Số tiền Mai phải trả cho 2 cái áo là:

\(110000\cdot2\cdot\left(1-20\%\right)=176000\left(đồng\right)\)

Số tiền Mai phải trả cho 1 đôi giày là:

\(480000\cdot\left(1-20\%\right)=384000\left(đồng\right)\)

Số tiền Mai còn lại là:

700000-176000-384000=140000(đồng)

còn lại số tiền là :
      700000-(110000+480000)=110000 ( đồng )

Giúp tui với 🥺

B. MN=5 cm

Ta có:

\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)

\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)

\(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\)

\(A=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)

\(A=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(A=2.\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(A=2.\left(\dfrac{4}{16}-\dfrac{1}{16}\right)\)

\(A=2.\dfrac{3}{16}\)

\(A=\dfrac{3}{8}\)

Vậy A = \(\dfrac{3}{8}\)

\(A=\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{120}\)

\(=\dfrac{2}{20}+\dfrac{2}{30}+...+\dfrac{2}{240}\)

\(=2\left(\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{15\cdot16}\right)\)

\(=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=2\cdot\dfrac{3}{16}=\dfrac{3}{8}\)

\(\dfrac{3}{8}=\dfrac{3\cdot5}{8\cdot5}=\dfrac{15}{40};\dfrac{2}{5}=\dfrac{2\cdot8}{5\cdot8}=\dfrac{16}{40}\)

mà 15<16

nên \(\dfrac{3}{8}< \dfrac{2}{5}\)

3/8= 3.5/ 8.5= 15/40

2/5= 2.8/ 5.8= 16/40

16/40 > 15/40 => 2/5 > 3/8  đẹp zai :)

ko gửi đc nek.

g. \(-\dfrac{2}{3}\times\dfrac{4}{5}+\dfrac{1}{5}\div\dfrac{9}{11}=-\dfrac{8}{15}+\dfrac{1}{5}\times\dfrac{11}{9}=-\dfrac{8}{15}+\dfrac{11}{45}=\dfrac{24}{45}+\dfrac{11}{45}=\dfrac{35}{45}=\dfrac{7}{9}\)

h.

\(\left(-6,2\div2+3,7\right)\div0,2=\left(-3,1+3,7\right)\div0,2=0,6\div0,2=3\)

k.

\(\dfrac{2}{3}+\dfrac{1}{5}\times\dfrac{10}{7}=\dfrac{2}{3}+\dfrac{10}{35}=\dfrac{2}{3}+\dfrac{2}{7}=\dfrac{14}{21}+\dfrac{6}{21}=\dfrac{20}{21}\)

m.

\(\dfrac{2}{7}+\dfrac{5}{7}\times\dfrac{14}{25}=\dfrac{2}{7}+\dfrac{1\times2}{1\times5}=\dfrac{2}{7}+\dfrac{2}{5}=\dfrac{10}{35}+\dfrac{14}{35}=\dfrac{24}{35}\)

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

...

\(\dfrac{1}{2021^2}< \dfrac{1}{2020\cdot2021}=\dfrac{1}{2020}-\dfrac{1}{2021}\)

Do đó: \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2021^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\)

=>\(B< 1-\dfrac{1}{2021}< 1\)

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12345678910

a: \(4n-5⋮n\)

=>\(-5⋮n\)

=>\(n\in\left\{1;-1;5;-5\right\}\)

b: \(-11⋮n-1\)

=>\(n-1\inƯ\left(-11\right)\)

=>\(n-1\in\left\{1;-1;11;-11\right\}\)

=>\(n\in\left\{2;0;12;-10\right\}\)

Tủ lạnh và ma

cụ thể là ma nữ

Ta có:

Sửa đề:

1/(3.5) + 1/(5.7) + 1/(7.9) + ... + 1/(99.101)

= 1/2 . (1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/99 - 1/101)

= 1/2 . (1/3 - 1/101)

= 1/2 . 98/303

= 49/303

Ta có:

303 > 300

⇒ 49/303 < 49/300 (1)

Mà 49 < 50

⇒ 49/300 < 50/300 = 1/6 (2)

Từ (1) và (2) ⇒ 49/303 < 1/6

Vậy 1/(3.5) + 1/(5.7) + 1/(7.9) + ... + 1/(99.101) < 1/6