Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{y+z-5}=\frac{y}{x+z+3}=\frac{z}{x+y+2}=\frac{x+y+z}{y+z-5+x+z+3+x+y+2}=\frac{x+y+z}{2x+2y+2z}=\frac12\)

=>\(\begin{cases}y+z-5=2x\\ x+z+3=2y\\ x+y+2=2z\end{cases}\Rightarrow\begin{cases}y+z=2x+5\\ y+z=2y-3\\ x+y=2z-2\end{cases}\)

\(\frac{x}{y+z-5}=\frac12\left(x+y+z\right)\)

=>\(\frac12\left(x+y+z\right)=\frac12\)

=>x+y+z=1

*Ta có: x+y+z=1

=>z+2z-2=1

=>3z-2=1

=>3z=3

=>z=1

*Ta có: x+y+z=1

=>y+2y-3=1

=>3y=4

=>\(y=\frac43\)

*Ta có: x+y+z=1

=>x+2x+5=1

=>3x+5=1

=>3x=-4

=>\(x=-\frac43\)

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; \(\frac{1}{\sqrt2}\)}

Theo tính chất của dãy tỉ số bằng nhau, ta có

\(\frac{y+z+1}{x}=\frac{x+y+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}=\frac{y+z+1+x+y+2+x+y-3+1}{x+y+z+x+y+z}\)

=\(\frac{\left(x+y+z\right)+\left(x+y+y1+2-3\right)}{\left(x+y+z\right)+\left(x+y+z\right)}=\frac{\left(x+y+z\right)+\left(x+y+y+1\right)}{\left(x+y+z\right)+\left(x+y+z\right)}\)

=>x+y+y+1=x+y+z

=>y+1=z

Vậy đáp số cần tìm là x,y,z khác 0

x tùy ý

y tùy ý

z=y+1

Đề là gì ????

Ta có :

\(\frac{x}{3}=\frac{y}{2};\frac{z}{5}=\frac{y}{4}\)

\(\Rightarrow\frac{x}{6}=\frac{y}{4}=\frac{z}{5}\)

Áp dụng tc của dãy tỉ số bằng nhau ta có :

\(\frac{x}{6}=\frac{y}{4}=\frac{z}{5}=\frac{-x-y+z}{-6-4+5}=\frac{-10}{-5}=2\)

\(\Rightarrow\begin{cases}x=12\\y=8\\z=10\end{cases}\)

\(\frac{x}{3}=\frac{y}{2};\frac{z}{5}=\frac{y}{4}\)

\(\Leftrightarrow\)\(\frac{x}{6}=\frac{y}{4};\frac{z}{5}=\frac{y}{4}\)

\(\Rightarrow\)\(\frac{x}{6}=\frac{y}{4}=\frac{z}{5}\)

Áp dụng tc của dãy tỉ số bằng nhau ta có:

\(\frac{x}{6}=\frac{y}{4}=\frac{z}{5}=\frac{-x-y+z}{-6-4+5}=\frac{-10}{-5}=2\)

=>\(\begin{cases}x=12\\y=8\\z=10\end{cases}\)

\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2=\frac{1}{x+y+z}\)

=> 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

lai 1 thag nua dug chieu xin lik e = cah do nua a T_T

Có: \(\frac{y-2}{3}=\frac{2y-4}{6}\)

\(\frac{z-3}{4}=\frac{3z-9}{12}\)

Suy ra\(\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}=\frac{\left(x-1\right)-\left(2y-4\right)+\left(3z-9\right)}{2-6+12}\)

\(=\frac{\left(x-2y+3z\right)-6}{8}=\frac{14-6}{8}=1\)

Vậy có \(\frac{x-1}{2};\frac{y-2}{3};\frac{z-3}{4}=1\)Thay vào có x=3; y=5; z=7