Tìm x ,y ,z biết: 2/3x =3/4y=5/6z và x^2+y^2+z^2=724 15/09/2021 Bởi Aubrey Tìm x ,y ,z biết: 2/3x =3/4y=5/6z và x^2+y^2+z^2=724
Đáp án: $\text {Vậy (x,y,z) = (±13,75; ±15,47; ±17,19)}$ Giải thích các bước giải: `Đặt` `2/(3x) = 3/(4y) = 5/(6z) = m ` `⇒ x = 2/(3m); y = 3/(4m); z = 5/(6m)` `Mà` `x² +y² +z² = 724` `⇒ (2/(3m))² + (3/(4m))² + (5/(6m))² = 724` `⇒ 4/(9m²) + 9/(16m²) + 25/(36m²) = 724` `⇒ 1/m².(4/9 + 9/16 + 25/36) = 724` `⇒ 1/m².245/144 = 724` `⇒ 1/m² = 104256/245` `⇒ 104256m² = 245` `⇒ m² = 245/104256` `⇒ m = ±(7√5)/(24√181)` $\text {+ Với m =}$ `(7√5)/(24√181):` `x = 2/(3.(7√5)/(24√181)) = 13,75; y = 3/(4.(7√5)/(24√181)) = 15,47; z = 5/(6.(7√5)/(24√181)) = 17,19` $\text {+ Với m =}$ `-(7√5)/(24√181):` `x = 2/(3.-(7√5)/(24√181)) = -13,75; y = 3/(4.(7√5)/(24√181)) = -15,47; z = 5/(6.(7√5)/(24√181)) = -17,19` $\text {Vậy (x,y,z) = (±13,75; ±15,47; ±17,19)}$ Bình luận
Đáp án: Áp dụng t.c dãy tỉ số bằng nhau ta có: $\begin{array}{l}\dfrac{2}{{3x}} = \dfrac{3}{{4y}} = \dfrac{5}{{6z}}\\ \Rightarrow {\left( {\dfrac{2}{{3x}}} \right)^2} = {\left( {\dfrac{3}{{4y}}} \right)^2} = {\left( {\dfrac{5}{{6z}}} \right)^2}\\ = \dfrac{{\dfrac{4}{9}}}{{{x^2}}} = \dfrac{{\dfrac{9}{{16}}}}{{{y^2}}} = \dfrac{{\dfrac{{25}}{{36}}}}{{{z^2}}}\\ = \dfrac{{\dfrac{4}{9} + \dfrac{9}{{16}} + \dfrac{{25}}{{36}}}}{{{x^2} + {y^2} + {z^2}}} = \dfrac{{\dfrac{{245}}{{144}}}}{{724}} = \dfrac{{245}}{{104256}}\\ \Rightarrow \left\{ \begin{array}{l}{x^2} = \dfrac{4}{9}:\dfrac{{245}}{{104256}} = \dfrac{{46336}}{{245}}\\{y^2} = \dfrac{9}{{16}}:\dfrac{{245}}{{104256}} = \dfrac{{58644}}{{245}}\\{z^2} = \dfrac{{14480}}{{49}}\end{array} \right.\\ \Rightarrow \left( {x;y;z} \right) = \left\{ { \pm 13,75; \pm 15,47; \pm 17,19} \right\}\end{array}$ Bình luận
Đáp án:
$\text {Vậy (x,y,z) = (±13,75; ±15,47; ±17,19)}$
Giải thích các bước giải:
`Đặt` `2/(3x) = 3/(4y) = 5/(6z) = m `
`⇒ x = 2/(3m); y = 3/(4m); z = 5/(6m)`
`Mà` `x² +y² +z² = 724`
`⇒ (2/(3m))² + (3/(4m))² + (5/(6m))² = 724`
`⇒ 4/(9m²) + 9/(16m²) + 25/(36m²) = 724`
`⇒ 1/m².(4/9 + 9/16 + 25/36) = 724`
`⇒ 1/m².245/144 = 724`
`⇒ 1/m² = 104256/245`
`⇒ 104256m² = 245`
`⇒ m² = 245/104256`
`⇒ m = ±(7√5)/(24√181)`
$\text {+ Với m =}$ `(7√5)/(24√181):`
`x = 2/(3.(7√5)/(24√181)) = 13,75; y = 3/(4.(7√5)/(24√181)) = 15,47; z = 5/(6.(7√5)/(24√181)) = 17,19`
$\text {+ Với m =}$ `-(7√5)/(24√181):`
`x = 2/(3.-(7√5)/(24√181)) = -13,75; y = 3/(4.(7√5)/(24√181)) = -15,47; z = 5/(6.(7√5)/(24√181)) = -17,19`
$\text {Vậy (x,y,z) = (±13,75; ±15,47; ±17,19)}$
Đáp án:
Áp dụng t.c dãy tỉ số bằng nhau ta có:
$\begin{array}{l}
\dfrac{2}{{3x}} = \dfrac{3}{{4y}} = \dfrac{5}{{6z}}\\
\Rightarrow {\left( {\dfrac{2}{{3x}}} \right)^2} = {\left( {\dfrac{3}{{4y}}} \right)^2} = {\left( {\dfrac{5}{{6z}}} \right)^2}\\
= \dfrac{{\dfrac{4}{9}}}{{{x^2}}} = \dfrac{{\dfrac{9}{{16}}}}{{{y^2}}} = \dfrac{{\dfrac{{25}}{{36}}}}{{{z^2}}}\\
= \dfrac{{\dfrac{4}{9} + \dfrac{9}{{16}} + \dfrac{{25}}{{36}}}}{{{x^2} + {y^2} + {z^2}}} = \dfrac{{\dfrac{{245}}{{144}}}}{{724}} = \dfrac{{245}}{{104256}}\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} = \dfrac{4}{9}:\dfrac{{245}}{{104256}} = \dfrac{{46336}}{{245}}\\
{y^2} = \dfrac{9}{{16}}:\dfrac{{245}}{{104256}} = \dfrac{{58644}}{{245}}\\
{z^2} = \dfrac{{14480}}{{49}}
\end{array} \right.\\
\Rightarrow \left( {x;y;z} \right) = \left\{ { \pm 13,75; \pm 15,47; \pm 17,19} \right\}
\end{array}$