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

Toán Lớp 9: x căn (1-y^2)+y căn (1-z^2)+ z căn (1-x^2)=3/2 tính x^2+y^2+z^2

Toán Lớp 9: x căn (1-y^2)+y căn (1-z^2)+ z căn (1-x^2)=3/2
tính x^2+y^2+z^2

Comments ( 2 )

  1. $\begin{array}{l}
    x\sqrt {1 – {y^2}}  + y\sqrt {1 – {z^2}}  + z\sqrt {1 – {x^2}}  = \dfrac{3}{2}\\
     \Leftrightarrow 2x\sqrt {1 – {y^2}}  + 2y\sqrt {1 – {z^2}}  + 2z\sqrt {1 – {x^2}}  = 3\\
     \Leftrightarrow 3 – 2x\sqrt {1 – {y^2}}  + 2y\sqrt {1 – {z^2}}  + 2z\sqrt {1 – {x^2}}  = 0\\
     \Leftrightarrow \left( {{x^2} – 2x\sqrt {1 – {y^2}}  + 1 – {y^2}} \right) + \left( {{y^2} – 2y\sqrt {1 – {z^2}}  + 1 – {z^2}} \right)\\
     + \left( {{z^2} – 2z\sqrt {1 – {x^2}}  + 1 – {x^2}} \right) = 0\\
     \Leftrightarrow {\left( {x – \sqrt {1 – {y^2}} } \right)^2} + {\left( {y – \sqrt {1 – {z^2}} } \right)^2} + {\left( {z – \sqrt {1 – {x^2}} } \right)^2} = 0\\
     \Leftrightarrow \left\{ \begin{array}{l}
    x = \sqrt {1 – {y^2}} \\
    y = \sqrt {1 – {z^2}} \\
    z = \sqrt {1 – {x^2}} 
    \end{array} \right. \Rightarrow \left\{ \begin{array}{l}
    {x^2} = 1 – {y^2}\\
    {y^2} = 1 – {z^2}\\
    {z^2} = 1 – {x^2}
    \end{array} \right.\\
     \Rightarrow 2\left( {{x^2} + {y^2} + {z^2}} \right) = 3\\
     \Rightarrow {x^2} + {y^2} + {z^2} = \dfrac{3}{2}
    \end{array}$

  2. Giải đáp:
    x^2+y^2+z^2=3/2 
    Lời giải và giải thích chi tiết:
    x\sqrt[1-y^2]+y\sqrt[1-z^2]+z\sqrt[1-x^2]=3/2
    ⇔ 2x\sqrt[1-y^2]+2y\sqrt[1-z^2]+2z\sqrt[1-x^2]=3
    ⇔ 3-2x\sqrt[1-y^2]+2y\sqrt[1-z^2]+2z\sqrt[1-x^2]=0
    ⇔ (x^2-2x\sqrt[1-y^2]+1-y^2)+(y^2-2y\sqrt[1-z^2]+1-z^2)+(z^2-2z\sqrt[1-x^2]+1-x^2)=0
    ⇔ (x-\sqrt[1-y^2])^2+(y-\sqrt[1-z^2])^2+(z-\sqrt[1-x^2])^2=0
    Nhận xét :
    (x-\sqrt[1-y^2])^2≥0$∀x,y∈\mathbb{R}$
    (y-\sqrt[1-z^2])^2≥0$∀y,z∈\mathbb{R}$
    (z-\sqrt[1-x^2])^2≥0$∀x,z∈\mathbb{R}$
    ⇒ (x-\sqrt[1-y^2])^2+(y-\sqrt[1-z^2])^2+(z-\sqrt[1-x^2])^2≥0
    Dấu = xảy ra ⇔ $\begin{cases} x=\sqrt{1-y^2}\\y=\sqrt{1-z^2}\\z=\sqrt{1-x^2} \end{cases}$
    ⇔ $\begin{cases} x^2=1-y^2\\y^2=1-z^2\\z^2=1-x^2 \end{cases}$
    ⇒ 2(x^2+y^2+z^2)=3
    ⇔ x^2+y^2+z^2=3/2

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

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