A=(24.47-23)/(24+47-23) . [3(1+1/7-1/11-1/13+1/1001)]/[9(1+1/7-1/11-1/13+1/1001)]

=1105/48 . 3/9 =1105/144

\(\dfrac{24\cdot47-23}{24+47\cdot23}\cdot\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{\dfrac{9}{1001}-\dfrac{9}{13}+\dfrac{9}{7}-\dfrac{9}{11}+9}\\ =\dfrac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\dfrac{3\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}{9\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}\\ =\dfrac{24\cdot24+24\cdot23-23\cdot1}{24+24\cdot23+23\cdot23}\cdot\dfrac{1}{3}\\ =\dfrac{23\left(24-1\right)+24\cdot24}{24\left(1+23\right)+23\cdot23}\cdot\dfrac{1}{3}=\dfrac{23\cdot23+24\cdot24}{24\cdot24+23\cdot23}\cdot\dfrac{1}{3}\\ =1\cdot\dfrac{1}{3}=\dfrac{1}{3}\)

\(A=\frac{24.47-23}{24+47-23}.\frac{3+\frac{3}{7}+\frac{3}{11}-\frac{3}{1001}+\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}+9}\)

\(A=\frac{1105}{48}.\frac{3.\left(1+\frac{1}{7}+\frac{1}{11}-\frac{1}{1001}+\frac{1}{13}\right)}{9.\left(\frac{1}{1001}-\frac{1}{11}+\frac{1}{7}+1\right)}\)

\(A=\frac{1105}{48}.\frac{3.\frac{1311}{1001}}{9.\frac{1054}{1001}}\)

\(A=\frac{1105}{48}.\frac{3933}{1001}:\frac{9468}{1001}\)

\(A=\frac{1105}{48}.\frac{437}{1052}\)

\(A\approx9,56\)

Chú ý : Dấu xấp xỉ \(\approx\)

10 nha anh k em nha

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BẠN XEM LẠI CÁI ĐỀ XEM ĐÚNG KO

\(\frac{11}{12}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9.\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}=\frac{11}{12}.\frac{1}{3}=\frac{11}{36}\)