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\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2=\frac{1}{x+y+z}\)
=> x+y+z =1/2
+y+z+1=2x => x+y+z +1 =3x => 3x =1/2 +1 =3/2 => x =1/2
+x+y+2 =2y => x+y+z+2 =3y => 3y = 1/2 +2 = 5/2 => y =5/6
+z =1/2 -x-y =1/2 -1/2 -5/6 =-5/6
để B thuộc Z
=> căn x - 15 chia hết 3
căn x - 15 thuộc B(3)
=> căn x - 15 = 3K (K thuộc Z)
căn x = 3K + 15
x = (3K + 15)2
\(\frac{\sqrt{x}-15}{3}\)=\(\frac{\sqrt{x}}{3}\)-\(\frac{15}{3}\)=\(\frac{\sqrt{x}}{3}\)- 5
vì B thuộc Z => \(\frac{\sqrt{x}}{3}\)- 5 thuộc Z
=> \(\frac{\sqrt{x}}{3}\)thuộc Z
=>\(\sqrt{x}\)chia hết cho 3
=> \(\sqrt{x}\)= 9
a, 2009; 0
b, x= 0.5 ; y= 0.4; z=0.9
sai thì thôi nhé
Giải:
\(x-5\sqrt{x}\) = 0 (\(x\) ≥ 0)
\(\sqrt{x}\) .(\(\sqrt{x}\) - 5) = 0
\(\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}-5=0\end{array}\right.\)
\(\left[\begin{array}{l}x=0\\ \sqrt{x}=5\end{array}\right.\)
\(\left[\begin{array}{l}x=0\\ x=25\end{array}\right.\)
Vậy \(x\in\) {0; 25}
\(x^5\) = 2\(x^7\)
\(x^5\) - 2\(x^7\) = 0
\(x^5\).(1 - 2\(x^2\)) = 0
\(\left[\begin{array}{l}x^5=0\\ 1-2x^2=0\end{array}\right.\)
\(\left[\begin{array}{l}x=0\\ 2x^2=1\end{array}\right.\)
\(\left[\begin{array}{l}x=0\\ x^2=\frac12\end{array}\right.\)
\(\left[\begin{array}{l}x=0\\ x=\pm\sqrt{\frac12}\end{array}\right.\)
Vậy \(x\) ∈ {- \(\sqrt{\frac12}\); 0; \(\sqrt{\frac12}\)}
Giải:
\(x-5\sqrt{x}\) = 0 (\(x\) ≥ 0)
\(\sqrt{x}\) .(\(\sqrt{x}\) - 5) = 0
\(\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}-5=0\end{array}\right.\)
\(\left[\begin{array}{l}x=0\\ \sqrt{x}=5\end{array}\right.\)
\(\left[\begin{array}{l}x=0\\ x=25\end{array}\right.\)
Vậy \(x\in\) {0; 25}
\(x^5\) = 2\(x^7\)
\(x^5\) - 2\(x^7\) = 0
\(x^5\).(1 - 2\(x^2\)) = 0
\(\left[\begin{array}{l}x^5=0\\ 1-2x^2=0\end{array}\right.\)
\(\left[\begin{array}{l}x=0\\ 2x^2=1\end{array}\right.\)
\(\left[\begin{array}{l}x=0\\ x^2=\frac12\end{array}\right.\)
\(\left[\begin{array}{l}x=0\\ x=-\frac{1}{\sqrt2}\\ x=\frac{1}{\sqrt2}\end{array}\right.\)
Vậy \(x\) \(\in\) {- \(\frac{1}{\sqrt2}\); 0; \(\frac{1}{\sqrt2}\)}