Bài 94 trang 27 SBT toán 6 tập 2

Đề bài

Tính giá trị các biểu thức :

\(\displaystyle A = {{{1^2}} \over {1.2}}.{{{2^2}} \over {2.3}}.{{{3^2}} \over {3.4}}.{{{4^2}} \over {4.5}}\) 

\(\displaystyle B = {{{2^2}} \over {1.3}}.{{{3^2}} \over {2.4}}.{{{4^2}} \over {3.5}}.{{{5^2}} \over {4.6}}\) 

Lời giải

\(\displaystyle A = {{{1^2}} \over {1.2}}.{{{2^2}} \over {2.3}}.{{{3^2}} \over {3.4}}.{{{4^2}} \over {4.5}} \)

\(\displaystyle= {1 \over 2}.{2 \over 3}.{3 \over 4}.{4 \over 5} = {{1.2.3.4} \over {2.3.4.5}} = {1 \over 5}\) 

\(\displaystyle B = {{{2^2}} \over {1.3}}.{{{3^2}} \over {2.4}}.{{{4^2}} \over {3.5}}.{{{5^2}} \over {4.6}} \)\(\displaystyle B = \dfrac{2.2.3.3.4.4.5.5}{1.3.2.4.3.5.4.6}\)

\(\displaystyle= {{(2.3.4.5).(2.3.4.5)} \over {(1.2.3.4).(3.4.5.6)}} = {{5.2} \over 6} = {5 \over 3}\)


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