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Câu hỏi: Tính:
a) \(\dfrac{7}{8} + \dfrac{7}{8}:\dfrac{1}{8} - \dfrac{1}{2}\)
b) \(\dfrac{6}{{11}} + \dfrac{{11}}{3}.\dfrac{3}{{22}}\)
a) \(\dfrac{7}{8} + \dfrac{7}{8}:\dfrac{1}{8} - \dfrac{1}{2}\)
b) \(\dfrac{6}{{11}} + \dfrac{{11}}{3}.\dfrac{3}{{22}}\)
Phương pháp giải
a)
Chuyển phép chia về phép nhân.
Sử dụng tính chất: \(a.b + a.c = a.\left( {b + c} \right)\)
b)
Thực hiện phép nhân trước rồi đến phép cộng.
Lời giải chi tiết
a) \(\dfrac{7}{8} + \dfrac{7}{8}:\dfrac{1}{8} - \dfrac{1}{2}\)
\(\begin{array}{l} = \dfrac{7}{8} + \dfrac{7}{8}.8 - \dfrac{1}{2}\\ = \dfrac{7}{8}.1 + \dfrac{7}{8}.8 - \dfrac{1}{2}\\ = \left( {\dfrac{7}{8}.1 + \dfrac{7}{8}.8} \right) - \dfrac{1}{2}\\ = \dfrac{7}{8}.\left( {1 + 8} \right) - \dfrac{1}{2} = \dfrac{7}{8}.9 - \dfrac{1}{2}\\ = \dfrac{{63}}{8} - \dfrac{1}{2} = \dfrac{{63}}{8} - \dfrac{4}{8} = \dfrac{{63 - 4}}{8} = \dfrac{{59}}{8}\end{array}\)
b) \(\dfrac{6}{{11}} + \dfrac{{11}}{3}.\dfrac{3}{{22}}\)
\(\begin{array}{l} = \dfrac{6}{{11}} + \dfrac{{11.3}}{{3.22}} = \dfrac{6}{{11}} + \dfrac{1}{2}\\ = \dfrac{{12}}{{22}} + \dfrac{{11}}{{22}} = \dfrac{{12 + 11}}{{22}} = \dfrac{{23}}{{22}}\end{array}\)
a)
Chuyển phép chia về phép nhân.
Sử dụng tính chất: \(a.b + a.c = a.\left( {b + c} \right)\)
b)
Thực hiện phép nhân trước rồi đến phép cộng.
Lời giải chi tiết
a) \(\dfrac{7}{8} + \dfrac{7}{8}:\dfrac{1}{8} - \dfrac{1}{2}\)
\(\begin{array}{l} = \dfrac{7}{8} + \dfrac{7}{8}.8 - \dfrac{1}{2}\\ = \dfrac{7}{8}.1 + \dfrac{7}{8}.8 - \dfrac{1}{2}\\ = \left( {\dfrac{7}{8}.1 + \dfrac{7}{8}.8} \right) - \dfrac{1}{2}\\ = \dfrac{7}{8}.\left( {1 + 8} \right) - \dfrac{1}{2} = \dfrac{7}{8}.9 - \dfrac{1}{2}\\ = \dfrac{{63}}{8} - \dfrac{1}{2} = \dfrac{{63}}{8} - \dfrac{4}{8} = \dfrac{{63 - 4}}{8} = \dfrac{{59}}{8}\end{array}\)
b) \(\dfrac{6}{{11}} + \dfrac{{11}}{3}.\dfrac{3}{{22}}\)
\(\begin{array}{l} = \dfrac{6}{{11}} + \dfrac{{11.3}}{{3.22}} = \dfrac{6}{{11}} + \dfrac{1}{2}\\ = \dfrac{{12}}{{22}} + \dfrac{{11}}{{22}} = \dfrac{{12 + 11}}{{22}} = \dfrac{{23}}{{22}}\end{array}\)