KHO THE

\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)

\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)

\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)

Ta có:

\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)

\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)

\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow6A=3+1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow4A=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}=3-\frac{203}{3^{100}}\)

\(\Rightarrow A=\frac{3-\frac{203}{3^{100}}}{4}=\frac{3}{4}-\frac{203}{3^{100}.4}< \frac{3}{4}\Rightarrowđpcm\)

Vậy \(A< \frac{3}{4}\)

\(\dfrac{3}{2}\)(\(x\) - \(\dfrac{5}{3}\)) - \(\dfrac{4}{5}\) = \(x\) + 1

\(\dfrac{3}{2}\) \(x\) - \(\dfrac{15}{6}\) - \(\dfrac{4}{5}\) = \(x\) + 1

\(\dfrac{3}{2}\)\(x\) - \(x\)          = 1 + \(\dfrac{15}{6}\) + \(\dfrac{4}{5}\)

\(\dfrac{1}{2}\)\(x\)                 =\(\dfrac{43}{10}\)

    \(x\)                 = \(\dfrac{43}{10}\) \(\times\) 2

     \(x\)                = \(\dfrac{43}{5}\)

\(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{4}{5}=x+1\\ \Rightarrow\dfrac{3.\left(x-\dfrac{5}{3}\right)}{2}-\dfrac{4}{5}=x+1\\ \Rightarrow\dfrac{3x-5}{2}-\dfrac{4}{5}=x+1\Rightarrow\dfrac{5\left(3x-5\right)}{10}-\dfrac{8}{10}=x+1\\ \Rightarrow\dfrac{15x-33}{10}=x+1\\ \Rightarrow\dfrac{15x-33}{10}-x=x+1\\ \Rightarrow\dfrac{15x-33}{10}=x+1-x\\ \Rightarrow5x-33=10\\ \Rightarrow5x=10+33\\\Rightarrow5x=43\\ \Rightarrow x=\dfrac{43}{5} \)

\(\dfrac{3}{2}\times\left(x-\dfrac{5}{3}\right)-\dfrac{4}{5}=x+1\)

\(\Leftrightarrow\dfrac{3}{2}\times x-\dfrac{5}{2}-\dfrac{4}{5}=x+1\)

\(\Leftrightarrow\dfrac{3}{2}\times x-x=1+\dfrac{5}{2}+\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}\times x=\dfrac{43}{10}\)

\(\Leftrightarrow x=\dfrac{43}{10}\div\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{43}{5}\)

(-75)+(-7)^2+|-75|+49-(-4)^3

= (-75)+49+75+49+64

=2.49+64

=98+64

=162

(-75)+(-7)^2+|-75|+49-(-4)^3

=(-75)+49+75+49+4^3          

= -75+49+75+49+64

= (-75+75)+49+49+64

= 0+49+49+64

= 49+49+64

= 2x49+64

= 98+64

= 162

`a)25/(x+1)-1 1/6=-1/3-0,5`

`=>25/(x+1)=-1/3-1/2+1+1/6`

`=>25/(x+1)=1/3`

`=>75=x+1`

`=>x=74`

Vậy `x=74`

`b)(2x+25 3/5)^2-9/25=0`

`=>(2x+128/5)=9/25`

`**2x+128/5=3/5`

`=>2x=-125/5=-25`

`=>x=-25/2`

`**2x+128/5=-3/5`

`=>2x=-131/5`

`=>x=-131/10`

Giải:

a) \(\dfrac{25}{x+1}-1\dfrac{1}{6}=\dfrac{-1}{3}-0,5\) 

              \(\dfrac{25}{x+1}=\dfrac{-5}{6}+\dfrac{7}{6}\) 

              \(\dfrac{25}{x+1}=\dfrac{1}{3}\) 

\(\Rightarrow1.\left(x+1\right)=25.3\)  

\(\Rightarrow x+1=75\) 

\(\Rightarrow x=75-1\) 

\(\Rightarrow x=74\) 

b) \(\left(2x+25\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\) 

              \(\left(2x+\dfrac{128}{5}\right)^2=0+\dfrac{9}{25}\) 

             \(\left(2x+\dfrac{128}{5}\right)^2=\dfrac{9}{25}\) 

\(\Rightarrow\left[{}\begin{matrix}\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{3}{5}\right)^2\\\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{128}{5}=\dfrac{3}{5}\\2x+\dfrac{128}{5}=\dfrac{-3}{5}\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-25}{2}\\x=\dfrac{-131}{10}\end{matrix}\right.\) 

Chúc bạn học tốt!

a, 25/12

b,  25/72

a) = 3/4 + 4/3 = 25/12

b) = 3/8 - 16/9 = -101/72 * hình như hơi sai sai:"> *

(7/111 - 4/33 + 9/37).(1/2 - 1/3 -1/6)

= (7/111 - 4/33 + 9/37).0

= 0

(7/111 - 4/33 + 9/37).(1/2 - 1/3 - 1/6)

= (7/111 - 4/33 + 9/37).0

= 0