[กิตติ]

ข้อ 7 ตอนที่ 1
$\frac{2xy+yz-4xz}{47xyz} =\frac{2}{47}(\frac{1}{z} )+ \frac{1}{47}(\frac{1}{x})-\frac{4}{47}(\frac{1}{y})$
$\frac{3xy}{x+y}=4\rightarrow \frac{1}{4} =\frac{x+y}{3xy}\rightarrow \frac{3}{4} =\frac{1}{x}+\frac{1}{y} $......(1)
$\frac{2yz}{y+z}=3\rightarrow \frac{1}{3} =\frac{y+z}{2yz}\rightarrow \frac{2}{3} =\frac{1}{z}+\frac{1}{y} $......(2)
$\frac{5xz}{x+z}=2\rightarrow \frac{1}{2} =\frac{x+z}{5xz}\rightarrow \frac{5}{2} =\frac{1}{x}+\frac{1}{z} $......(3)
(1)-(2) $\frac{1}{12} =\frac{1}{x}-\frac{1}{z}$.....(4)
(4)+(3) $\frac{31}{12}=\frac{2}{x}\rightarrow \frac{1}{x}=\frac{31}{24}$
แทน $\frac{1}{x} $ ในสมการ (4)
$\frac{1}{z}=\frac{1}{x}-\frac{1}{12}=\frac{31}{24}-\frac{1}{12}=\frac{29}{24} $
$\frac{1}{y}=\frac{2}{3}- \frac{1}{z}=\frac{2}{3}-\frac{29}{24}=-\frac{13}{24} $
$\frac{2xy+yz-4xz}{47xyz} =\frac{2}{47}(\frac{1}{z} )+ \frac{1}{47}(\frac{1}{x})-\frac{4}{47}(\frac{1}{y})$
$=\frac{2}{47}(\frac{29}{24} )+ \frac{1}{47}(\frac{31}{24})-\frac{4}{47}(-\frac{13}{24})$
$=\frac{1}{47\times 24}\left(\,58+31+42\right) $
$=\frac{131}{47\times 24} =\frac{1}{8} $

$\frac{47xyz}{2xy+yz-4xz} =8$