Toán Lớp 8: tìm x a. x^3-25x=0 b. 4x^2-12x+4=0 c. 4x^2-9-x(2x-3)=0 d. (4x-3)^2-3x(3-4x)=0 e cần gấp:(

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Toán Lớp 8:  tìm x a. x^3-25x=0 b. 4x^2-12x+4=0 c. 4x^2-9-x(2x-3)=0 d. (4x-3)^2-3x(3-4x)=0 e cần gấp:(

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Giải đáp: a)S={-5;0;5} b)S={(3- sqrt{5})/(2);(3+ sqrt{5})/(2)} c)S={-3;3/2} d)S={3/7;3/4} Lời giải và giải thích chi tiết: a) x^3-25x=0 <=>x.(x^2-25)=0 <=>x.(x^2-5^2)=0 <=>x(x-5).(x+5)=0 <=> ( left[ begin{array}{l}x=0 x-5=0 x+5=0 end{array} right. ) <=> ( left[ begin{array}{l}x=0 x=5 x=-5 end{array} right. ) Vậy S={-5;0;5} b) 4x^2-12x+4=0 <=>4x^2-12x+9-5=0 <=>(2x)^2-2.2x.3+3^2=5 <=>(2x-3)^2=5 <=>(2x-3)^2=(+- sqrt{5})^2 <=> ( left[ begin{array}{l}2x-3=- sqrt{5} 2x-3= sqrt{5} end{array} right. ) <=> ( left[ begin{array}{l}2x=3- sqrt{5} 2x=3+ sqrt{5} end{array} right. ) <=> ( left[ begin{array}{l}x= dfrac{3- sqrt{5}}{2} x= dfrac{3+ sqrt{5}}{2} end{array} right. ) Vậy S={(3- sqrt{5})/(2);(3+ sqrt{5})/(2)} c) 4x^2-9-x(2x-3)=0 <=>(2x)^2-3^2-x(2x-3)=0 <=>(2x-3).(2x+3)-x(2x-3)=0 <=>(2x+3-x).(2x-3)=0 <=>(x+3).(2x-3)=0 <=> ( left[ begin{array}{l}x+3=0 2x-3=0 end{array} right. ) <=> ( left[ begin{array}{l}x=-3 2x=3 end{array} right. ) <=> ( left[ begin{array}{l}x=-3 x= dfrac{3}{2} end{array} right. ) Vậy S={-3;3/2} d) (4x-3)^2-3x(3-4x)=0 <=>(4x-3).(4x-3)+3x(4x-3)=0 <=>(4x-3+3x).(4x-3)=0 <=>(7x-3).(4x-3)=0 <=> ( left[ begin{array}{l}7x-3=0 4x-3=0 end{array} right. ) <=> ( left[ begin{array}{l}7x=3 4x=3 end{array} right. ) <=> ( left[ begin{array}{l}x= dfrac{3}{7} x= dfrac{3}{4} end{array} right. ) Vậy S={3/7;3/4}