a)  \(\left(\frac{2}{5}-\frac{1}{2}\right)^2-\frac{11}{5}:\frac{-11}{5}=\left(-\frac{1}{10}\right)^2+1=1\frac{1}{100}\)

b)  \(\left(-\frac{5}{7}\right)^2+8.\left(0,5\right)^2+\left(-1\right)^{2010}=\frac{25}{49}+2+1=3\frac{25}{49}\)

c)  \(\frac{9999^2}{3333^2}+\left(0,5\right)^2.\left(-2\right)^4-\left(-\frac{4}{3}\right)^2=9+1-\frac{16}{9}=8\frac{2}{9}\)

d) \(\left|-\frac{2}{5}+\frac{1}{7}\right|:\frac{-3}{35}+\frac{-3}{7}.\frac{7}{5}=\frac{9}{35}.\frac{35}{-3}-\frac{3}{5}=-3\frac{3}{5}\)

e) \(\frac{1}{2}-\left(-0,4\right)+\frac{1}{3}+\frac{1}{5}-\frac{-1}{6}+\frac{-4}{35}+\frac{1}{41}\)

\(=\frac{1}{2}+\frac{2}{5}+\frac{1}{3}+\frac{1}{5}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}=1\frac{732}{1435}\)

ick mình

Phải làm đúng mới k

vì x/5=y/4=> 4x=5y mà x.y=20=>

x=5,y=4

Đặt 

x/5=y/4=k

khi đó:

x=5k

y=4k

Ta lại có:

x.y=4k.5k=20k^2=20

=> K=+-1

Khi k=1

Khi k=-1

Giải ra nhé

a) \(\frac{45^{10}.5^{20}}{75^{15}}\)

=
\(\frac{\left(5.9\right)^{10}.5^{20}}{\left(5.15\right)^{15}}\)

=  \(\frac{5^{10}.9^{10}.5^{20}}{5^{15}.15^{15}}\)

=    \(\frac{5^{10}.3^{20}.5^{20}}{5^{15}.15^{15}}\)

=    \(\frac{5^{10}.15^{20}}{5^{15}.15^{15}}\)

=     \(\frac{15^5}{5^5}\)

=     \(\frac{3^5.5^5}{5^5}\)

= \(3^5\)

b) \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}\)

=   \(\frac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}\)

=    \(\frac{2^5}{0,4}\)

= \(2^5\) : 0,4

(=) 32 : \(\frac{2}{5}\)

= 90

c) \(\frac{2^{15}.9^4}{6^6.8^3}\)

=  \(\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}\)

=    \(\frac{2^{15}.3^8}{2^6.3^6.2^9}\)

=   \(3^2\)

M<1 => \(\frac{x-3}{x+2}\)<1

       <=> \(\frac{x-3}{x+2}\)- 1 < 0

       <=> \(\frac{x-3}{x+2}\)-\(\frac{x+2}{x+2}\)< 0

       <=> \(\frac{x-3-x-2}{x+2}\)< 0

       <=>              -5         < 0

=> Vô nghiệm

các bạn ơi đề bài của bài trên là THỰC HIỆN PHÉP TÍNH HỢP LÝ ( NẾU CÓ ) nhé . Lúc tạo câu hỏi mình không để ý nên ghi sai

Ta có :

\(C=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{n-1}{n!}\)

\(=\left(\frac{2}{2!}+\frac{3}{3!}+\frac{4}{4!}+...+\frac{n}{n!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{n!}\right)\)

\(=\left(1+\frac{1}{2!}+\frac{1}{3!}+....+\frac{1}{\left(n-1\right)!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+....+\frac{1}{n!}\right)\)

\(=1+\frac{1}{2!}+\frac{1}{3!}+....+\frac{1}{\left(n-1\right)!}-\frac{1}{2!}-\frac{1}{3!}-\frac{1}{4!}-....-\frac{1}{n!}\)

\(=1-\frac{1}{n!}=\frac{n!-1}{n!}\)

A=\([\)\(\frac{2}{7}\)\(\times\)(\(\frac{1}{4}-\frac{1}{3}\))\(]\)\(\div\)\([\)(\(\frac{2}{7}\times\)(\(\frac{3}{9}-\frac{2}{5}\))\(]\)
  =(\(\frac{2}{7}\times\)\(\frac{-1}{12}\))\(\div(\)\(\frac{2}{7}\times\)\(\frac{-1}{15}\))
=\(\frac{-1}{42}\)\(\div\)\(\frac{-2}{35}\)
=\(\frac{-1}{42}\)\(\times\)\(\frac{35}{-2}\)
=\(\frac{5}{12}\)