A =             1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\)+ \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\)+ \(\dfrac{1}{64}\)+ \(\dfrac{1}{128}\)

A\(\times\)2 = 2 + 1 + \(\dfrac{1}{2}\) +  \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)

A \(\times\) 2 - A = 2 - \(\dfrac{1}{128}\)

A \(\times\)( 2-1) = \(\dfrac{255}{128}\)

A = \(\dfrac{255}{128}\)

Gọi \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\) là T

\(T=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)

\(2T=2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{64}\)

\(2T-T=\left(2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{64}\right)-\left(1+\dfrac{1}{2}+....+\dfrac{1}{64}+\dfrac{1}{128}\right)\)

\(T=2+\left(1-1\right)+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+....+\left(\dfrac{1}{64}-\dfrac{1}{64}\right)-\dfrac{1}{128}\)

\(T=2+0+0+...-\dfrac{1}{128}\)

\(T=\dfrac{256}{128}-\dfrac{1}{128}\)

\(T=\dfrac{255}{128}\)

\(VT=1.\left(2+1\right)\left(2^2+1\right)...\left(2^{16}+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{16}+1\right)\)

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

\(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2009^2}\)

\(=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}\right)\)

\(>1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2008.2009}\right)\)

\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}\right)\)

\(=1-\left(1-\frac{1}{2009}\right)\)

\(=\frac{1}{2009}\)

1+2+4+8+16+32+64+128+256+512+1024+2048

=1+(2+8)+(4+16)+(32+128)+(64+256)+(512+2048)+1024

=1+10+20+160+320+2560+1024

=4095

 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048 = 4095 

k nha      công chúa nụ cười    =_=   ^_^

= 4095 bạn nhé ! k mk nha !

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

- Vì : 

 \(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

...................

\(\frac{1}{n^2}< \frac{1}{n\left(n-1\right)}\)

Cộng vế với vế , ta suy ra 

A < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.....+\frac{1}{n-1}-\frac{1}{n}\)

= \(1-\frac{1}{n}< 1\)

=> A<1 ( đpcm )

Ta có:\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)>\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)=\(\frac{1}{1}-\frac{1}{n}\)<1 => \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\)

\(x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\) (ĐKXĐ : \(x\ge1;y\ge2;z\ge3\))

\(\Leftrightarrow\left(x-1-2\sqrt{x-1}+1\right)+\left(y-2-4\sqrt{y-2}+4\right)+\left(z-3-6\sqrt{z-3}+9\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

Vì \(\left(\sqrt{x-1}-1\right)^2\ge0;\left(\sqrt{y-2}-2\right)^2\ge0;\left(\sqrt{z-3}-3\right)^2\ge0\)

nên phương trình tương đương với : \(\hept{\begin{cases}\left(\sqrt{x-1}-1\right)^2=0\\\left(\sqrt{y-2}-2\right)^2=0\\\left(\sqrt{z-3}-3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=12\end{cases}}}\)(TMĐK)

Vậy nghiệm của phương trình :  \(\left(x;y;z\right)=\left(2;6;12\right)\)

dùng hàng đẳng thức A^2-B^2=(A-B)(A+B) nhé còn phần b chuyển vế sang rồi dùng HĐT là được

a) \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

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

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

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

b) \(100^2+103^2+105^2+94^2=101^2+98^2+96^2+107^2\)

\(\Leftrightarrow\left(100^2-98^2\right)+\left(103^2-101^2\right)+\left(105-107^2\right)+\left(94^2-96^2\right)=0\)

\(\Leftrightarrow2\left(100+98+103+101-105-107-94-96\right)=0\)

\(\Leftrightarrow2\times0=0\)(ĐPCM)

a) \(\frac{1}{a\left(a+1\right)}=\frac{\left(a+1\right)-a}{a\left(a+1\right)}=\frac{a+1}{a\left(a+1\right)}-\frac{a}{a\left(a+1\right)}=\frac{1}{a}-\frac{1}{a+1}\)

b) \(\frac{2}{a\left(a+1\right)\left(a+2\right)}=\frac{\left(a+2\right)-a}{a\left(a+1\right)\left(a+2\right)}=\frac{a+2}{a\left(a+1\right)\left(a+2\right)}-\frac{a}{a\left(a+1\right)\left(a+2\right)}\)

\(=\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\)

a, Ta có : \(\frac{1}{a}-\frac{1}{a+1}=\frac{a+1}{a\left(a+1\right)}-\frac{a}{a\left(a+1\right)}=\frac{a+1-a}{a\left(a+1\right)}\)

\(VT=\frac{1}{a\left(a+1\right)}\left(đpcm\right)\)

b, Ta có : \(\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}=\frac{a+2}{a\left(a+1\right)\left(a+2\right)}-\frac{a}{a\left(a+1\right)\left(a+2\right)}\)

\(VT=\frac{2}{a\left(a+1\right)\left(a+2\right)}\left(đpcm\right)\)