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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 7: Tìm x, y, z: $\frac{x}{2}$ = $\frac{y}{3}$, $\frac{y}{4}$ = $\frac{z}{5}$ và $x^{2}$ – $y^{2}$ = -16

Toán Lớp 7: Tìm x, y, z:
$\frac{x}{2}$ = $\frac{y}{3}$, $\frac{y}{4}$ = $\frac{z}{5}$ và $x^{2}$ – $y^{2}$ = -16

Comments ( 2 )

  1. $\text{Ta có:}$
    x/2 = y/3 ; y/4 = z/5
    => x/2 . 1/4 = y/3 . 1/4 ; y/4 . 1/3 = z/5 . 1/3
    => x/8 = y/12 ; y/12 = z/15
    => x/8 = y/12 = z/15
    $\text{Đặt}$ x/8 = y/12 = z/15 = k
    => {(x=8k),(y=12k),(z=15k):}
    $\text{Mà}$ x^{2} – y^{2} = -16
    => (8k)^{2} – (12k)^{2} = -16
    => 64k^{2} – 144k^{2} = -16
    => -80k^{2} = -16
    => k^{2} = (-16) : (-80)
    => k^{2} = 1/5
    $\text{Vì}$ k^{2} \ne 1/5 $\text{nên}$ x,y,z \in \emptyset
    $\text{Vậy}$ x,y,z \in \emptyset

  2. $\\$
    Giải đáp + giải thích các bước giải :
    $\dfrac{x}{2}=\dfrac{y}{3}⇒\dfrac{x}{8}=\dfrac{y}{12}(1)$
    $\dfrac{y}{4}=\dfrac{z}{5}⇒\dfrac{y}{12}=\dfrac{z}{15}(2)$
    $(1)(2)⇒\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}$
    Đặt $\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=k$
    $⇒x=8k,y=12k,z=15k$
    $x^2-y^2=-16$
    $⇒ 64k^2 – 144k^2=-16$
    $⇒k^2=\dfrac{1}{5}$
    $⇒k=±\sqrt{\dfrac{1}{5}}$
    Với $k=\sqrt{\dfrac{1}{5}}$
    $x=\dfrac{8\sqrt{5}}{5}$
    $y=\dfrac{12\sqrt{5}}{5}$
    $z=3\sqrt{5}$
    Với $k=-\sqrt{\dfrac{1}{5}}$
    $x=\dfrac{-8\sqrt{5}}{5}$
    $y=\dfrac{-12\sqrt{5}}{5}$
    $z=-3\sqrt{5}$
    Vậy $(x;y;z)=(\dfrac{8\sqrt{5}}{5};\dfrac{12\sqrt{5}}{5};3\sqrt{5}),(\dfrac{-8\sqrt{5}}{5};\dfrac{-12\sqrt{5}}{5};-3\sqrt{5})$
     

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222-9+11+12:2*14+14 = ? ( )