(36 + 52) x 46 + (36 + 52) x 53 + 88

=   88      x 46 +        88     x     141

=      4048      +              12408

=                    16456

\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}.\)

\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10\cdot\frac{4}{55}=\frac{3}{11}\)

\(x-\frac{8}{11}=\frac{3}{11}\)

 x = 1

a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{47}-\dfrac{4}{47}+\dfrac{9}{53}\right)}=\dfrac{3}{4}\)

b: \(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(=2\cdot\dfrac{1004}{2010}=\dfrac{2008}{2010}=\dfrac{1004}{1005}\)

c: \(S=\dfrac{1}{3\cdot6}+\dfrac{1}{6\cdot9}+...+\dfrac{1}{30\cdot33}\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)

Bài 2: 

Ta có: (x-3)(x+4)>0

=>x>3 hoặc x<-4

Bài 3:

a: \(5S=5-5^2+...+5^{99}-5^{100}\)

\(\Leftrightarrow6S=1-5^{100}\)

hay \(S=\dfrac{1-5^{100}}{6}\)

  \(\frac{\left(72\times1995-1996\right)\times\left(18-36\div2\right)\times\left(148\div5-7\right)}{1+3+5+...+1993+1995}\)

= \(\frac{\left(72\times1995-1996\right)\times\left(18-18\right)\times\left(148\div5-7\right)}{1+3+5+...+1993+1995}\)

= \(\frac{\left(72\times1995-1996\right)\times0\times\left(148\div5-7\right)}{1+3+5+...+1993+1995}\)

= \(\frac{0}{1+3+5+...+1993+1995}\)

= 0 : (1 + 3 + 5 + ... + 1993 + 1995)

= 0

(72x1995-1996)x(18-36:2)x(148:5-7) / 1+3+5+....+1993+1995

=(72x1995-1996)x0x(148:5-7) / 1+3+5+...+1993+1995

=0 / 1+3+5+...+1993+1995

=0

Tích nha

lêu lêu

Gọi 2 số tự nhiên cần tìm là x,y.Mà 36 là ƯCLN của x,y => x = 36m;y=36n

Theo bài ra: x+y = 132 => 36m+36n=132 => 36(m+n) = 132 => m+n=... Đề bị làm sao ý -_-