Toán 9 Hệ phương trình

Ann Lee

Cựu Mod Toán
Thành viên
14 Tháng tám 2017
1,782
2,981
459
Hưng Yên
{1x+1y+z=121y+1z+x=131z+1x+y=14{x+y+zx(y+z)=12x+y+zy(z+x)=13x+y+zz(x+y)=14{2(x+y+z)xyz=x(y+z)xyz=1y+1z3(x+y+z)xyz=y(z+x)xyz=1z+1x4(x+y+z)xyz=z(x+y)xyz=1x+1y(1)\left\{\begin{matrix} \frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}\\\frac{1}{y}+\frac{1}{z+x}=\frac{1}{3} \\\frac{1}{z}+\frac{1}{x+y}=\frac{1}{4} \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{x+y+z}{x(y+z)}=\frac{1}{2}\\\frac{x+y+z}{y(z+x)}=\frac{1}{3} \\\frac{x+y+z}{z(x+y)} =\frac{1}{4} \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{2(x+y+z)}{xyz}=\frac{x(y+z)}{xyz}=\frac{1}{y}+\frac{1}{z}\\\frac{3(x+y+z)}{xyz}=\frac{y(z+x)}{xyz}=\frac{1}{z}+\frac{1}{x} \\\frac{4(x+y+z)}{xyz}=\frac{z(x+y)}{xyz}=\frac{1}{x}+\frac{1}{y} \end{matrix}\right.(1)
Đặt x+y+zxyz=1a()\frac{x+y+z}{xyz}=\frac{1}{a}(*)
Thay vào (1)(1) ta được {2a=1y+1z3a=1z+1x4a=1x+1y{1x=52a1y=32a1z=a2{x=2a5y=2a3z=2a()\left\{\begin{matrix} \frac{2}{a}=\frac{1}{y}+\frac{1}{z}\\ \frac{3}{a}=\frac{1}{z}+\frac{1}{x} \\\frac{4}{a}=\frac{1}{x}+\frac{1}{y} \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{1}{x}=\frac{5}{2a}\\\frac{1}{y}= \frac{3}{2a} \\\frac{1}{z}=\frac{a}{2} \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{2a}{5}\\y=\frac{2a}{3} \\z=\frac{2}{a} \end{matrix}\right.(**)
Thay trở lại ()(*), tìm được aa rồi tìm được x,y,zx,y,z
 
Top Bottom