a: \(=\dfrac{5}{9}\cdot\dfrac{7}{4}+\dfrac{13}{9}\cdot\dfrac{7}{4}=\dfrac{7}{4}\cdot2=\dfrac{7}{2}\)

b: \(=\dfrac{14}{13}\cdot2+\dfrac{3}{13}\cdot2-2\cdot\dfrac{7}{13}=\dfrac{28}{13}+\dfrac{6}{13}-\dfrac{14}{13}=\dfrac{20}{13}\)

48/49

Đặt \(B=\frac{1}{1.2}+\frac{5}{2.7}+\frac{8}{7.15}+\frac{13}{15.28}+\frac{21}{28.49}+\frac{32}{49.81}\)

   \(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}+\frac{1}{28}-\frac{1}{49}+\frac{1}{49}-\frac{1}{81}\)

   \(\Rightarrow B=1-\frac{1}{81}\)

      \(\Rightarrow B=\frac{80}{81}\)

Giải:

\(A^2-\left(\dfrac{3}{5}\right)^2=\dfrac{1}{1.2}+\dfrac{1}{2.7}+\dfrac{1}{7.5}+\dfrac{1}{5.13}+\dfrac{1}{13.8}+\dfrac{1}{8.19}+\dfrac{1}{19.11}+\dfrac{1}{11.25}\) 

Gọi: \(B=\dfrac{1}{1.2}+\dfrac{1}{2.7}+\dfrac{1}{7.5}+\dfrac{1}{5.13}+\dfrac{1}{13.8}+\dfrac{1}{8.19}+\dfrac{1}{19.11}+\dfrac{1}{11.25}\) 

\(B=\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+\dfrac{1}{13.16}+\dfrac{1}{16.19}+\dfrac{1}{19.22}+\dfrac{1}{22.25}\right):\dfrac{1}{2}\) \(B=\left[\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{19.22}+\dfrac{3}{22.25}\right)\right]:\dfrac{1}{2}\) 

\(B=\left[\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{25}\right)\right]:\dfrac{1}{2}\) 

\(B=\left[\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{25}\right)\right]:\dfrac{1}{2}\) 

\(B=\left[\dfrac{1}{3}.\dfrac{24}{25}\right]:\dfrac{1}{2}\) 

\(B=\dfrac{8}{25}:\dfrac{1}{2}\) 

\(B=\dfrac{16}{25}\) 

\(\Rightarrow A^2-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}\) 

                  \(A^2=\dfrac{16}{25}+\dfrac{9}{25}\) 

                  \(A^2=1\) 

\(\Rightarrow A^2=1^2\) hoặc \(A^2=\left(-1\right)^2\) 

      \(A=1\) hoặc \(A=-1\) 

Chúc bạn học tốt!

\(a^2-\frac{3}{5^2}=\frac{1}{1.2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}+\frac{1}{19.11}+\frac{1}{11.25}\)

\(a^2-\frac{3}{5^2}=2.\left(\frac{1}{2.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}+\frac{1}{19.22}+\frac{1}{22.25}\right)\)

\(a^2-\frac{3}{5^2}=2.\frac{1}{3}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{22}-\frac{1}{25}\right)\)

\(a^2-\frac{3}{5^2}=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{25}\right)\)

=> \(a^2-\frac{3}{25}=\frac{2}{3}.\frac{23}{50}=\frac{23}{75}\)

=> \(a^2=\frac{23}{75}+\frac{3}{25}=\frac{32}{75}\)

=> \(a=\sqrt{\frac{32}{75}}\)(Nếu thế thì đây phải là đề của lớp 7 chứ nhỉ)

Sửa đề: \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot7}+\frac{1}{7\cdot5}+\cdots+\frac{1}{50\cdot97}\)

\(\frac{A}{2}=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\cdots+\frac{1}{97\cdot100}\)

=>\(\frac12A=\frac13\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\cdots+\frac{3}{97\cdot100}\right)\)

\(=\frac13\left(1-\frac14+\frac14-\frac17+\cdots+\frac{1}{97}-\frac{1}{100}\right)\)

=>\(\frac12\cdot A=\frac13\left(1-\frac{1}{100}\right)=\frac13\cdot\frac{99}{100}=\frac{33}{100}\)

=>\(A=\frac{33}{100}:\frac12=\frac{33}{100}\cdot2=\frac{33}{50}\)

a) 3.2x+(-1.2)x+2.7= -4.9

⇔(3.2-1.2)x= -4.9-2.7

⇔2x= -7.6

⇔x= -7.6:2

⇔x= -3.8

b) \(\dfrac{37-x}{x+13}\)=\(\dfrac{3}{7}\)

⇔(37-x)*7=(x+13)*3

⇔259-7x=3x+39

⇔259-39=3x+7x

⇔220=10x

⇔x= 220:10

⇔x=22