Ta có :2²A=1+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{n^2}\)

            2²A-A=1-\(\dfrac{1}{\left(2n\right)^2}\)

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

          3A<\(\dfrac{1}{2}\) suy ra A<\(\dfrac{1}{2}\)

A   = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\)+.......+\(\dfrac{1}{\left(2.n\right)^2}\)

A  =  \(\dfrac{1}{2^2}\) + \(\dfrac{1}{\left(2.2\right)^2}\)+ \(\dfrac{1}{\left(2.3\right)^2}\) +....+\(\dfrac{1}{\left(2.n\right)^2}\)

A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^2.2^2}\) + \(\dfrac{1}{2^2.3^2}\)+......+ \(\dfrac{1}{2^2.n^2}\)

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

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

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

     4A = 1 + \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + ...+\(\dfrac{1}{n.n}\)

       1   = 1

     \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\)

      \(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\)

     ...................

 \(\dfrac{1}{n.n}\) < \(\dfrac{1}{\left(n-1\right).n}\)

Cộng vế với vế ta có: 

4A = 1+\(\dfrac{1}{2.2}\)+\(\dfrac{1}{3.3}\)+....+\(\dfrac{1}{n.n}\) <1+ \(\dfrac{1}{1.2}\)+ \(\dfrac{1}{2.3}\)+ ......+ \(\dfrac{1}{\left(n-1\right).n}\)

4A < 1+ \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+....+\(\dfrac{1}{n-1}\) - \(\dfrac{1}{n}\) = 2 - \(\dfrac{1}{n}\)

A < ( 2 - \(\dfrac{1}{n}\)): 4 

A < 2 : 4 - \(\dfrac{1}{n}\) : 4

A < \(\dfrac{1}{2}\) - \(\dfrac{1}{4n}\) < \(\dfrac{1}{2}\)

Vậy A < \(\dfrac{1}{2}\) 

hai số trước cộng vào bằng số sau rồi tiếp tục

cs ai chỉ dc ko đang cần gấp í ạ

Bạn nên viết đề bằng công thức tán để được hỗ trợ tốt hơn. 

2021 + \(\dfrac{3}{(3+\dfrac{3}{3-x})}\)     =    2023

            \(\dfrac{3}{(3+\dfrac{3}{3-x})}\)    = 2023 - 2021 

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

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

                     \(\dfrac{3}{3-x}\) = \(\dfrac{3}{2}\) - 3

                     \(\dfrac{3}{3-x}\) = - \(\dfrac{3}{2}\)

                       3 - \(x\) = 3 : ( -\(\dfrac{3}{2}\))

                     3 - \(x\) = -2

                          \(x\) = 3  - ( -2) 

                           \(x\) = 5

Vậy \(x\) = 5

ai biết ko cứu tôi với

( \(\dfrac{2}{123}\) + \(\dfrac{2023}{2022}\) )( \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) - \(\dfrac{2}{15}\))

=( \(\dfrac{2}{123}\) + \(\dfrac{2023}{2022}\) )( \(\dfrac{5}{15}\) - \(\dfrac{3}{15}\) - \(\dfrac{2}{15}\))

=( \(\dfrac{2}{123}\) + \(\dfrac{2023}{2022}\))( \(\dfrac{5-3-2}{15}\))

=( \(\dfrac{1}{123}\) + \(\dfrac{2023}{2022}\)). \(\dfrac{0}{15}\)

= ( \(\dfrac{1}{123}+\dfrac{2023}{2022}\)).0

= 0

\(x+xy+y=1\)

\(2x+2xy+2y=2\)

\(2x\left(1+y\right)+2y=2\)

\(2x\left(y+1\right)+2y+2=4\)

\(2x\left(y+1\right)+2\left(y+1\right)=4\)

\(\left(2x+2\right)\left(y+1\right)=4\)

\(2\left(x+1\right)\left(y+1\right)=4\)

\(\left(x+1\right)\left(y+1\right)=2\)

\(TH1:\left\{{}\begin{matrix}x+1=1\\y+1=2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)

\(TH2:\left\{{}\begin{matrix}x+1=2\\y+1=1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)

\(TH3:\left\{{}\begin{matrix}x+1=-1\\y+1=-2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)

\(TH4:\left\{{}\begin{matrix}x+1=-2\\y+1=-1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)

\(Vậy...\)

x+xy+y=1⇔x(y+1)+y+1=2⇔(x+1)(y+1)=2

⇒(x+1;y+1)=(-1;-2),(-2;-1),(1;2),(2;1)

sau tự tính nhé :3

Lời giải:

$2^x+2^{x+1}+2^{x+2}+...+2^{x+2019}=2^{x+2023}-8$

$2^x(1+2+2^2+...+2^{2019})=2^{x+2023}-8$

Xét:

$A=1+2+2^2+...+2^{2019}$

$2A=2+2^2+2^3+...+2^{2020}$

$\Rightarrow A=2A-A=2^{2020}-1$

Khi đó:

$2^x.A=2^{x+2023}-8$

$2^x(2^{2020}-1)=2^{x+2023}-2^3$

$2^x(2^{2023}-2^{2020}+1)-2^3=0$

$2^x(2^{2020}.7+1)=2^3$

$x$ ra số sẽ khá xấu. Bạn coi lại.

\(2ab-a+b=3\)

\(4ab-2a+2b=6\)

\(2a\left(2b-1\right)+2b=6\)

\(2a\left(2b-1\right)+\left(2b-1\right)=7\)

\(\left(2a+1\right)\left(2b-1\right)=7\)

\(TH1:\left\{{}\begin{matrix}2a+1=1\\2b-1=7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=0\\b=4\end{matrix}\right.\)

\(TH2:\left\{{}\begin{matrix}2a+1=7\\2b-1=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=\dfrac{1}{2}\end{matrix}\right.\)

\(TH3:\left\{{}\begin{matrix}2a+1=-1\\2b-1=-7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=-3\end{matrix}\right.\)

\(TH4:\left\{{}\begin{matrix}2a+1=-7\\2b-1=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-4\\b=0\end{matrix}\right.\)

Rồi em tự tính ab ra nha