Bài 1:Áp dụng C-S dạng engel

\(\frac{3}{xy+yz+xz}+\frac{2}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{2}{x^2+y^2+z^2}\)

\(\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{2}\right)^2>14\)

Ta có: \(x+y+z=0\)

=> \(\left(x+y+z\right)^2=0\)

<=> \(x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)

<=> \(x^2+y^2+z^2=0\) ( Dô \(xy+yz+xz=0\) )

=> \(x=y=z=0\) (1)

Thay (1) vào Q ta được:

Q = \(\left(-1\right)^{2017}+0^{2018}+1^{2019}=0\)

M+2019=2xy−yz−zx+2020M+2019=2xy−yz−zx+2020

=2xy−yz−zx+x2+y2+z2=2xy−yz−zx+x2+y2+z2

=(x+y−z2)2+3z24≥0=(x+y−z2)2+3z24≥0

⇒Mmin=0⇒Mmin=0 khi ⎧⎩⎨⎪⎪⎪⎪x+y−z2=03z24=0x2+y2+z2=2020{x+y−z2=03z24=0x2+y2+z2=2020

⇔⎧⎩⎨⎪⎪x+y=0z=0x2+y2=2020⇔{x+y=0z=0x2+y2=2020 ⇒⎧⎩⎨⎪⎪x=±1010−−−−√y=−xz=0

mình không hiểu ạ

x² + y² + z² = xy + yz + zx

⇔ 2( x² + y² + z² ) = 2( xy + yz + zx )

⇔ 2x² + 2y² + 2z² = 2xy + 2yz + 2zx

⇔ 2x² + 2y² + 2z² = 2xy + 2yz + 2zx

⇔ 2x² + 2y² + 2z² - 2xy - 2yz - 2zx = 0

⇔ ( x² - 2xy + y² ) + ( y² - 2yz + z² ) + ( z² - 2xz + x² ) = 0

⇔ ( x - y )² + ( y - z )² + ( z - x )² = 0

Vì : \(\hept{\begin{cases}\left(x-y\right)^2\\\left(y-z\right)^2\\\left(z-x\right)^2\end{cases}}\ge0\forall x,y,z\)=> ( x - y )² + ( y - z )² + ( z - x )² ≥ 0 ∀ x, y, z

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}}\Leftrightarrow x=y=z\)

Khi đó M = ( x - y + 1 )²⁰¹⁹ + ( y - z + 1 )²⁰²⁰ < đã sửa >

               = ( x - x + 1 )²⁰¹⁹ + ( y - y + 1 )²⁰²⁰

               = 1²⁰¹⁹ + 1²⁰²⁰

               = 1 + 1 = 2

Đk: $x\geq \frac{1}{2}$

Pt $\Leftrightarrow 4x^2+3x-7=4(\sqrt{x^3+3x^2}-2)+2(\sqrt{2x-1}-1)$

$\Leftrightarrow +4\frac{(x-1)(x+2)^2}{\sqrt{x^3+3x^2}+2}+4\frac{x-1}{\sqrt{2x-1}+1}-(x-1)(4x+7)=0$

$\Leftrightarrow (x-1)[\frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-(4x+7)]=0$

$\Leftrightarrow x=1\vee \frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7=0$ $(*)$

Xét hàm số $f(x)=\frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7,x\in [\frac{1}{2};+\infty )$ thì $f(x)>0,\forall x\in [\frac{1}{2};+\infty )$

$\Rightarrow $ Pt $(*)$ vô nghiệm

\(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\Rightarrow x^2+y^2+z^2+2xy+2yz+2zx=0\)

\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)

Mà \(xy+yz+zx=0\)(theo đề) nên \(2\left(xy+yz+zx\right)=0\)

\(\Rightarrow x^2+y^2+z^2=0\)

Vì \(\hept{\begin{cases}x^2\ge0\\y^2\ge0\\z^2\ge0\end{cases}}\) (với mọi x;y;z) nên \(x^2+y^2+z^2\ge0\) (với mọi x;y;z)

Để \(x^2+y^2+z^2=0\) \(\Leftrightarrow\) \(\hept{\begin{cases}x^2=0\\y^2=0\\z^2=0\end{cases}\Leftrightarrow}x=y=z=0\)

Vậy \(A=\left(0-1\right)^{2016}+0^{2017}+\left(0+1\right)^{2018}=\left(-1\right)^{2016}+0+1^{2018}=2\)

ta có : xy + yz +zx = 0

        * yz = -xy-zx

\(\Rightarrow\)*xy = - yz - zx

         *zx= -xy-yz

ta có : M = \(\frac{xy}{z}+\frac{zx}{y}+\frac{yz}{x}\)

          M = \(\frac{-yz-zx}{z}+\frac{-xy-yz}{y}+\frac{-xy-zx}{x}\)

          M = \(\frac{z\times\left(-y-x\right)}{z}+\frac{y\times\left(-x-z\right)}{y}+\frac{x\times\left(-y-z\right)}{x}\)

          M = -y - x - x - z - y - z

         M = -2y - 2x - 2z

         M = -2( x+y+z )

   mà x+y+z=-1

         M = (-2) . (-1)

         M =2

 Quản lý

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