[Đại số 9] Giải hệ phương trình

K

khai221050

L

lp_qt

{1x+1y+1z=22xy1z2=4\left\{\begin{matrix} \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2\\ \dfrac{2}{xy}-\dfrac{1}{z^2}=4 \end{matrix}\right.

xét: 2xy1z2=4\dfrac{2}{xy}-\dfrac{1}{z^2}=4

\Leftrightarrow 2xy=1z2+4>0\dfrac{2}{xy}=\dfrac{1}{z^2}+4 >0

\Rightarrowxy>0xy>0

xét: 1x+1y+1z=2\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2

\Leftrightarrow 1x+1y=21z\dfrac{1}{x}+\dfrac{1}{y}=2-\dfrac{1}{z}

•TH1: Cả 2 vế của pt đều âm

21z<02-\dfrac{1}{z}<0 \Rightarrow z>0z>0

khi đó 1x+1y=21z\dfrac{1}{x}+\dfrac{1}{y}=2-\dfrac{1}{z}

\Leftrightarrow 1x1y=1z2\dfrac{-1}{x}-\dfrac{1}{y}= \dfrac{1}{z}-2

\Leftrightarrow 1x2+1y2+2xy=1z24z+4\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{xy}= \dfrac{1}{z^2}-\frac{4}{z}+4

\Leftrightarrow 1x2+1y2=4z(vl)\dfrac{1}{x^2}+\dfrac{1}{y^2}=-\dfrac{4}{z} (vl)

• TH2: Cả 2 vế của pt đều dương

1x+1y>0\dfrac{1}{x}+\dfrac{1}{y} > 0 \Leftrightarrow x+yxy>0\dfrac{x+y}{xy} >0 \Leftrightarrow x+y>0(vıˋxy>0)x+y>0 (vì xy>0) \Rightarrow x;y>0x;y>0

khi đó 1x+1y2xy\dfrac{1}{x}+\dfrac{1}{y} \ge \dfrac{2}{\sqrt{xy}}

\Leftrightarrow (1x+1y)24xy(\dfrac{1}{x}+\dfrac{1}{y})^2 \ge \dfrac{4}{xy} (bình phương 2 vế ko âm)

\Leftrightarrow (21z)22.(1z2+4)(2-\dfrac{1}{z})^2 \ge 2.(\dfrac{1}{z^2}+4)

\Leftrightarrow 1z24z+42z2+8\dfrac{1}{z^2}-\dfrac{4}{z}+4 \ge \dfrac{2}{z^2}+8

\Leftrightarrow 1z2+4z+40\dfrac{1}{z^2}+\dfrac{4}{z}+4 \le 0

\Leftrightarrow (1z+2)20(\dfrac{1}{z}+2)^2 \le 0

\Leftrightarrow 1z+2=0\dfrac{1}{z}+2=0

\Leftrightarrow z=12z=\dfrac{-1}{2}

\Leftrightarrow x=y=12x=y=\dfrac{1}{2}
 
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