Xét tử số : Đặt B = $1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}$

Số số hạng của tử số là : ( 99 - 1 ) : 2 + 1 = 50 ( số )

=> Tử số có 50 phân số

Ta có : $B=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)$

$=\left(\frac{99}{99}+\frac{1}{99}\right)+\left(\frac{97}{3.97}+\frac{3}{3.97}\right)+\left(\frac{95}{5.95}+\frac{5}{5.95}\right)+...+\left(\frac{51}{49.51}+\frac{49}{49.51}\right)$

$=\frac{100}{1.99}+\frac{100}{3.97}+\frac{100}{5.95}+...+\frac{100}{49.51}$

Xét mẫu số : Đặt C = $\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{97.3}+\frac{1}{99.1}$

$=\left(\frac{1}{1.99}+\frac{1}{99.1}\right)+\left(\frac{1}{3.97}+\frac{1}{97.3}\right)+...+\left(\frac{1}{49.51}+\frac{1}{51.49}\right)$

$=2.\frac{1}{1.99}+2.\frac{1}{3.97}+...+2.\frac{1}{49.51}$

$=2\left(\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}\right)$

Thay B và C vào A ta có :

$A=\frac{100\left(\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}\right)}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}\right)}$

$\Rightarrow A=\frac{100}{2}=50$

Vậy A = 50

đề lỗi sorry

thoog cảm !!!

\(A=\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\)

\(=\frac{1}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{44.49}\right)\)

\(=\frac{1}{5}\left(\frac{9-4}{4.9}+\frac{14-9}{9.14}+\frac{19-14}{14.19}+...+\frac{49-44}{44.49}\right)\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\)

\(=\frac{9}{196}\)

\(B=1-3-5-7-...-49\)

\(=1-\left(3+5+7+...+49\right)\)

\(=1-\frac{\left(49+3\right).\left[\left(49-3\right)\div2+1\right]}{2}=-623\)

\(S=\frac{A.B}{89}=\frac{9}{196}.\left(-7\right)=-\frac{9}{28}\)

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

Đặt \(d=\left(x-y,2x+2y+1\right)\).

Suy ra \(y^2=\left(x-y\right)\left(2x+2y+1\right)⋮d^2\)

\(\Rightarrow y⋮d\).

\(\left(2x+2y+1\right)-2\left(x-y\right)=4y+1⋮d\Rightarrow1⋮d\)

Suy ra \(d=1\).

Do đó \(x-y,2x+2y+1\)đồng thời là các số chính phương. 

a.\(5^{-3}.5^{2n}=5^{3n}\Leftrightarrow5^{-3+2n}=5^{3n}\Leftrightarrow-3+2n=3n\Leftrightarrow n=-3\)

b.\(a^{\left(2n+6\right)\left(3n-9\right)}=1=a^0\Leftrightarrow\left(2n+6\right)\left(3n-9\right)=0\Leftrightarrow\orbr{\begin{cases}2n+6=0\\3n-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}n=-3\\n=3\end{cases}}}\)

c.\(\left(\frac{2}{3}\right)^{3n}=\left(\frac{2}{3}\right)^{-12}\Leftrightarrow3n=-12\Leftrightarrow n=-4\)

d.\(\frac{1}{3}3^n=7.3^2.3^4-2.3^n\Leftrightarrow\frac{7}{3}3^n=7.3^6\Leftrightarrow3^n=3^7\Leftrightarrow n=7\)

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Đặt \(d=\left(a+b+2,2a+b+1\right)\).

\(\Rightarrow a^2=\left(a+b+2\right)\left(2a+b+1\right)⋮d^2\)

\(\Rightarrow a⋮d\).

\(\left(2a+b+1\right)-\left(a+b+2\right)=a-1⋮d\Rightarrow1⋮d\).

Do đó \(d=1\).

Suy ra \(a+b+2,2a+b+1\)đồng thời là các số chính phương.