giúp em với em đang cần gấp lắm ạ tìm x ,y,z a) $\frac{2x}{3}$ = $\frac{3y}{4}$ = $\frac{4z}{5}$ và x+y+z= 38 15/07/2021 Bởi Savannah giúp em với em đang cần gấp lắm ạ tìm x ,y,z a) $\frac{2x}{3}$ = $\frac{3y}{4}$ = $\frac{4z}{5}$ và x+y+z= 38
$\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}$ $↔\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}$ Áp dụng tính chất dãy tỉ số bằng nhau: $\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{38}{\dfrac{49}{12}}=\dfrac{456}{49}$ $→\dfrac{x}{\dfrac{3}{2}}=\dfrac{456}{49}→x=\dfrac{684}{49}$ $\dfrac{y}{\dfrac{4}{2}}=\dfrac{456}{49}→y=\dfrac{608}{49}$ $\dfrac{z}{\dfrac{4}{5}}=\dfrac{456}{49}→z=\dfrac{570}{49}$ Vậy $(x,y,z)=(\dfrac{684}{49},\dfrac{608}{49},\dfrac{570}{49})$ Bình luận
Giải thích các bước giải: Áp dụng tính chất dãy tỉ số bằng nhau ta có: \(\begin{array}{l}\dfrac{{2x}}{3} = \dfrac{{3y}}{4} = \dfrac{{4z}}{5}\\ \Leftrightarrow \dfrac{{\dfrac{{2x}}{2}}}{{\dfrac{3}{2}}} = \dfrac{{\dfrac{{3y}}{3}}}{{\dfrac{4}{3}}} = \dfrac{{\dfrac{{4z}}{4}}}{{\dfrac{5}{4}}}\\ \Leftrightarrow \dfrac{x}{{\dfrac{3}{2}}} = \dfrac{y}{{\dfrac{4}{3}}} = \dfrac{z}{{\dfrac{5}{4}}} = \dfrac{{x + y + z}}{{\dfrac{3}{2} + \dfrac{4}{3} + \dfrac{5}{4}}} = \dfrac{{38}}{{\dfrac{{49}}{{12}}}} = \dfrac{{456}}{{49}}\\ \Rightarrow \left\{ \begin{array}{l}\dfrac{x}{{\dfrac{3}{2}}} = \dfrac{{456}}{{49}}\\\dfrac{y}{{\dfrac{4}{3}}} = \dfrac{{456}}{{49}}\\\dfrac{z}{{\dfrac{5}{4}}} = \dfrac{{456}}{{49}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \dfrac{{456}}{{49}}.\dfrac{3}{2} = \dfrac{{684}}{{49}}\\y = \dfrac{{456}}{{49}}.\dfrac{4}{3} = \dfrac{{608}}{{49}}\\z = \dfrac{{456}}{{49}}.\dfrac{5}{4} = \dfrac{{570}}{{49}}\end{array} \right.\end{array}\) Bình luận
$\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}$
$↔\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}$
Áp dụng tính chất dãy tỉ số bằng nhau:
$\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{38}{\dfrac{49}{12}}=\dfrac{456}{49}$
$→\dfrac{x}{\dfrac{3}{2}}=\dfrac{456}{49}→x=\dfrac{684}{49}$
$\dfrac{y}{\dfrac{4}{2}}=\dfrac{456}{49}→y=\dfrac{608}{49}$
$\dfrac{z}{\dfrac{4}{5}}=\dfrac{456}{49}→z=\dfrac{570}{49}$
Vậy $(x,y,z)=(\dfrac{684}{49},\dfrac{608}{49},\dfrac{570}{49})$
Giải thích các bước giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\begin{array}{l}
\dfrac{{2x}}{3} = \dfrac{{3y}}{4} = \dfrac{{4z}}{5}\\
\Leftrightarrow \dfrac{{\dfrac{{2x}}{2}}}{{\dfrac{3}{2}}} = \dfrac{{\dfrac{{3y}}{3}}}{{\dfrac{4}{3}}} = \dfrac{{\dfrac{{4z}}{4}}}{{\dfrac{5}{4}}}\\
\Leftrightarrow \dfrac{x}{{\dfrac{3}{2}}} = \dfrac{y}{{\dfrac{4}{3}}} = \dfrac{z}{{\dfrac{5}{4}}} = \dfrac{{x + y + z}}{{\dfrac{3}{2} + \dfrac{4}{3} + \dfrac{5}{4}}} = \dfrac{{38}}{{\dfrac{{49}}{{12}}}} = \dfrac{{456}}{{49}}\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{x}{{\dfrac{3}{2}}} = \dfrac{{456}}{{49}}\\
\dfrac{y}{{\dfrac{4}{3}}} = \dfrac{{456}}{{49}}\\
\dfrac{z}{{\dfrac{5}{4}}} = \dfrac{{456}}{{49}}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{{456}}{{49}}.\dfrac{3}{2} = \dfrac{{684}}{{49}}\\
y = \dfrac{{456}}{{49}}.\dfrac{4}{3} = \dfrac{{608}}{{49}}\\
z = \dfrac{{456}}{{49}}.\dfrac{5}{4} = \dfrac{{570}}{{49}}
\end{array} \right.
\end{array}\)