\(S=\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{600}+\dfrac{1}{650}\)

\(=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{24\cdot25}+\dfrac{1}{25\cdot26}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}\)

\(=\dfrac{1}{2}-\dfrac{1}{26}=\dfrac{13-1}{26}=\dfrac{12}{26}=\dfrac{6}{13}\)

(đề bài)

=1-1/2+1-1/6+1-1/12+...+1-1/650

=(1+1+1+...+1)-(1/2+1/6+1/12+...+1/650)

=25-(1/1.2+1/2.3+1/3.4+...+1/25.26)

25-(1-1/2+1/2-1/3+1/3-1/4+...+1/25-1/26)

=25-(1-1/26)

=25/1-25/26

=625/26

    ĐS:625/26

 Hi!Hi!Kết bạn nhé!!!!!!!!!!

tại sao:(1+1+1+...+1)=25

C=1/2+5/6+11/12+...+649/650

  =(1-1/2)+(1-1/6)+(1-1/12)+...+(1-1/650)

  =(1+1+1+...+1)-(1/2+1/6+1/12+...+1/650)

  =109+(1/1.2+1/2.3+1/3.4+...+1/25.26)

  =109+(1-1/2+1/2-1/3+1/3-1/4+...=1/25-1/26)

 =109+(1-1/26)=109+25/26

a) (2x-1)³ = 27
(2x-1)³ = 9³
2x-1 = 9
2x = 9+1
2x = 10
x = 10:5
x = 2
Vậy x = 2

b) (2x-1)⁴ = 81
(2x-1)⁴ = (\(\pm\)3⁴)
2x-1 = \(\pm\)3
Trường hợp 1:
2x-1 = 3
2x = 3+1
2x = 4
x = 4:2
x = 2
Trường hợp 2:
2x-1 = -3
2x = -3+1
2x = -2
x = -2:2
x = -1
Vậy x \(\in[_{ }2;-1]\)
Vì không tìm thấy ngoặc nhọn nên mình dùng tạm ngoặc vuông nhé

À phần b) bạn sửa dòng (2x-1)⁴ = (\(\pm\)3⁴) thành (2x-1)⁴ = (\(\pm\)3)⁴ nhé
Mình vừa viết nhầm

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{650}\)  = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{25.26}\)= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{25}-\frac{1}{26}\)= \(1-\frac{1}{26}=\frac{25}{26}\)

~Học tốt~

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{650}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{25.26}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{25}-\frac{1}{26}\)

\(=1-\frac{1}{26}=\frac{25}{26}\)

1/2+1/6+1/12+....+1/650

=1/1.2+1/2.3+....+1/25.26

=1-1/2+1/2+.....+1/25-1/26

=1-1/26

=25/26

a) \(2\dfrac{3}{4}-x=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{11}{4}-x=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{11}{4}-\dfrac{3}{4}=\dfrac{8}{4}=2\)

b) \(x:\dfrac{5}{6}=-\dfrac{3}{5}\)

\(\Rightarrow x=-\dfrac{3}{5}.\dfrac{5}{6}=-\dfrac{15}{30}=-\dfrac{1}{2}\)

c) \(1\dfrac{1}{3}+\dfrac{2}{3}:x=1\)

\(\Rightarrow\dfrac{2}{3}:x=1-1\dfrac{1}{3}\)

\(\Rightarrow\dfrac{2}{3}:x=-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{3}:-\dfrac{1}{3}\)

\(\Rightarrow x=-2\)

d) \(x-\dfrac{1}{9}=\dfrac{8}{3}\)

\(\Rightarrow x=\dfrac{8}{3}+\dfrac{1}{9}\)

\(\Rightarrow x=\dfrac{25}{9}\)

e) \(\dfrac{1}{2}x+650\%x-x=-6\)

\(\Rightarrow\dfrac{1}{2}x+\dfrac{13}{2}x-x=-6\)

\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{13}{2}-1\right)-6\)

\(\Rightarrow6x=-6\)

\(\Rightarrow x=\dfrac{-6}{6}=-1\)

g) \(2\left(x-\dfrac{1}{2}\right)+3\left(-1+\dfrac{x}{3}\right)=x\left(\dfrac{2}{x}-1\right)\) \(\text{Đ}K:x\ne0\)

\(\Rightarrow2x-1-3+x=2-x\)

\(\Rightarrow3x-4=2-x\)

\(\Rightarrow3x+x=2+4\)

\(\Rightarrow4x=6\)

\(\Rightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)

h) \(x-\dfrac{2}{20}=-\dfrac{5}{2}-x\)

\(\Rightarrow x+x=-\dfrac{5}{2}+\dfrac{2}{20}\)

\(\Rightarrow2x=-\dfrac{12}{5}\)

\(\Rightarrow x=-\dfrac{12}{5}:2=-\dfrac{6}{5}\)

i) \(\left(\dfrac{x}{2}-1\right)^3+2=-\dfrac{11}{8}\)

\(\Rightarrow\left(\dfrac{x}{2}-1\right)^3=-\dfrac{11}{8}-2\)

\(\Rightarrow\dfrac{x}{2}-1=\sqrt[3]{-\dfrac{27}{8}}\)

\(\Rightarrow\dfrac{x}{2}-1=-\dfrac{3}{2}\)

\(\Rightarrow\dfrac{x}{2}=-\dfrac{3}{2}+1\)

\(\Rightarrow x=-\dfrac{1}{2}.2=-1\)

k) \(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\\dfrac{3}{4}-1\dfrac{1}{2}x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\\dfrac{3}{2}x=\dfrac{3}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}.3=-\dfrac{3}{2}\\x=\dfrac{3}{4}:\dfrac{3}{2}=\dfrac{1}{2}\end{matrix}\right.\)

\(2^2+4^2+6^2+...+24^2=2^2.\left(1^2+2^2+3^2+...+12^2\right)\)