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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