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

Toán Lớp 9: cho x,y,z>0 thỏa mãn x+y+z=11. Tìm GTNN của P=x^3+4y^3+3z^3

Tháng Hai 6, 2023 Toán học 1 Comment 0 views

Toán Lớp 9: cho x,y,z>0 thỏa mãn x+y+z=11. Tìm GTNN của P=x^3+4y^3+3z^3

Comments ( 1 )

  1. dinhcongson
    dinhcongson
    6 Tháng Hai, 2023 at 10:42 chiều
    Reply
    Giải đáp:
     
    Lời giải và giải thích chi tiết:
    x^3+(66/(9+2\sqrt{3}))^3+(66/(9+2\sqrt{3}))^3>=3\root{3}{x^3.(66/(9+2\sqrt{3}))^3.(66/(9+2\sqrt{3}))^3}
    =>x^3+(66/(9+2\sqrt{3}))^3+(66/(9+2\sqrt{3}))^3>=3x(66/(9+2\sqrt{3}))^2
    4[y^3+(33/(9+2\sqrt{3}))^3+(33/(9+2\sqrt{3}))^3]>=4.3\root{3}{y^3.(33/(9+2\sqrt{3}))^3.(33/(9+2\sqrt{3}))^3}
    =>4[y^3+(33/(9+2\sqrt{3}))^3+(33/(9+2\sqrt{3}))^3]>=12y.(33/(9+2\sqrt{3}))^2
    =>4[y^3+(33/(9+2\sqrt{3}))^3+(33/(9+2\sqrt{3}))^3]>=3y.(66/(9+2\sqrt{3}))^2
    3[z^3+((66\sqrt{3}-44)/23)^3+((66\sqrt{3}-44)/23)^3]>=3.3\root{3}{z^3.((66\sqrt{3}-44)/23)^3.((66\sqrt{3}-44)/23)^3}
    =>3[z^3+((66\sqrt{3}-44)/23)^3+((66\sqrt{3}-44)/23)^3]>=3z(66/(9+2\sqrt{3}))^2
    =>x^3+(66/(9+2\sqrt{3}))^3+(66/(9+2\sqrt{3}))^3+4[y^3+(33/(9+2\sqrt{3}))^3+(33/(9+2\sqrt{3}))^3]+3[z^3+((66\sqrt{3}-44)/23)^3+((66\sqrt{3}-44)/23)^3]>=3x(66/(9+2\sqrt{3}))^2+3y(66/(9+2\sqrt{3}))^2+3z(66/(9+2\sqrt{3}))^2
    =>x^3+4y^3+3z^3+2(66/(9+2\sqrt{3}))^3+8(33/(9+2\sqrt{3}))^3+6((66\sqrt{3}-44)/23)^3>=3(x+y+z)(66/(9+2\sqrt{3}))^2
    =>x^3+4y^3+3z^3+2(66/(9+2\sqrt{3}))^3+8(33/(9+2\sqrt{3}))^3+6((66\sqrt{3}-44)/23)^3>=3(x+y+z)(66/(9+2\sqrt{3}))^2
    =>x^3+4y^3+3z^3+2(66/(9+2\sqrt{3}))^3+8(33/(9+2\sqrt{3}))^3+6((66\sqrt{3}-44)/23)^3>=3.11.(66/(9+2\sqrt{3}))^2
    =>x^3+4y^3+3z^3>=33(66/(9+2\sqrt{3}))^2-2(66/(9+2\sqrt{3}))^3-8(33/(9+2\sqrt{3}))^3-6((66\sqrt{3}-44)/23)^3
    Dấu = xảy ra 
    <=>$\begin{cases} x=\dfrac{66}{9+2\sqrt{3} } \\y=\dfrac{33}{9+2\sqrt{3} } \\z= \dfrac{66\sqrt{3}-44 }{23 } \end{cases}$

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

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