ĐKXĐ: x khác 0 

1/(1 ×2)+  1/(2 ×3)+  1/(3 ×4 )+⋯.+  1/(x ×( x+1))=  30/31

1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/x - 1/x +1 = 30/31

1 - 1/x+1 = 30/31

x+1-1/x+1 = 30/31

x/x+1 = 30/31

31x = 30x + 30

31x - 30x = 30

x = 30(TM)

Vậy x là 30.

Bạn ơi dấu . là dấu nhân nhé!!!

a. 15 -(x - 8) = 2

          (x - 8)=15-2

          (x - 8)=13

            x     =13+8

            x=21

b. 2.x-19=31

    2x     =31+19

    2x     =50

      x     =50:2

      x     =25

c. 30-3.(x+1)=15

        3.(x+1)=30-15

        3.(x+1)=15

           (x+1)=15:3

           (x+1)=5

            x     =5-1

            x=4 

T mk nhé bạn ^...^ ^_^

Gạch dưới là dấu trừ?

" _ " là dấu trừ à 

\(A=\dfrac{x^{98}+x^{97}+x^{96}+...+x+1}{x^{32}+x^{31}+x^{30}+...+x+1}\\ x=2\\ \Rightarrow A=\dfrac{2^{98}+2^{97}+2^{96}+...+2+1}{2^{32}+2^{31}+2^{30}+...+2+1}\)

Đặt

\(B = 2^{98} + 2^{97} + 2^{96} + ... + 2 + 1 \\ C = 2^{32} + 2^{31} + 2^{30} + ... + 2 + 1\)

\(B=2^{98}+2^{97}+2^{96}+...+2+1\\ =\left(2-1\right)\left(2^{98}+2^{97}+2^{96}+...+2+1\right)\\ =2^{99}-1\\ =\left(2^{33}-1\right)\left(2^{66}+2^{33}+1\right)\\ C=2^{32}+2^{31}+2^{30}+...+2+1\\ =\left(2-1\right)\left(2^{32}+2^{31}+2^{30}+...+2+1\right)\\ =2^{33}-1\\ A=\dfrac{B}{C}=\dfrac{\left(2^{33}-1\right)\left(2^{66}+2^{33}+1\right)}{2^{33}-1}=2^{66}+2^{33}+1\)

a) Ta có: \(\left(x-1\right)^{x+2}-\left(x-1\right)^{x+4}=0\)

\(\Leftrightarrow\left(x-1\right)^x\cdot\left(x-1\right)^2-\left(x-1\right)^x\cdot\left(x-1\right)^4=0\)

\(\Leftrightarrow\left(x-1\right)^{x+2}\cdot\left[1-\left(x-1\right)^2\right]=0\)

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

b) Ta có: \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{15}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\)

\(\Leftrightarrow2x=\dfrac{1}{64}\)

hay \(x=\dfrac{1}{128}\)

\(A\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\\\Leftrightarrow \left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\\ \Leftrightarrow\left(x-1\right)^{x+2}\left(\left(x-1\right)^{x+2}+1\right)=0\\ \Leftrightarrow\left(x-1\right)^{x+2}=0hoac\left(x-1\right)^{x+2}+1=0\)

Giả tiếp đc x=1

Rút gọn.

\(B=\dfrac{x^{39}x^{36}x^{33}...x^31}{x^{40}x^{38}x^{36}...x^21}=\dfrac{x^{\left(39+36+33+...+3\right)}}{x^{\left(40+38+36+...+2\right)}}\)

ta có: \(39+36+33+...+3=\dfrac{\left(39+3\right)\left(\dfrac{39-3}{3}+1\right)}{2}=273\)

\(40+38+36+....+2=\dfrac{\left(40+2\right)\left(\dfrac{40-2}{2}+1\right)}{2}=420\)

=> \(B=\dfrac{x^{273}}{x^{420}}=\dfrac{1}{x^{147}}\)

Tương tự như B => \(A=\dfrac{x^{4560}}{x^{496}}=x^{4064}\)

Ta có:

\(B=\dfrac{x^{\left(39+36+33+....+3\right)}}{x^{\left(40+38+36+....+2\right)}}\)

\(39+36+33+....+3=\dfrac{\left(39+3\right)\left(\dfrac{39-3}{3}+1\right)}{2}=273\)

\(40+38+36+....+2=\dfrac{\left(40+2\right)\left(\dfrac{40-2}{2}+1\right)}{2}=420\)

\(\Rightarrow B=\dfrac{x^{273}}{x^{420}}=\dfrac{1}{x^{147}}\)

tương tự => \(A=\dfrac{x^{4560}}{x^{496}}=x^{4064}\)