áp dụng tính chất dãy tỉ số bằng nhau, tìm x,y,z biết:
$\frac{6x}{1}$ = $\frac{9y}{2}$ = $\frac{12z}{3}$ và$ x^2+2y^2+3z^2=1$
áp dụng tính chất dãy tỉ số bằng nhau, tìm x,y,z biết:
$\frac{6x}{1}$ = $\frac{9y}{2}$ = $\frac{12z}{3}$ và$ x^2+2y^2+3z^2=1$
Đáp án:
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Giải thích các bước giải:
`(6x)/1=(9y)/2=(12z)/3`
`=>`$\dfrac{x}{\dfrac{1}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{1}{4}}$`
`=>`$\dfrac{x^2}{{(\dfrac{1}{6}})^2}=\dfrac{2y^2}{{2.(\dfrac{2}{9}})^2}=\dfrac{3z^2}{3.{(\dfrac{1}{4}})^2}$
`=>` $\dfrac{x^2}{{\dfrac{1}{36}}}=\dfrac{2y^2}{{\dfrac{8}{81}}}=\dfrac{3z^2}{{\dfrac{3}{16}}}$
`=`$\dfrac{x^2+2y^2+3z^2}{\dfrac{1}{36}+\dfrac{8}{81}+\dfrac{3}{16}}$`=`$\dfrac{1}{\dfrac{407}{1296}}=\dfrac{1296}{407}$
`=>`
`x^2=1296/407 . 1/36=36/407 => x = +- \sqrt{\frac{36}{407}}`
`y^2=1296/407 . 8/81 = 128/407 => y = +-\sqrt{\frac{128}{407}}`
`z^2=1296/407 . 3/16=243/407 => z = +-\sqrt{\frac{243}{407}}`
Vậy `….`
$\dfrac{6x}{1}=\dfrac{9y}{2}=\dfrac{12z}{3}$
$→\dfrac{x}{\dfrac{1}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{1}{4}}$
$→\dfrac{x²}{\dfrac{1}{36}}=\dfrac{y²}{\dfrac{4}{81}}=\dfrac{z²}{\dfrac{1}{16}}$
$→\dfrac{x²}{\dfrac{1}{36}}=\dfrac{2y²}{\dfrac{8}{81}}=\dfrac{3z²}{\dfrac{3}{16}}$
Áp dụng tính chất dãy tỉ số bằng nhau:
$\dfrac{x²}{\dfrac{1}{36}}=\dfrac{2y²}{\dfrac{8}{81}}=\dfrac{3z²}{\dfrac{3}{16}}=\dfrac{x²+2y²+3z²}{\dfrac{1}{36}+\dfrac{8}{81}+\dfrac{3}{16}}=\dfrac{1}{\dfrac{407}{1296}}=\dfrac{1296}{407}$
$→\begin{cases}x²=\dfrac{36}{407}\\y²=\dfrac{64}{407}\\z²=\dfrac{81}{407}\end{cases}$
$→\begin{cases}x=±\dfrac{6}{\sqrt{407}}\\y=±\dfrac{8}{\sqrt{407}}\\z=±\dfrac{9}{\sqrt{407}}\end{cases}$