\(B=7^5:7^3-6^2\cdot2+2^3\cdot2^2\)
\(B=7^2-36\cdot2+2^5\)
\(B=49-36\cdot2+32\)
\(B=49-72+32\)
\(B=9\)

a,17.131+17.169

= 17 . (131 + 169)

= 17 . 300

= 5100

b,5⁶:5³+2³.2²

= 5³ + 2⁵

= 125 + 32

= 157

# HOK TỐT #

\(A=4^3-6^3:6^2+11\cdot3^2\\ =64-6+11\cdot9\\ =58+99\\ =157\\ B=5\cdot35-5^2\cdot2\\ =5\cdot\left(35-10\right)\\ =5\cdot25\\ =125\\ C=\left(7-3^3:3^2\right):2^2+99\\ =\left(7-3\right):4+99\\ =4:4+99\\ =1+99=100\\ D=2^7:2^2+5^4:5^3\cdot2^4-3\cdot2^5\\ =2^5+5\cdot2^4-6\cdot2^4\\ =2^4\cdot\left(2+5-6\right)\\ =2^4\\ =16\)

đợi tý

a) 157

b) 125

c) 100

d) 16

\(3^6:3^2+2^3.2^2-3^3.3\)

\(=3^4+2^5-3^4\)

\(=2^5=32\)

     A = 1/99 - 1/99.98 - 1/98.97 - ............... - 1/3.2 - 1/2.1

\(A=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

đặt \(B=\frac{1}{99.98}+\frac{1}{97.87}+...+\frac{1}{3.2}+\frac{1}{2.1}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\)

\(B=1-\frac{1}{99}\)

\(B=\frac{98}{99}\)

\(\Rightarrow A=\frac{1}{99}-\frac{98}{99}=\frac{-97}{99}\)

1. a) 40

    b) 3

    c) 18

    d) 3

2. a) 291

    b) 37

    c) 61

    d) 284

\(\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}\)

\(=\frac{2\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}+\frac{2\left(\sqrt{6}-2\right)}{\left(\sqrt{6}+2\right)\left(\sqrt{6}-2\right)}+\frac{5\sqrt{6}}{\sqrt{6}.\sqrt{6}}\)

\(=\frac{2\sqrt{6}+4}{6-4}+\frac{2\sqrt{6}-4}{6-4}+\frac{5\sqrt{6}}{6}\)

\(=\frac{2\sqrt{6}+4}{2}+\frac{2\sqrt{6}-4}{2}+\frac{5\sqrt{6}}{6}\)

\(=\frac{6\sqrt{6}+12+6\sqrt{6}-12+5\sqrt{6}}{6}\)

\(=\frac{17\sqrt{6}}{6}\)

a; - \(\dfrac{2}{3}\) + \(\dfrac{3}{4}\) - (- \(\dfrac{1}{6}\)) + (- \(\dfrac{2}{5}\))

= - \(\dfrac{2}{3}\) + \(\dfrac{3}{4}\) + \(\dfrac{1}{6}\)  - \(\dfrac{2}{5}\)

=  \(-\dfrac{40}{60}\) + \(\dfrac{45}{60}\) + \(\dfrac{10}{60}\) - \(\dfrac{24}{60}\)

=      \(\dfrac{5}{60}\) + \(\dfrac{10}{60}\) - \(\dfrac{24}{60}\)

  =      \(\dfrac{15}{60}\) - \(\dfrac{24}{60}\)

 =  - \(\dfrac{3}{20}\) 

b; (- \(\dfrac{2}{3}\)) + (- \(\dfrac{1}{5}\)) + \(\dfrac{3}{4}\) - \(\dfrac{5}{6}\) - \(\dfrac{-7}{10}\)

 = - \(\dfrac{2}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{3}{4}\) - \(\dfrac{5}{6}\) + \(\dfrac{7}{10}\)

= - \(\dfrac{40}{60}\) - \(\dfrac{12}{60}\) + \(\dfrac{45}{60}\) - \(\dfrac{50}{60}\) + \(\dfrac{42}{60}\)

= - \(\dfrac{52}{60}\) + \(\dfrac{45}{60}\) - \(\dfrac{50}{60}\) + \(\dfrac{42}{60}\)

= - \(\dfrac{7}{60}\) - \(\dfrac{50}{60}\) + \(\dfrac{42}{60}\)

= - \(\dfrac{57}{60}\) + \(\dfrac{42}{60}\)

= - \(\dfrac{1}{4}\)

+ d; \(\dfrac{1}{100.99}\) - \(\dfrac{1}{99.98}\) - \(\dfrac{1}{98.97}\) - ... - \(\dfrac{1}{3.2}\) - \(\dfrac{1}{2.1}\)

  = \(\dfrac{1}{100.99}\) - (\(\dfrac{1}{99.98}\) + \(\dfrac{1}{98.97}\) + ... + \(\dfrac{1}{3.2}\) + \(\dfrac{1}{2.1}\))

 = \(\dfrac{1}{100.99}\) - (\(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ... + \(\dfrac{1}{98.97}\) + \(\dfrac{1}{98.99}\))

=  \(\dfrac{1}{100.99}\) - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ...+ \(\dfrac{1}{98}\) - \(\dfrac{1}{97}\) + \(\dfrac{1}{98}\) - \(\dfrac{1}{99}\))

=  \(\dfrac{1}{100.99}\) - (\(\dfrac{1}{1}\) - \(\dfrac{1}{99}\))

=  \(\dfrac{1}{100.99}\) - \(\dfrac{98}{99}\)

= - \(\dfrac{9799}{9900}\)