Sửa đề: \(5^9\cdot49^2\)

\(=\dfrac{5^{10}\cdot7^3-5^9\cdot7^4}{5^9\cdot7^3+5^9\cdot14^3}-\dfrac{2^{12}-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}\)

\(=\dfrac{5^9\cdot7^3\left(5-7\right)}{5^9\cdot7^3\left(1+8\right)}-\dfrac{2^{12}\left(1-3^4\right)}{2^{12}\left(3^6+3^5\right)}=\dfrac{-2}{9}+\dfrac{80}{972}\)

=-34/243

2.3/7+(2/9-10/7)-5/3:1/9=-967/63

kết quả lớn lắm nè: -967/63 

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Ta có:

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2+10^2}\)

\(=\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)+...+\left(\dfrac{1}{9^2}-\dfrac{1}{10^2}\right)\)

\(=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)

\(=\dfrac{1}{1^2}-\dfrac{1}{10^2}\)

\(=1-\dfrac{1}{100}\)

Vì \(1-\dfrac{1}{100}< 1\)

Nên \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2+10^2}< 1\) (Đpcm)

\(vt:\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^2+10^2}\)

=\(\dfrac{1}{1}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+..+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)

=\(\dfrac{1}{1}-\dfrac{1}{10^2}\)

=>A<1

Lời giải chi tiết:

2=1+1          6=2+4          8=5+3          10=8+2

3=1+2          6=3+3          8=4+4          10=7+3

4=3+1          7=6+1          9=8+1          10=6+4

4=2+2          7=5+2          9=7=2          10=5+5

5=4+1          7=4+3         9=6+3           10=10+0

5=3+2          8=7+1         9=5=4           10=0+10

6=5+1          8=6=2         10=9+1         1=0+1

a: \(x-\frac25+\frac{7}{12}=\frac12\cdot\frac34\cdot\frac{-9}{16}\)

=>\(x-\frac{24}{60}+\frac{35}{60}=\frac38\cdot\frac{-9}{16}=\frac{-27}{128}\)

=>\(x=-\frac{27}{128}-\frac{11}{60}=\frac{-27\cdot15-11\cdot32}{1920}=-\frac{757}{1920}\)

b: \(\frac49x-\frac12=-\frac59\)

=>\(\frac49x=-\frac59+\frac12=\frac{-10+9}{18}=\frac{-1}{18}\)

=>\(x=-\frac{1}{18}:\frac49=-\frac{1}{18}\cdot\frac94=\frac{-1}{2\cdot4}=-\frac18\)

c: \(-\frac23x+\frac58=\frac{7}{12}\)

=>\(-\frac23x=\frac{7}{12}-\frac58=\frac{14}{24}-\frac{15}{24}=-\frac{1}{24}\)

=>\(x=\frac{1}{24}:\frac23=\frac{1}{24}\cdot\frac32=\frac{1}{8\cdot2}=\frac{1}{16}\)

d: \(\frac25x:\frac73\cdot\frac18=\frac{29}{15}\)

=>\(x\cdot\frac25\cdot\frac37\cdot\frac18=\frac{29}{15}\)

=>\(x\cdot\frac{3}{4\cdot5\cdot7}=\frac{29}{15}\)

=>\(x=\frac{29}{15}:\frac{3}{140}=\frac{29}{15}\cdot\frac{140}{3}=\frac{29\cdot28}{3\cdot3}=\frac{812}{9}\)