Có: A=\(\dfrac{3}{1.5}+\dfrac{3}{5.10}+...+\dfrac{3}{100.105}\)

=> A=\(3.\dfrac{5}{5}\left(\dfrac{1}{1.5}+\dfrac{1}{5.10}+...+\dfrac{1}{100.105}\right)\)

=> A= \(3.\dfrac{1}{5}\left(\dfrac{5}{1.5}+\dfrac{5}{5.10}+...+\dfrac{5}{100.105}\right)\)

=> A=\(\dfrac{3}{5}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{105}\right)\)

=> A= \(\dfrac{3}{5}\left(1-\dfrac{1}{105}\right)\)=\(\dfrac{3}{5}.\dfrac{104}{105}=\dfrac{312}{525}\)

Ta có:

\(A=\frac{3}{1\cdot5}+\frac{3}{5\cdot10}+...+\frac{3}{100\cdot105}\)

\(=\frac{3}{5}\cdot\left(\frac{5}{1\cdot5}+\frac{5}{5\cdot10}+...+\frac{5}{100\cdot105}\right)\)

\(=\frac{3}{5}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{105}\right)\)

\(=\frac{3}{5}\left(1-\frac{1}{105}\right)=\frac{3}{5}\cdot\frac{104}{105}=\frac{312}{525}\)

Bằng 1 nha :)

25\(^{10}\).(\(\frac15\))\(^{20}\)

= (5\(\)\(^2\))\(^{10}\).(\(\frac15\))\(^{20}\)

= 5\(^{20}\).(\(\frac15\))\(^{20}\)

= (5.\(\frac15\))\(^{20}\)

= 1\(^{20}\)

= 1

mọi người giúp mình nha

a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.5^8}=7\)

b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3^9.5^2.5^3}{3.5.5^4.3^8}=\frac{3^9.5^5}{3^9.5^5}=1\)

c) \(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{91}.3^{16}}=\frac{2^{50}.3^{16}\left(3^{45}+2^{40}\right)}{2^{51}.3^{16}\left(3^{45}+2^{40}\right)}=\frac{1}{2}\)

d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2\)

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

\(=\frac{1}{100}+\frac{121}{100}=\frac{122}{100}=\frac{61}{50}\)

\(A=\frac{25^3.5^5}{6.5^{10}}\)

\(A=\frac{\left(5^2\right)^3.5^5}{6.5^{10}}\)

\(A=\frac{5^6.5^5}{6.5^{10}}\)

\(A=\frac{5^{11}}{6.5^{10}}\)

\(A=\frac{5}{6}\)

(Dùng phương pháp giảm ước)

\(=\frac{\left(5^2\right)^3.5^5}{6.5^{10}}\)

\(=\frac{5^6.5^5}{6.5^{10}}\)

\(=\frac{5^{11}}{6.5^{10}}\)

\(=\frac{5}{6}\)

VẬY \(A=\frac{5}{6}\)

nhờ 2 câu nha

tim x 

\(1.\) sai đề rồi nha. dề đúng phải là  \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)
  đặt \(\left(x^2+1\right)=k\)   \(\Rightarrow\)biểu thức trên có dạng là  \(k^2+3xk+2x^2=0\)
                                       \(\Leftrightarrow\)                                 \(\left(k+x\right)\left(k+2x\right)=0\)     
suy ra             \(k+x=0\)              hoặc                      \(k+2x=0\)
            \(x^2+1+x=0\)                                \(x^2+1+2x=0\)
           bấm máy tình không ra                                     \(\left(x+1\right)^2=0\)     
           nên ko có giá trị của x                                              \(x+1=0\)
                                                                                                  \(x=-1\)
             Vậy \(x=-1\)

S = 1.3 + 3.5 + 5.7 + ...+ 99.101

=>6S = 1.3.6 + 3.5.6 + 5.7.6 + ...+ 99.101.6

6S = 1.3.(5+1) + 3.5.(7-1) + 5.7.(9-3) + ...+ 99.101.(103-97)

6S = 1.3.5 + 1.3 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ...+ 99.101.103 - 97.99.101

6S = 1.3 + 99.101.103

S = 171 650

S = 1.3 + 3.5 + 5.7 + ...+ 99.101

=>6S = 1.3.6 + 3.5.6 + 5.7.6 + ...+ 99.101.6

6S = 1.3.(5+1) + 3.5.(7-1) + 5.7.(9-3) + ...+ 99.101.(103-97)

6S = 1.3.5 + 1.3 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ...+ 99.101.103 - 97.99.101

6S = 1.3 + 99.101.103

S = 171 650

Gọi A = 1.3+3.5+5.7+...+21.23

=> A = 1.(1+2)+3.(3+2)+5.(5+2)+...+21.(21+2)

=> A = 1²+1.2+3²+2.3+5²+2.5+...+21²+2.21

=> A = 1²+3²+5²+...+21²+(1.2+3.2+5.2+...+2.21)

Gọi B = 1²+3²+5²+...+21²

=> B = (21.22.23)/3

Gọi C = 1.2+2.3+5.2+...+2.21

=> C = 2(1+3+5+...+21)

=> C = 2{(21+1).[(21-1):2+1]}/2

=> C = 22x11=242

Vậy A = (21.22.23)/3+242

B = ( 1/ 3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ......+ 1/2015 - 1/ 2017 ) x3 + 1

B =(1/3 - 1/ 2017 ) x3 +1

B= ? ? ?