1.tính giá trị của biểu thức.
A= (1+ $\frac{1}{1X3}$ )X(1+ $\frac{1}{2X4}$)X(1+$\frac{1}{3X5}$)X(1+$\frac{1}{99X101}$ )
B= $\frac{1}{1X3}$ X$\frac{9}{2X4}$ X$\frac{16}{3X5}$ X…X$\frac{10000}{99X101}$
1.tính giá trị của biểu thức.
A= (1+ $\frac{1}{1X3}$ )X(1+ $\frac{1}{2X4}$)X(1+$\frac{1}{3X5}$)X(1+$\frac{1}{99X101}$ )
B= $\frac{1}{1X3}$ X$\frac{9}{2X4}$ X$\frac{16}{3X5}$ X…X$\frac{10000}{99X101}$
Đáp án:
$A=\dfrac{200}{101}\\
B=\dfrac{50}{101}$
Giải thích các bước giải:
$A=\left ( 1+\dfrac{1}{1.3} \right ).\left ( 1+\dfrac{1}{2.4} \right ).\left ( 1+\dfrac{1}{3.5} \right )….\left ( 1+\dfrac{1}{99.101} \right )\\
=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}….\dfrac{10000}{99.101}\\
=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}….\dfrac{100.100}{99.101}\\
=\dfrac{2.3.4….100}{1.2.3…99}.\dfrac{2.3.4…100}{3.4.5…101}\\
=\dfrac{100}{1}.\dfrac{2}{101}\\
=\dfrac{200}{101}\\
B=\dfrac{1}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}…\dfrac{10000}{99.101}\\
=\dfrac{1}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}…\dfrac{100.100}{99.101}\\
=\dfrac{1.3.4…100}{1.2.3…99}.\dfrac{1.3.4…100}{3.4.5…101}\\
=\dfrac{100}{2}.\dfrac{1}{101}\\
=\dfrac{50}{101}$