Toán Lớp 8: a,X^2-6x+8=0 b,x ³-9x ²+27x-27=0 c,3x ³-6x ²+9x=0

Question

Toán Lớp 8:  a,X^2-6x+8=0 b,x ³-9x ²+27x-27=0 c,3x ³-6x ²+9x=0

truongthanhhung · Answer

$ #phnn$ a)x^2-6x+8=0 <=>x^2-2x-4x+8=0 <=>x(x-2)-4(x-2)=0 <=>(x-4)(x-2)=0 <=> $ left[ begin{matrix} x-4=0 x-2=0 end{matrix} right.$ <=> $ left[ begin{matrix} x=4 x=2 end{matrix} right.$ text{Vậy S = {4;2}} b)x^3-9x^2+27x-27=0 <=>x^3-3.x^2.3+3.x.3^2-3^3=0 <=>(x-3)^3=0 =>x-3=0 <=>x=3 text{Vậy S={3}} c)3x^3-6x^2+9x=0 <=>3x(x^2-2x+3)=0 <=>3x(x^2-2.x.1+1^2+2)=0 <=>3x[(x+1)^2+2]=0 text{Vì} (x+1)^2+2>0∀x =>3x=0 <=>x=0 text{Vậy S={0}}

maingocquynhnhu2k1 · Answer

Giải đáp + Lời giải và giải thích chi tiết ! to Tìm x: a) x^2-6x+8 = 0 <=> (x^2-6x+9)-1 = 0 <=> (x-3)^2 = 1 <=> ( left[ begin{array}{l}x-3=1 x-3=-1 end{array} right. ) <=> ( left[ begin{array}{l}x=4 x=2 end{array} right. ) Vậy S= {4; 2} b) x^3-9x^2+27x-27 = 0 <=> x^3-3. x^2. 3+3. x. 3^2-3^3 = 0 <=> (x-3)^3 = 0 <=> x-3 = 0 <=> x = 3 Vậy S= {3} Áp dụng: (a-b)^3 = a^3-3a^2b+3a^2b-b^3 c) 3x^3-6x^2+9x = 0 <=> 3x(x^2-2x+3) = 0 <=> 3x[(x^2-2x+1)+2] = 0 <=> 3x[(x-1)^2+2] = 0 <=> ( left[ begin{array}{l}3x=0 (x-1)^2+2=0 end{array} right. ) <=> ( left[ begin{array}{l}x=0 (x-1)^2=-2 end{array} right. ) Vì (x-1)^2 >= 0 AA x Mà (x-1)^2 = -2 (vô lý) <=>x=0 Vậy S= {0}