đầy đủ câu trả lời mới đc nhé các bạn!

1.b

2.d

3.c

4.a

5.a

6.a

7.b

8.c

9.a

10.c

Trả lời:

\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2005}{2006}\)

\(\Rightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2005}{2006}\)

\(\Rightarrow1-\frac{1}{5x+6}=\frac{2005}{2006}\)

\(\Rightarrow\frac{1}{5x+6}=1-\frac{2005}{2006}\)

\(\Rightarrow\frac{1}{5x+6}=\frac{1}{2006}\)

\(\Rightarrow5x+6=2006\)

\(\Rightarrow5x=2000\)

\(\Rightarrow x=400\)

Vậy x = 400

Trả lời:

\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2008}-\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\right)=\frac{5}{8}\)\(\frac{5}{8}\)

Đặt \(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\), ta được : \(\frac{x}{2008}-A=\frac{5}{8}\) (*)

\(\Rightarrow A=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)

\(\Rightarrow A=2\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)

\(\Rightarrow A=2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)

\(\Rightarrow A=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)

\(\Rightarrow A=2\left(\frac{1}{4}-\frac{1}{16}\right)=2.\frac{3}{16}=\frac{3}{8}\)

Thay A vào (*) , ta có: 

\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2008}=1\)

\(\Rightarrow x=2008\)

Vậy x = 2008 

:v lớp 10

D

a, 230 + [32 + (x - 5)] =315

32 + (x-5) = 315 - 230

32 + (x-5) =85

x-5=85-32

x-5=53

x=53+5

x=58

A = { 0 ; 1 ; 2 ; 3 ; ...}

B = { -5 }

C = { 8 }

D = { 400}