Question

BT1: Tìm x\(\in\)N * :

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\cdot\left(2x+3\right)}\)=\(\frac{49}{99}\)

Answer 1

\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left\{\left(2x+1\right).\left(2x+3\right)\right\}}\right)=\frac{49}{99}\)

\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)=\frac{49}{99}\)

\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2x+3}\right)=\frac{49}{99}\)

\(\frac{1}{2}.\cdot\left(\frac{2x+3}{2x+3}-\frac{1}{2x+3}\right)=\frac{49}{99}\)

\(\frac{1}{2}.\frac{2x+2}{2x+3}=\frac{49}{99}\)

\(\frac{2x+2}{2x+3}=\frac{49}{99}:\frac{1}{2}\)

\(\frac{2x+2}{2x+3}=\frac{98}{99}\)

=) \(2x+2=98\)và \(2x+3=99\)

TH1 : \(2x+2=98\)

\(2x=98-2\)

\(2x=96\)

\(x=96:2\)

\(x=48\)( THỎa mãn )

TH2 : 
\(2x+3=99\)

\(2x=99-3\)

\(2x=96\)

\(x=96:2\)

\(x=48\)( THỎa mãn )

Vậy x = 48

Answer 2

Đặt A=

Answer 3

Tk mk nha mk làm đúng đó ~!!

Answer 4

\(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{\left(2x+1\right)\times\left(2x+3\right)}=\frac{49}{99}\)

<=> \(\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{3\times7}+...+\frac{2}{\left(2x+1\right)\times\left(2x+3\right)}\right)=\frac{49}{99}\)

<=> \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{49}{99}\div\frac{1}{2}\)

<=> \(1-\frac{1}{2x+3}=\frac{98}{99}\)

<=> \(\frac{1}{2x+3}=\frac{1}{99}\)

<=> \(2x+3=99\)

<=> \(2x=96\)

<=> \(x=48\)