) \(\dfrac{7}{3}\)

    \(\dfrac{x}{5}\) = \(\dfrac{-1}{2}\)

    \(x\) = \(\dfrac{-1}{2}\) \(\times\) 5

   \(x\) = \(\dfrac{-5}{2}\)

Vậy \(x\) = \(\dfrac{-5}{2}\);

c; \(x\) : \(\dfrac{4}{11}\) = \(\dfrac{11}{4}\) \(\times\) 2

   \(x\) : \(\dfrac{4}{11}\) = \(\dfrac{11}{2}\)

   \(x\) = \(\dfrac{11}{2}\) \(\times\) \(\dfrac{4}{11}\)

   \(x\) = 2

Vậy \(x\) = 2

d; \(x^2\) + \(\dfrac{9}{-25}\)  = \(\dfrac{2}{5}\) : \(\dfrac{5}{8}\)

   \(x^2\) - \(\dfrac{9}{25}\)      =  \(\dfrac{16}{25}\)

   \(x^2\)              = \(\dfrac{16}{25}\) + \(\dfrac{9}{25}\)

   \(x^2\)             = \(\dfrac{25}{25}\)

   \(x^2\)             = 1

  \(\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

Vậy \(x\)\(\in\) {-1; 1}

Bài 3: 

a; A = \(\dfrac{2}{13}\)\(\times\) \(\dfrac{5}{9}\)+ \(\dfrac{2}{13}\)\(\times\)\(\dfrac{4}{9}\) + \(\dfrac{11}{13}\)

   A = \(\dfrac{2}{13}\) \(\times\)(\(\dfrac{5}{9}\) + \(\dfrac{4}{9}\)) + \(\dfrac{11}{13}\)

  A = \(\dfrac{2}{13}\) \(\times\) \(\dfrac{9}{9}\) + \(\dfrac{11}{13}\) 

A = \(\dfrac{2}{13}\) + \(\dfrac{11}{13}\)

A = 1 

b; B = \(\dfrac{1}{10}\).\(\dfrac{4}{11}\) + \(\dfrac{1}{10}\).\(\dfrac{8}{11}\) - \(\dfrac{1}{10}\).\(\dfrac{1}{11}\)

   B =   \(\dfrac{1}{10}\) x (\(\dfrac{4}{11}\) + \(\dfrac{8}{11}\) - \(\dfrac{1}{11}\))

  B =   \(\dfrac{1}{10}\) x (\(\dfrac{12}{11}\) - \(\dfrac{1}{11}\))

  B =     \(\dfrac{1}{10}\) x  \(\dfrac{11}{11}\)

 B = \(\dfrac{1}{10}\)

a) \(\dfrac{5}{11}\cdot\dfrac{5}{7}+\dfrac{5}{11}\cdot\dfrac{2}{7}+\dfrac{6}{11}=\dfrac{5}{11}\cdot\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}=\dfrac{5}{11}\cdot1+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=\dfrac{11}{11}=1\) 

b) \(\dfrac{3}{13}\cdot\dfrac{6}{11}+\dfrac{3}{13}\cdot\dfrac{9}{11}-\dfrac{3}{13}\cdot\dfrac{4}{11}=\dfrac{3}{13}\cdot\left(\dfrac{6}{11}+\dfrac{9}{11}-\dfrac{4}{11}\right)=\dfrac{3}{13}\cdot\dfrac{11}{11}=\dfrac{3}{13}\cdot1=\dfrac{3}{13}\) 

c) \(\dfrac{-5}{6}\cdot\dfrac{4}{19}+\dfrac{7}{12}\cdot\dfrac{4}{-19}-\dfrac{40}{57}=\dfrac{-5}{6}\cdot\dfrac{4}{19}+\dfrac{-7}{12}\cdot\dfrac{4}{19}-\dfrac{40}{57}=\dfrac{4}{19}\cdot\left(\dfrac{-5}{6}+\dfrac{-7}{12}\right)-\dfrac{40}{57}\)

\(=\dfrac{4}{19}\cdot\dfrac{-17}{12}-\dfrac{40}{47}=\dfrac{-17}{57}-\dfrac{40}{57}=\dfrac{-57}{57}=-1\)

d) \(\left(\dfrac{11}{4}\cdot\dfrac{-5}{9}+\dfrac{4}{9}\cdot\dfrac{11}{-4}\right)\cdot\dfrac{8}{33}=\left(\dfrac{11}{4}\cdot\dfrac{-5}{9}+\dfrac{-4}{9}\cdot\dfrac{11}{4}\right)\cdot\dfrac{8}{33}=\dfrac{11}{4}\cdot\dfrac{8}{33}\cdot\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)\)

\(=\dfrac{11}{4}\cdot\dfrac{8}{33}\cdot1=\dfrac{11\cdot8}{4\cdot33}=\dfrac{2}{3}\) 

e) \(\left(\dfrac{12}{61}-\dfrac{31}{22}+\dfrac{14}{91}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)=\left(\dfrac{12}{61}-\dfrac{31}{22}+\dfrac{14}{91}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\)

\(=\left(\dfrac{12}{61}-\dfrac{31}{22}+\dfrac{14}{91}\right)\cdot0=0\)

 giúp mik với gấp quá

helpp mee huhuhuhu

a:

b: Các điểm thuộc đường thẳng c là M,Q

Các điểm không thuộc đường thẳng c là P,N

c: Các đoạn thẳng có trong hình là MQ,MP,PN,MN,PQ

a:

b: Các điểm thuộc đường thẳng c là M,Q

Các điểm không thuộc đường thẳng c là P,N

c: Các đoạn thẳng có trong hình là MQ,MP,PN,MN,PQ