Bài 60 trang 31 SGK Toán 7 tập 1

Đề bài

Tìm $x$ trong các tỉ lệ thức sau:

$\eqalign{ & a)\,\,\left( {{1 \over 3}x} \right):{2 \over 3} = 1{3 \over 4}:{2 \over 5} \cr & b)\,\,4,5:0,3{\rm{ }} = {\rm{ }}2.25:\left( {0,1.x} \right) \cr & c)\,\,8:\left( {{1 \over 4}.x} \right) = 2:0,02 \cr & d)\,\,3:2{1 \over 4} = {3 \over 4}:(6.x) \cr}$

Lời giải chi tiết

$\eqalign{ & a)\,\,\left( {{1 \over 3}x} \right):{2 \over 3} = 1{3 \over 4}:{2 \over 5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,{x \over 3}:{2 \over 3} = {7 \over 4}:{2 \over 5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,{x \over 3}.{3 \over 2} = {7 \over 4}.{5 \over 2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,{x \over 2} = {{35} \over 8} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\;x = {{35.2} \over 8} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\;x = {{35} \over 4} \cr}$.

Vậy $x={{35} \over 4}$

$\begin{array}{l} b)\;4,5:0,3 = 2,25:\left( {0,1.x} \right)\\ \;\;\;15 = 2,25:\left( {0,1.x} \right)\\ \;\;\;0,1.x = 2,25:15\\ \;\;\;0,1x = 0,15\\\;\;\;x=0,15:0,1\\ \;\;\;x = 1,5 \end{array}$

Vậy x=1,5 

$\eqalign{ & c)\,\,8:\left( {{1 \over 4}.x} \right) = 2:0,02 \cr & \,\,\,\,\,\,\,8:\left( {{1 \over 4}.x} \right) = 100 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{1 \over 4}x = 8:100 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{1 \over 4}x = 0,08 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0,08:{1 \over 4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0,08.4 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0,32 \cr}$

Vậy x=0,32

$\begin{array}{l} d)\,3:2\dfrac{1}{4} = \dfrac{3}{4}:\left( {6.x} \right)\\ 3:\dfrac{9}{4} = \dfrac{3}{4}:\left( {6.x} \right)\\ 3.\dfrac{4}{9} = \dfrac{3}{4}:\left( {6.x} \right)\\ \dfrac{4}{3} = \dfrac{3}{4}:\left( {6.x} \right)\\ 6.x = \dfrac{3}{4}:\dfrac{4}{3}\\ 6.x = \dfrac{3}{4}.\dfrac{3}{4}\\ 6.x = \dfrac{9}{{16}}\\ x = \dfrac{9}{{16}}:6\\ x = \dfrac{9}{{16}}.\dfrac{1}{6}\\ x = \dfrac{3}{{32}} \end{array}$

Vậy $x = \frac{3}{{32}}$

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