Toán Lớp 8: tìm x,y biết : a)x^3-16x=0 b) 9x^2+6x+4y^2-8y+5=0

Question

Toán Lớp 8:  tìm x,y biết : a)x^3-16x=0             b) 9x^2+6x+4y^2-8y+5=0

thanhthuythu · Answer

a. $x^{3}-16x=0->x.(x^{2}-16)=0$   -> $x.(x^{2}-4^{2})=0->x.(x+4).(x-4)=0$   -> $x=0$ hoặc $x+4=0$ hoặc $x-4=0$   -> $x=0$ hoặc $x=-4$ hoặc $x=4$   b. $9x^{2}+6x+4y^{2}-8y+5=0$   $->(9x^{2}+6x+1)+(4y^{2}-8y+4)=0$   $->((3x)^{2}+2.3x+1^{2})+4.(y^{2}-2.y.1+1^{2})=0$   $->(3x+1)^{2}+4(y-1)^{2}=0$   -> $3x+1=0$ v&agrave; $y-1=0$   -> $x= frac{1}{3}$ v&agrave; $y=1$

behocgioitoan · Answer

Giải đáp:    a) x = 0 hoặc x = +- 4   b) x = -1/3 v&agrave; y = 1   Lời giải và giải thích chi tiết:    a) x^3 - 16x = 0   &hArr; x (x^2-16) = 0   &hArr; ( left[  begin{array}{l}x=0  x^2-16=0 end{array}  right. )    &hArr; ( left[  begin{array}{l}x=0  x^2=16 end{array}  right. )    &hArr; ( left[  begin{array}{l}x=0  x= pm4 end{array}  right. )    b) 9x^2 + 6x + 4y^2 - 8y + 5 = 0   &hArr; 9x^2 + 6x + 1 + 4y^2 - 8y + 4 = 0   &hArr; (9x^2+6x+1) + 4 (y^2-2y+1) = 0   &hArr; (3x+1)^2 + 4 (y-1)^2 = 0   V&igrave; a^2 &ge; 0 &forall; a   &rArr; (3x+1)^2 &ge; 0 &forall; x v&agrave; 4 (y-1)^2 &ge; 0 &forall; y   &rArr; ( left[  begin{array}{l}(3x+1)^2=0  4(y-1)^2=0 end{array}  right. )    &hArr; ( left[  begin{array}{l}3x+1=0  (y-1)^2=0 end{array}  right. )    &hArr; ( left[  begin{array}{l}3x=-1  y-1=0 end{array}  right. )    &hArr; ( left[  begin{array}{l}x=- dfrac{1}{3}  y=1 end{array}  right. )