\(728:\left(73x3.54+0.17x25+3.54x27+0.23x25\right)\)
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=\(3054.73+\frac{23}{100}.25+\frac{177}{50}.27+\frac{3}{20}.25\)
=\(3054.73+\frac{23}{4}+\frac{177.27}{50}+\frac{15}{4}\)
=\(3054.73+\frac{177.27}{50}+\frac{19}{2}\)
=\(222942+\frac{2627}{25}\)
ĐKXĐ: ...
Lấy pt cuối trừ 3 lần pt đầu ta được:
\(\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^3+\left(\sqrt{y}-\frac{1}{\sqrt{y}}\right)^3+\left(\sqrt{z}-\frac{1}{\sqrt{z}}\right)^3=\frac{512}{27}\)
Pt (2) tương đương:
\(x+\frac{1}{x}-2+y+\frac{1}{y}-2+z+\frac{1}{z}-2=\frac{64}{9}\)
\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2+\left(\sqrt{y}-\frac{1}{\sqrt{y}}\right)^2+\left(\sqrt{z}-\frac{1}{\sqrt{z}}\right)^2=\frac{64}{9}\)
Đặt \(\left(\sqrt{x}-\frac{1}{\sqrt{x}};\sqrt{y}-\frac{1}{\sqrt{y}};\sqrt{z}-\frac{1}{\sqrt{z}}\right)=\left(a;b;c\right)\)
Hệ trở thành:
\(\left\{{}\begin{matrix}a+b+c=\frac{8}{3}\\a^2+b^2+c^2=\frac{64}{9}\\a^3+b^3+c^3=\frac{512}{27}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b+c=\frac{8}{3}\\ab+bc+ca=0\\a^3+b^3+c^3=\frac{512}{27}\end{matrix}\right.\)
Ta có: \(a^3+b^3+c^3-3abc=\frac{512}{27}-3abc\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=\frac{512}{27}-3abc\)
\(\Leftrightarrow\frac{8}{3}.\left(\frac{64}{9}-0\right)=\frac{512}{27}-3abc\)
\(\Rightarrow abc=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b+c=\frac{8}{3}\\ab+bc+ca=0\\abc=0\end{matrix}\right.\) \(\Leftrightarrow\left(a;b;c\right)=\left(0;0;\frac{8}{3}\right)\) và hoán vị
Hay \(\left(x;y;z\right)=\left(1;1;9\right)\) và hoán vị
2 × 6 = 12 6 × 2 = 12 12 : 2 = 6 12 : 6 = 2 | 7 × 4 = 28 4 × 7 = 28 28 : 7 = 4 28 : 4 = 7 | 9 × 3 = 27 3 × 9 = 27 27 : 9 = 3 27 : 3 = 9 | 8 × 5 = 40 5 × 8 = 40 40 : 8 = 5 40 : 5 = 8 |
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728 : (73 x 3,54 + 0,17 x25 +3,5 x27 +0,23x25 )
728 :( 258 ,42 +4,25 +95,58+5,75 )
728 : 364
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