Đề bài
Tìm \[x\] trong các tỉ lệ thức sau:
\[\eqalign{
& a]\,\,\left[ {{1 \over 3}x} \right]:{2 \over 3} = 1{3 \over 4}:{2 \over 5} \cr
& b]\,\,4,5:0,3{\rm{ }} = {\rm{ }}2.25:\left[ {0,1.x} \right] \cr
& c]\,\,8:\left[ {{1 \over 4}.x} \right] = 2:0,02 \cr
& d]\,\,3:2{1 \over 4} = {3 \over 4}:[6.x] \cr} \]
Phương pháp giải - Xem chi tiết
Áp dụng:
- Tính chất tỉ lệ thức.
\[\dfrac{a}{b} =\dfrac{c}{d} \Leftrightarrow ad = bc\]
- Quy tắc nhân, chia số hữu tỉ.
\[\begin{array}{l}
\dfrac{A}{B}.\dfrac{C}{D} =\dfrac{{AC}}{{BD}}\\
\dfrac{A}{B}:\dfrac{C}{D} =\dfrac{A}{B}.\dfrac{D}{C} =\dfrac{{AD}}{{BC}}
\end{array}\]
Lời giải chi tiết
\[\begin{array}{l}
a]{\kern 1pt} {\kern 1pt} \left[ {\dfrac{1}{3}x} \right]:\dfrac{2}{3} = 1\dfrac{3}{4}:\dfrac{2}{5}\\
\Rightarrow \left[ {\dfrac{1}{3}x} \right].\dfrac{2}{5} = \dfrac{2}{3}.1\dfrac{3}{4}\\
\Rightarrow \dfrac{1}{3}x = \dfrac{{\dfrac{2}{3}.\dfrac{7}{4}}}{{\dfrac{2}{5}}}\\
\Rightarrow \dfrac{1}{3}x = \dfrac{{35}}{{12}}\\
\Rightarrow x = \dfrac{{35}}{{12}}:\dfrac{1}{3} = \dfrac{{35}}{4}
\end{array}\]
\[\begin{array}{l}
b]\;\,4,5:0,3 = 2,25:\left[ {0,1.x} \right]\\
\Rightarrow 4,5.\left[ {0,1.x} \right] = 0,3.2,25\\
\Rightarrow 0,1.x = \dfrac{{0,3.2,25}}{{4,5}}\\
\Rightarrow x = \dfrac{{0,15}}{{0,1}} = 1,5
\end{array}\]
\[\begin{array}{l}
c]{\kern 1pt} {\kern 1pt} \,8:\left[ {\dfrac{1}{4}.x} \right] = 2:0,02\\
\Rightarrow 2.\left[ {\dfrac{1}{4}.x} \right] = 8.0,02\\
\Rightarrow \dfrac{1}{4}x = \dfrac{{8.0,02}}{2}\\
\Rightarrow x = 0,08:\dfrac{1}{4} = 0,32
\end{array}\]
\[\begin{array}{l}
d]\,\,{\kern 1pt} {\kern 1pt} 3:2\dfrac{1}{4} = \dfrac{3}{4}:[6.x]\\
\Rightarrow 3.\left[ {6.x} \right] = 2\dfrac{1}{4}.\dfrac{3}{4}\\
\Rightarrow 6.x = \dfrac{{\dfrac{9}{4}.\dfrac{3}{4}}}{3}\\
\Rightarrow x = \dfrac{9}{{16}}:6 = \dfrac{3}{{32}}
\end{array}\]