Bài 64 trang 34 SGK Toán 6 tập 2

Đề bài

Hoàn thành các phép tính:

a) \(\displaystyle {7 \over 9} – { \ldots  \over 3} = {1 \over 9}\)

b) \(\displaystyle {1 \over  \ldots } – {{ – 2} \over {15}} = {7 \over {15}}\)

c) \(\displaystyle {{ – 11} \over {14}} – {{ – 4} \over  \ldots } = {{ – 3} \over {14}}\)

d) \(\displaystyle { \ldots  \over {21}} – {2 \over 3} = {5 \over {21}}\) 

Lời giải chi tiết

Thay dấu ba chấm bởi x rồi tìm x.

\(\displaystyle a)\;{7 \over 9} – {x \over 3} = {1 \over 9}\)

Suy ra: \(\displaystyle {x \over 3} = {7 \over 9} – {1 \over 9}={7 \over 9} + {-1 \over 9}={7+(-1) \over 9}\)

Hay \(\displaystyle {x \over 3} = {6 \over 9} = {2 \over 3}\)

Vậy x = 2 

b)

\(\dfrac{1}{x} – \dfrac{{ – 2}}{{15}} = \dfrac{7}{{15}} \)

\(\dfrac{1}{x} = \dfrac{7}{{15}} + \dfrac{{ – 2}}{{15}}\)

\(\dfrac{1}{x}= \dfrac{5}{{15}} \)

\(\dfrac{1}{x}= \dfrac{1}{3}\)
Suy ra \( x = 3\)

c)

\(\dfrac{{ – 11}}{{14}} – \dfrac{{ – 4}}{x} = \dfrac{{ – 3}}{{14}} \)

\(\Rightarrow \dfrac{{ – 4}}{x} = \dfrac{{ – 11}}{{14}} – \dfrac{{ – 3}}{{14}} \)

\(\dfrac{{ – 4}}{x}= \dfrac{{ – 11}}{{14}} + \dfrac{3}{{14}} \)

\(\dfrac{{ – 4}}{x}= \dfrac{{ – 11 + 3}}{{14}} \)

\(\dfrac{{ – 4}}{x}= \dfrac{{ – 8}}{{14}}\)

\(\dfrac{{ – 4}}{x}= \dfrac{{ – 4}}{7}\)

\( \Rightarrow x = 7\) 

d)\(\dfrac{x}{{21}} – \dfrac{2}{3} = \dfrac{5}{{21}}\)

\( \Rightarrow \dfrac{x}{{21}} = \dfrac{5}{{21}} + \dfrac{2}{3} \)

\(\dfrac{x}{{21}} = \dfrac{5}{{21}} + \dfrac{{14}}{{21}}\)

\(\dfrac{x}{{21}} = \dfrac{{19}}{{21}}\)

\( \Rightarrow x = 19\) 

Các số cần điền là: a) 2 ; b) 3; c) 7; d) 19.

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