\(A=2^{100}-2^{99}+2^{98}-2^{97}+....+2^2-2\)

\(2A=2^{101}-2^{100}+2^{99}-2^{98}+....+2^3-2^2\)

\(2A+A=2^{101}-2\)

\(A=\frac{2^{101}-2}{3}\)

b) tương tự

\(B=\frac{3^{101}+1}{4}\)

b,     B        =                       \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\)  + \(\dfrac{1}{2^3}\) -   \(\dfrac{1}{2^4}\)+.....+ \(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)

2 \(\times\)  B       =                 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) -  \(\dfrac{1}{2^3}\) + \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)

2 \(\times\) B + B  =                1  -  \(\dfrac{1}{2^{100}}\)

3B             =              ( 1 - \(\dfrac{1}{2^{100}}\)) 

             B =               ( 1 - \(\dfrac{1}{2^{100}}\)) : 3

       A              =          1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+......+ \(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\) 

A\(\times\)  3             =   3 +  1 + \(\dfrac{1}{3}\) +  \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^2}\)+....+  \(\dfrac{1}{3^{n-1}}\) 

A \(\times\) 3 - A        = 3 - \(\dfrac{1}{3^n}\)

       2A           = 3  - \(\dfrac{1}{3^n}\)

         A           = ( 3 - \(\dfrac{1}{3^n}\)) : 2

C = \(\dfrac{3}{2^2}\) \(\times\) \(\dfrac{8}{3^2}\) \(\times\) \(\dfrac{15}{4^2}\) \(\times\) ...........\(\times\) \(\dfrac{899}{30^2}\)

C = \(\dfrac{1\times3}{2^2}\) \(\times\) \(\dfrac{2\times4}{3^2}\) \(\times\) \(\dfrac{3\times5}{4^2}\) \(\times\)........\(\times\) \(\dfrac{29\times31}{30^2}\)

C = \(\dfrac{1\times2\times\left(3\times4\times5\times....\times29\right)^2\times30\times31}{2^2\times\left(3\times4\times5\times.......\times29\right)^2\times30^2}\)

C =  \(\dfrac{2\times\left(3\times4\times5\times.....\times29\right)^2\times30}{2\times\left(3\times4\times5\times.....\times29\right)^2\times30}\) \(\times\) \(\dfrac{1\times31}{2\times30}\)

C = 1 \(\times\) \(\dfrac{31}{60}\)

C = \(\dfrac{31}{60}\)

       A =          1 +   \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) +.......+\(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\)  

3\(\times\) A  =  3  +  \(\dfrac{1}{3}\) +  \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+........+ \(\dfrac{1}{3^{n-1}}\)

3A - A =  3 + \(\dfrac{1}{3}\) - 1 - \(\dfrac{1}{3^n}\) 

    2A  = \(\dfrac{7}{3}\) - \(\dfrac{1}{3^n}\)

      A  = ( \(\dfrac{7}{3}\) - \(\dfrac{1}{3^n}\)): 2

     A =   \(\dfrac{7.3^{n-1}-1}{3^n}\) : 2

     A = \(\dfrac{7.3^{n-1}-1}{2.3^n}\)

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

2B    =  2 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) -  \(\dfrac{1}{2^3}\)+ \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)

2B + B = 2 - \(\dfrac{1}{2^{100}}\)

  3B     =  2 - \(\dfrac{1}{2^{100}}\)

    B     =   ( 2 - \(\dfrac{1}{2^{100}}\)): 3

    B     =     \(\dfrac{2.2^{100}-1}{2^{100}}\) : 3

    B     = \(\dfrac{2^{101}-1}{3.2^{100}}\)

Ta xét biểu thức sau : 

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left[\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}\right)^2\right]}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)(với n > 0)

Áp dụng : \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)

\(=\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+...+\left(\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\right)\)

\(=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}=1-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)

why the heck difficult

Nhanh nhé các bn, mai mik phải nộp r...huhu

a) | x² + 2 | + | x² + 1 | = x² + 2 + x² + 1 = 2x² + 3

b) | 2x - 3 | + | 3x - 2 | = 2x - 3 + 3x - 2 = 5x - 5 = 5( x - 1 ) với x > 2

c) | x - 4 | + | 5 - x | = -( x - 4 ) + 5 - x = -x + 4 + 5 - x = -2x + 9 ( với 4 > x )

d) | 1 - x² | - | 1 + x² | = -( 1 - x² ) - ( 1 + x² ) = -1 + x² - 1 - x² = -2 ( với x > 1 )

a. Vì \(x^2\ge0\forall x\)

\(\Rightarrow\left|x^2+2\right|+\left|x^2+1\right|=x^2+2+x^2+1=2x^2+3\)

b. Vì \(x>2\)\(\Rightarrow2x-3>1;3x-2>4\)

\(\Rightarrow\left|2x-3\right|+\left|3x-2\right|=2x-3+3x-2=5x-5\)

c. Vì \(4>x\)\(\Rightarrow x-4< 0;5-x>1\)

\(\Rightarrow\left|x-4\right|+\left|5-x\right|=-x+4+5-x=-2x+9\)

d. Vì \(x>1\)\(\Rightarrow1-x^2< 0;1+x^2>2\)

\(\Rightarrow\left|1-x^2\right|+\left|1+x^2\right|=-x+x^2+1+x^2=2x^2-x+1\)