\(M=1\times\dfrac{1}{2}+\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+...+\dfrac{1}{99}\times\dfrac{1}{100}\)

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

\(M=1-\dfrac{1}{100}\)

\(M=\dfrac{99}{100}\)

??????????

Ko bit làm 

3) \(...\Rightarrow2^x\left(2^3+1\right)=36\)

\(\Rightarrow2^x.9=36\)

\(\Rightarrow2^x=4\)

\(\Rightarrow2^x=2^2\Rightarrow x=2\)

4) \(...\Rightarrow4^{x+1}-4^x=12\)

\(\Rightarrow4^x\left(4-1\right)=12\)

\(\Rightarrow4^x.3=12\)

\(\Rightarrow4^x=4=4^1\Rightarrow x=1\)

5) \(...\Rightarrow5^{x+1}\left(5^2-1\right)=3000\)

\(\Rightarrow5^{x+1}.24=3000\)

\(\Rightarrow5^{x+1}=125\)

\(\Rightarrow5^{x+1}=5^3\)

\(\Rightarrow x+1=3\)

\(\Rightarrow x=2\)

6) Bạn xem lại đề

a. \(2^x.2^3+2^x=36\)

\(2^x\left(2^3+1\right)=36\)

\(2^x.9=36\)

\(2^x=4\Rightarrow x=2\)

b. \(4^x.4^1-\left(2^2\right)^x=12\)

\(4^x.4-4^x=12\)

\(4^x\left(4-1\right)=12\)

\(4^x.3=12\)

\(4^x=4\)

x = 1

c. \(5^x.5^3-5^x.5^1=3000\)

\(5^x\left(5^3-5^1\right)=3000\)

\(5^x.120=3000\)

\(5^x=25\)

x = 2

d. \(4^{x+1}=2^{2x}\)

\(4^x.4=\left(2^2\right)^x\)

\(4^x.4=4^x\)

Có vẻ như câu 4 này để bài thiếu 

\(\sqrt{x-4\sqrt{x-4}}=1\) (ĐKXĐ: \(x\ge4\))

\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}\right)^2-2\cdot\sqrt{x-4}\cdot2+2^2}=1\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}-2\right)^2}=1\)

\(\Leftrightarrow\left|\sqrt{x-4}-2\right|=1\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-4=9\\x-4=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=13\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{13;5\right\}\).

#\(Toru\)

\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2003}\right).\left(1-\dfrac{1}{2004}\right)\)

\(A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}....\dfrac{2002}{2003}.\dfrac{2003}{2004}\)

\(A=\dfrac{1}{2004}\)

(1-1/2)x(1-1/3)x(1-1/4)x(1-1/5)x.......x (1-1/2003)x(1-1/2004)

=1/2 x 2/3 x 3/4 x 4/5 x.....x2002/2003 x 2003/2004

=\(\frac{1\times2\times3\times4\times...\times2002\times2003}{2\times3\times4\times5....\times2003\times2004}\)

=\(\frac{1}{2004}\)

\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)x...x\left(1-\frac{1}{2003}\right)x\left(1-\frac{1}{2004}\right)\)

\(=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x...x\frac{2002}{2003}x\frac{2003}{2004}\)

 \(=\frac{1x2x3x4x....x2002x2003}{2x3x4x5x...x2003x2004}\)

   \(=\frac{1}{2004}\)

\(=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{2002}{2003}x\frac{2003}{2004}=\frac{1x2x3x...x2002x2003}{2x3x4x...x2003x2004}=\frac{1}{2004}\)

\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2002}{2003}.\frac{2003}{2004}\)

\(=\frac{1.2.3...2002.2003}{2.3.4...2003.2004}=\frac{1}{2004}\)

ta sẽ ra được kết quả qua cách giảm ước của cả tử và mẫu . vậy cuối cùng nhìn lại trên tử còn 1 mẫu thì còn 2004 vậy phân số ra được là 1/2004

Tính nhanh : 

( 1 - 1/2 ) × ( 1 - 1/3 ) × ( 1 - 1/4 ) × ... × ( 1 - 1/2003 ) × ( 1 - 1/2004 ) 

= 1/2 × 2/3 × 3/4 × ... × 2002/2003 × 2003/2004

= 1 × 2 × 3 × ... × 2002 × 2003 / 2 × 3 × 4 × ... × 2003 × 2004 

= 1/2004

(1 - 1/ 2) x (1 - 1/3) x ( 1 - 1/4 ) x ... x (1 - 1/2003) × ( 1 - 1/ 2004 )

                     Giải

( 1 - 1/2 ) × ( 1 - 1/3 ) × ( 1 - 1/4 ) × ... × ( 1 - 1/2003 ) × ( 1 - 1/2004 ) 

= 1/2 × 2/3 × 3/4 × ... × 2002/2003 × 2003/2004

= 1 × 2 × 3 × ... × 2002 × 2003 / 2 × 3 × 4 × ... × 2003 × 2004 

= 1/2004

7) \(A=1^2-2^2+3^2-4^2+...-2004^2+2005^2\)

\(A=\left(-1\right)\left(1^{ }+2\right)+\left(-1\right)\left(3+4\right)+...+\left(-1\right)\left(2003+2004\right)+2005^2\)

\(A=-\left(1+2+3+...+2004\right)+2005^2\)

\(A=-\dfrac{2004.\left(2004+1\right)}{2}+2005^2\)

\(A=-1002.2005+2005^2\)

\(A=2005\left(2005-1002\right)=2005.1003=2011015\)

8) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(B=\dfrac{\left(2^2-1\right)}{2-1}\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)

\(B=\left(2^{64}-1\right)-2^{64}\)

\(B=-1\)

7. 2011015

8. -1