Toán Lớp 8: 25-4x^2=0 tìm x biết 9x^2-1=0 X^3-9x=0 X^2+x+1/4=0

Question

Toán Lớp 8:  25-4x^2=0 tìm x biết 9x^2-1=0 X^3-9x=0 X^2+x+1/4=0

behocgioitoan · Answer

Giải đáp: S={5/2;-5/2} S={1/3;-1/3} S={0;3;-3} S={-1/2} Lời giải và giải thích chi tiết: 25-4x^2=0 <=>5^2-(2x)^2=0 <=>(5-2x)(5+2x)=0 <=>[(5-2x=0),(5+2x=0):} <=> ( left[ begin{array}{l}x= dfrac{5}{2} x=- dfrac{5}{2} end{array} right. ) Vậy S={5/2;-5/2} 9x^2-1=0 <=>(3x)^2-1^2=0 <=>(3x-1)(3x+1)=0 <=>[(3x-1=0),(3x+1=0):} <=> ( left[ begin{array}{l}x= dfrac{1}{3} x=- dfrac{1}{3} end{array} right. ) Vậy S={1/3;-1/3} x^3-9x=0 <=>x(x^2-9)=0 <=>x(x-3)(x+3)=0 <=>[(x=0),(x-3=0),(x+3=0):} <=>[(x=0),(x=3),(x=-3):} Vậy S={0;3;-3} x^2+x+1/4=0 <=>x^2+2*x*1/2+(1/2)^2=0 <=>(x+1/2)^2=0 <=>x+1/2=0<=>x=-1/2 Vậy S={-1/2}.

dongoctuan · Answer

a) 25-4x^2=0   &hArr;(5-2x)(5+2x)=0   &hArr; ( left[  begin{array}{l}5-2x=0  5+2x=0 end{array}  right. )    &hArr; ( left[  begin{array}{l}x= dfrac{5}{2}  x=- dfrac{5}{2} end{array}  right. )    Vậy S={5/2,-5/2}   b) 9x^2-1=0   &hArr;(3x-1)(3x+1)=0   &hArr; ( left[  begin{array}{l}3x-1=0  3x+1=0 end{array}  right. )    &hArr; ( left[  begin{array}{l}x= dfrac{1}{3}  x=- dfrac{1}{3} end{array}  right. )    Vậy S={1/3,-1/3}   c) x^3-9x=0   &hArr;x(x^2-9)=0   &hArr;x(x-3)(x+3)=0   &hArr; ( left[  begin{array}{l}x=0  x-3=0  x+3=0 end{array}  right. )    &hArr; ( left[  begin{array}{l}x=0  x=3  x=-3 end{array}  right. )    Vậy S={0,3,-3}   d) x^2+x+1/4=0   &hArr;(x+1/2)^2=0   &hArr;x+1/2=0   &hArr;x=-1/2   Vậy S={-1/2}