a, \(\dfrac{15\times14-1}{13\times15+14}\)

 = \(\dfrac{15\times\left(13+1\right)-1}{13\times15+14}\) 

= \(\dfrac{13\times15+15-1}{13\times15+14}\) 

= \(\dfrac{13\times15+14}{13\times15+14}\)

= 1

c, \(\dfrac{1+2+3+4+5+6+7+8+9}{10}\)

= \(\dfrac{\left(9+1\right)\times\left\{\left(9-1\right):1+1\right\}:2}{10}\)

= \(\dfrac{10\times9:2}{10}\)

= \(\dfrac{9}{2}\)

bài 2 

a] = 3 x \(\frac{4343}{7171}\)= \(\frac{17372}{7171}\)= \(\frac{172}{71}\)

b] = \(\frac{1}{33}\)x \(\frac{44}{7}\)= \(\frac{1}{3}\)x \(\frac{4}{7}\)=\(\frac{4}{21}\)

bài 1 

a] y là 9

b] <=> 64y + 36y = 700 - 75 - 225

<=> 100y = 400

<=> y = 4

trên lớp cô sửa rồi nên mình giải luôn:

1) Tìm y

a) y3 + 3y = 12 x 11

    y3 + 3y = 132

    y x 10 + 3 + 3 x 10 + y = 132

   ( y x 10 + y ) + ( 3 x 10 + 3 ) = 132

     11 x y + 33 = 132 

     11 x y = 132 - 33

     11 x y = 99

            y = 99 : 11

            y = 9

b) 64 x y + 225 = 700 - 75 - 36 x y

    64 x y + 225 = 625 - 36 x y

    64 x y + 36 x y = 625 -225

    64 x y + 36 x y = 400

    ( 64 + 36 ) x y = 400

    100 x y = 400

              y =  400 : 100

              y = 4

2) Tính

a) \(\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}\)

\(=\frac{4343}{7171}\times4\)

\(=\frac{43}{71}\times4\)

\(=\frac{172}{71}\)

b) A = \(\frac{1}{33}\times\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)

   Ta có:

   \(\frac{3333}{2020}=\frac{3333:101}{2020:101}=\frac{33}{20}\)

   \(\frac{333333}{303030}=\frac{333333:10101}{303030:10101}=\frac{33}{30}\)

  \(\frac{33333333}{42424242}=\frac{33333333:1010101}{42424242:1010101}=\frac{33}{42}\)

   A = \(\frac{1}{33}\times\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

   A = \(\frac{1}{33}\times33\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

  A = 1 x \(\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

  A = 1 x \(\left(\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\right)\)

  A = 1 x \(\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

  A = 1 x \(\left(\frac{1}{3}-\frac{1}{7}\right)\)

  A = 1 x \(\left(\frac{7}{21}-\frac{3}{21}\right)\)

  A = 1 x \(\frac{4}{21}\)

  A = \(\frac{4}{21}\)

\(\dfrac{121212}{232323}=\dfrac{12x10101}{23x10101}=\dfrac{12}{23}=\dfrac{12.54}{23.54}=\dfrac{648}{1242}\)

\(\dfrac{4343}{5454}=\dfrac{43x101}{54x101}=\dfrac{43}{54}=\dfrac{43.23}{23.54}=\dfrac{989}{1242}\)

mà \(\dfrac{989}{1242}>\dfrac{648}{1242}\)

Nên \(\dfrac{121212}{232323}< \dfrac{4343}{5454}\)

\(\Rightarrow-\dfrac{121212}{232323}>-\dfrac{4343}{5454}\)

\(\Rightarrow\dfrac{-121212}{232323}>\dfrac{4343}{-5454}\)

Đính chính, có thể so sánh cách 2 nhanh hơn.

...\(\dfrac{12}{23}< \dfrac{12+31}{23+31}=\dfrac{43}{54}\)

\(\Rightarrow\dfrac{121212}{232323}< \dfrac{4343}{5454}\)

\(...\Rightarrow\dfrac{-121212}{232323}>\dfrac{4343}{-5454}\)

\(3737\times43-4343\times37\)

\(=160691-160691\)

\(=0\)

\(3737.43-4343.37=101.37.43-101.43.37=0\)

16091-16091=0

nhớ kết bạn với mik nha

\(43^{43}-17^{17}\)

\(=43^{40}.43^3-17^{16}.17\)

\(=\overline{.....1}.\overline{.....7}-\overline{.....1}.7\)

\(=\overline{.....7}-\overline{.....7}\)

\(=\overline{.....0⋮}10\)

\(\Rightarrow dpcm\)

15

\(=15x\left(\dfrac{21}{43}+\dfrac{22}{43}\right)=15x1=15\)

15 nhé bạn

\(\left(-8686\right)^5:\left(4343\right)^5\)

\(=\left(\frac{-8686}{4343}\right)^5\)

\(=\left(\frac{-2}{1}\right)^5\)

\(=\left(-2\right)^5=-32\)

3737 x 43 - 4343 x 37 / 2 + 4 + 6 + ... +100
= 3737 x 43 - 4343 x 37 / 2 + 4 + 6+ ....+ 100
= 0/ 2 + 4+ 6 + ...+ 100
= 0
het rui