Đặt \(A=1+2+2^2+...+2^{2009}\)

=>\(2A=2+2^2+2^3+...+2^{2010}\)

=>\(2A-A=2+2^2+...+2^{2010}-1-2-...-2^{2009}\)

=>\(A=2^{2010}-1\)

\(\dfrac{A}{1-2^{2010}}=\dfrac{2^{2010}-1}{1-2^{2010}}=-1\)

Câu hỏi đâu bạn ? 

$A=1.21+3.41+...+49.501$ hiển nhiên $>1$ rồi mà bạn. Bạn xem lại đề.

A = ^{\(\dfrac{1}{1.2}\)} + ^{\(\dfrac{1}{3.4}\)} + ^{\(\dfrac{1}{5.6}\)} + ... + ^{\(\dfrac{1}{49.50}\)} 

A <  ^{\(\dfrac{1}{1.2}\)} + ^{\(\dfrac{1}{3.4}\)} + ^{\(\dfrac{1}{5.6}\)} + ... + ^{\(\dfrac{1}{49.50}\)} +^{\(\dfrac{1}{2.3}+\dfrac{1}{4.5}+\dfrac{1}{6.7}\)+...+\(\dfrac{1}{48.49}\)}

A < ^{\(\dfrac{1}{1.2}\)} + ^{\(\dfrac{1}{2.3}\)} + ^{\(\dfrac{1}{3.4}\)} + ^{\(\dfrac{1}{4.5}\)}+^{\(\dfrac{1}{5.6}\)} +^{\(\dfrac{1}{6.7}\)}+.. +^{\(\dfrac{1}{47.48}\)}+ ^{\(\dfrac{1}{48.49}\)}+ ^{\(\dfrac{1}{49.50}\)} 

A < ^{\(\dfrac{1}{1}\)}-^{\(\dfrac{1}{2}\)} + ^{\(\dfrac{1}{2}\)} - ^{\(\dfrac{1}{3}\)} + ^{\(\dfrac{1}{3}\)} - ^{\(\dfrac{1}{4}\)} + ... + ^{\(\dfrac{1}{49}\)} - ^{\(\dfrac{1}{50}\)}

A < ^{\(\dfrac{1}{1}\)} - ^{\(\dfrac{1}{50}\)} < 1 (đpcm)

Lời giải:

$\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{60^2}$

$=\frac{1}{9}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{60^2}

$< \frac{1}{9}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{59.60}$

$= \frac{1}{9}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{60-59}{59.60}$

$=\frac{1}{9}+\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}$

$=\frac{1}{9}+\frac{1}{3}-\frac{1}{60}=\frac{4}{9}-\frac{1}{60}< \frac{4}{9}$

Ta có đpcm.

Lời giải:

$\frac{-7}{x}+\frac{8}{15}=\frac{-1}{20}$

$\frac{-7}{x}=\frac{-1}{20}-\frac{8}{15}=\frac{-7}{12}$

$x=-7: \frac{-7}{12}=12$

-7/x + 8/15 = -1/20

-7/x = -1/20 - 8/15

-7/x  = -7/12

=> x = 12

Vậy x = 12 . 

\(45\%\cdot x-0,385=-1,685\)

=>\(x\cdot45\%=-1,685+0,385=-1,3\)
=>\(x=-1,3:\dfrac{9}{20}=-1,3\cdot\dfrac{20}{9}=-\dfrac{26}{9}\)

45% x - 0,385 = -1,685

45% x             = -1,685 + 0,385

45% x             = -1,3

0,45 x             = -1,3

        x             = -1,3 : 0,45

        x             = -2,8888

3xy+x-15y-10=0

=>3xy-15y+x-5-5=0

=>3y(x-5)+(x-5)=5

=>(x-5)(3y+1)=5

=>\(\left(x-5;3y+1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(6;\dfrac{4}{3}\right);\left(10;0\right);\left(4;-2\right);\left(0;-\dfrac{2}{3}\right)\right\}\)

3xy + x - 15y - 10 = 0

Suy ra ta có 2 TH

\(\Leftrightarrow3xy+x=0\) (TH1)  

\(\Leftrightarrow15y-10=0\) (TH2)

Đề chưa đúng em nhé 

Tử số phải là: 200 - (3 + 2/3 +...+ 2/100) 

Lời giải:

Nếu $p,q$ cùng là snt lẻ thì $p^2-q=1$ chẵn (vô lý)

Do đó trong 2 số $p,q$ tồn tại ít nhất 1 số chẵn.

Nếu $p$ chẵn $\Rightarrow p=2$ (do $p$ nguyên tố)

$\Rightarrow q=p^2-1=2^2-1=3$ (thỏa mãn)

Nếu $q$ chẵn $\Rightarrow q=2$ (do $q$ nguyên tố)

$\Rightarrow p^2=q+1=2+1=3$ (loại)

Vậy $(p,q)=(2,3)$

Vố số nghiệm nha

\(\left(\dfrac{1}{2}x-\dfrac{3}{4}\right)\left(x+\dfrac{1}{2}\right)=0\)

=>\(\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{4}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{3}{4}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

=-1,5=-3/2