Toán Lớp 7: Tìm `x,y,z` thuộc Q biết: `b, y.(y-z-x)= 25` `c, (x+2)^2 + ( y-3) ^4 + ( z-5) ^6 = 0` Giup mk với, dạng này mk mới học nên chưa hiểu lắ

Question

Toán Lớp 7:  Tìm x,y,z thuộc Q biết: b, y.(y-z-x)= 25 c, (x+2)^2 + ( y-3) ^4 + ( z-5) ^6 = 0 Giup mk với, dạng này mk mới học nên chưa hiểu lắm

thanhmaianh · Answer

$ a)x(x&minus;y+z)=&minus;11;y(y&minus;x&minus;z)=25;z(z+x&minus;y)=35 $   $ &rArr;x(x&minus;y+z)+y(y&minus;z&minus;x)+z(z+x&minus;y)=49 $   $ &rArr;x(x&minus;y+z)&minus;y(x&minus;y+z)+z(x&minus;y+z)=49 $   $ &rArr;(x&minus;y+z)(x&minus;y+z)=49=(x&minus;y+z)^2 $   $ &rArr;(x&minus;y+z)=&plusmn;7 $   $ (x+2)^2;(y&minus;3)^4+(z&minus;5)^6&ge;0;(x+2)^2;(y&minus;3)^4+(z&minus;5)^6=0 $   $ &rArr; $  ( left[  begin{array}{l}x+2=0;x=0-2=-2  y-3=0;y=0+3=3  z-5=0;z=0+5=5 end{array}  right. )

maianh67 · Answer

$ a)x(x&minus;y+z)=&minus;11;y(y&minus;x&minus;z)=25;z(z+x&minus;y)=35 $     $ &rArr;  x(x&minus;y+z)+y(y&minus;z&minus;x)+z(z+x&minus;y)=49 $     $ &rArr; x(x&minus;y+z)&minus;y(x&minus;y+z)+z(x&minus;y+z)=49 $     $ &rArr; (x&minus;y+z)(x&minus;y+z)=49 = (x&minus;y+z)^2 $      $ &rArr; ( x-y+z)=&plusmn;7 $     $ x(x-y+z) = -11 $      $ &rArr; $  ( left[  begin{array}{l}x.7=-11&rArr;x=-11:7=-11/7  x.(-7)=-11&rArr;x=-11/-7=11/7 end{array}  right. )           $ b) ( x+2)^2 ; ( y-3)^4 + ( z-5)^6 &ge; 0 ; ( x+2)^2 ; ( y-3)^4 + ( z-5)^6 = 0 $     $ &rArr; $  ( left[  begin{array}{l}x+2=0&rArr;x=0-2=-2   y-3=0&rArr; y=0+3=3  z-5=0&rArr;z=0+5=5 end{array}  right. )      $ &rArr; ( x;y;z)=(-2;3;5) $