\(\text{1)}4.2^3-36:9\Rightarrow\text{4*2^3-36/9}\)

\(\text{2)}52.63+52.18-19.52\Rightarrow\text{52*63+52*18-19*52}\)

\(\text{3)}5^3.5^2.\left[42-\left(15-5\right)^2\right]\Rightarrow\text{5^3*5^2*(42-(15-5)^2}\left(\text{do trong excel không dùng dấu [ ]}\right)\)

\(\text{4)}\frac{1}{2}.\frac{1}{3}+\frac{1}{4}:\frac{1}{5}+\left(\frac{1}{2}\right)^3\Rightarrow\text{1/2*1/3+(1/4)/(1/5)+(1/2)^3}\)

a) = \(\frac{1+2+3+..+11}{3}=\frac{\left(1+11\right)+\left(2+9\right)+..+\left(5+7\right)+6}{3}=\frac{12.5+6}{3}=\frac{66}{3}=22\)

b)          A = \(9+\frac{9}{2}+\frac{9}{4}+\frac{9}{8}+...+\frac{9}{128}+\frac{9}{256}\)

2 x A = 18 + 9 + \(\frac{9}{2}+\frac{9}{4}+\frac{9}{8}+...+\frac{9}{128}\)

Lấy 2 x A - A = 18 - \(\frac{9}{256}\)

A = 4599/256

c) \(B=\frac{5}{1\times3}+\frac{5}{3\times5}+\frac{5}{5\times7}+...+\frac{5}{101\times103}\)

\(B=\frac{5}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{25}{101\times103}\right)\)

\(B=\frac{5}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)=\frac{5}{2}\times\left(1-\frac{1}{103}\right)\)= 255/103

a) 32 - 6 . (8 - 2³) + 18 =  32 - 6 . (8 - 8) + 18

= 32 - 6 . 0 + 18 = 32 + 18 = 50

b) (3 . 5 - 9)³ . (1 + 2 . 3)² + 4²

= (15 - 9)³ . (1 + 6)² + 4²

= 6³ . 7² + 4² = 216 . 49 + 16 = 10 584 + 16 = 10 600

a) 32 - 6 . (8 - 2³) + 18 =  32 - 6 . (8 - 8) + 18

= 32 - 6 . 0 + 18 = 32 + 18 = 50

b) (3 . 5 - 9)³ . (1 + 2 . 3)² + 4²

= (15 - 9)³ . (1 + 6)² + 4²

= 6³ . 7² + 4² = 216 . 49 + 16 = 10 584 + 16 = 10 600

a: \(=\dfrac{1}{2}\cdot\dfrac{2\cdot21+3\cdot27-5\cdot7}{189}\)

\(=\dfrac{1}{2}\cdot\dfrac{88}{189}=\dfrac{44}{189}\)

b: \(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}=\dfrac{1}{2}+\dfrac{4}{5}=\dfrac{13}{10}\)

c: \(=\dfrac{5}{21}-\dfrac{7}{14}\cdot\dfrac{36}{27}=\dfrac{5}{21}-\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{5}{21}-\dfrac{2}{3}=\dfrac{5-14}{21}=\dfrac{-9}{21}=\dfrac{-3}{7}\)

\(\frac{1}{2}\cdot\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{27}\right)=\frac{1}{2}\cdot\left(\frac{6}{27}-\frac{5}{27}+\frac{3}{7}\right)=\frac{1}{2}\cdot\frac{3}{7}+\frac{1}{2}\cdot\frac{1}{27}=\frac{3}{14}+\frac{1}{54}=\frac{44}{189}\)

\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)=\frac{1}{2}+\frac{3}{4}-\frac{3}{4}+\frac{4}{5}=\frac{1}{2}+\frac{4}{5}=\frac{13}{10}\)

\(\frac{1}{3}\cdot\frac{5}{7}-\frac{7}{27}\cdot\frac{36}{14}=\frac{1}{3}\cdot\frac{5}{7}-\frac{7}{27}\cdot\frac{18}{7}=\frac{1}{3}\cdot\frac{5}{7}-\frac{6}{7}=\frac{5}{21}-\frac{14}{21}=\frac{-3}{7}\)

Đặt \(A=\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{99.101}\)

\(\Rightarrow\frac{1}{2}A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{3}-\frac{1}{101}\)

\(\Rightarrow\frac{1}{2}A=\frac{101}{303}-\frac{3}{303}\)

\(\Rightarrow\frac{1}{2}A=\frac{98}{303}\)

\(\Rightarrow A=\frac{98}{303}\div\frac{1}{2}\)

\(\Rightarrow A=\frac{199}{303}\)

\(x+\frac{3}{22}=\frac{27}{121}.\frac{11}{9}\)

\(\Leftrightarrow x+\frac{3}{22}=\frac{27.11}{121.9}\)

\(\Leftrightarrow x+\frac{3}{22}=\frac{3.1}{11.1}\)

\(\Leftrightarrow x+\frac{3}{22}=\frac{3}{11}\)

\(\Leftrightarrow x=\frac{3}{11}-\frac{3}{22}\)

\(\Leftrightarrow x=\frac{6}{22}-\frac{3}{22}\)

\(\Leftrightarrow x=\frac{3}{22}\)