tìm x bt a/2x²(x-4)+3=2x²-5 b/5x(x-3)=(x-2)(5x-1)-5 c/(x-5)(x-4)-(x+1)(x-2)=7 d/(2x-1)(x-2)-(x+3)(2x-7)=3 e/(4x+1)(x-3)-(x-1)(4x-1)=15 f/5(x-3)(x-7)-(

0 bình luận về “tìm x bt a/2x²(x-4)+3=2x²-5 b/5x(x-3)=(x-2)(5x-1)-5 c/(x-5)(x-4)-(x+1)(x-2)=7 d/(2x-1)(x-2)-(x+3)(2x-7)=3 e/(4x+1)(x-3)-(x-1)(4x-1)=15 f/5(x-3)(x-7)-(”

1. Đáp án: 

    @nguyencuong18082006

   Giải thích các bước giải:

   a) $2x^{2}(x-4)+3=2x^{2}-5_{}$ 

   ⇔ $2x^{3}-8x^{2}+3=2x^{2}-5_{}$ 

   ⇔ $2x^{3}-10x^{2}+8=0_{}$ 

   ⇔ $2x^{3}-2x^{2}-8x^{2}+8=0_{}$ 

   ⇔ $2x^{2}(x-1)-8(x-1)=0_{}$ 

   ⇔ $(x-1)(2x^{2}-8)=0_{}$ 

   ⇔ \(\left[ \begin{array}{l}x-1=0\\2x^{2}-8=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=1\\x=±2\end{array} \right.\) 

    b) $5x.(x-3)=(x-2)(5x-1)-5_{}$  

   ⇔ $5x^{2}-15x=5x^{2}-x-10x+2-5_{}$ 

   ⇔ $-15x=-x-10x+2-5_{}$ 

   ⇔ $-15x=-11x-3_{}$ 

   ⇔ $-15x+11x=-3_{}$ 

   ⇔ $-4x=-3_{}$ 

   ⇔ $x=\frac{3}{4}_{}$

   c) $x^{2}-4x-5x+20-(x^{2}-2x+x-2)=7_{}$ 

   ⇔ $x^{2}-4x-5x+20-(x^{2}-x-2)=7_{}$ 

   ⇔ $x^{2}-4x-5x+20-x^{2}+x+2=7_{}$ 

   ⇔ $-8x+22=7_{}$ 

   ⇔ $-8x=-15_{}$ 

   ⇔ $x=_{}$ $\frac{15}{8}$ 

   d) $(2x-1)(x-2)-(x+3)(2x-7)=3_{}$ 

   ⇔ $2x^{2}-3x-x+2-(2x^{2}-7x+6x-21)=3_{}$ 

   ⇔ $2x^{2}-4x-x+2-2x^{2}+x+21=3_{}$ 

   ⇔ $-4x+23=3_{}$ 

   ⇔ $-4x=-20_{}$ 

   ⇔ $x=5_{}$ 

   e) $(4x+1)(x-3)-(x-1)(4x-1)=15_{}$ 

   ⇔ $4x^{2}-12x+x-3-(4x^{2}-x-4x+1)=15_{}$ 

   ⇔ $4x^{2}-12x+x-3-4x^{2}+5x-1=15_{}$ 

   ⇔ $-6x-4=15_{}$ 

   ⇔ $-6x=19_{}$ 

   ⇔ $x=\frac{16}{9}_{}$ 

   f) $5(x-3)(x-7)-(5x+1)(x-2)=8_{}$ 

   ⇔ $(2x-15)(x-7)-(5x^{2}-10x+x-2)=8_{}$ 

   ⇔ $5x^{2}-35x-15x+105-(5x^{2}-9x-2)=8_{}$ 

   ⇔ $5x^{2}-35x-15x+105-5x^{2}+9x+2=8_{}$ 

   ⇔ $-41x+107=8_{}$ 

   ⇔ $-41x=-99_{}$ 

   ⇔ $x=_{}$ $\frac{99}{41}$ 

   g) $3(x-7)(x+7)-(x+1)(3x+2)=13_{}$ 

   ⇔ $3(x^{2}-49)-(3x^{2}+2x+3x+2)=13_{}$ 

   ⇔ $3x^{2}-147-3x^{2}-5x-2=13_{}$ 

   ⇔ $-149-5x=13_{}$ 

   ⇔ $-5x=162_{}$ 

   ⇔ $x=_{}$ $\frac{162}{5}$ 

   #Chúc bạn học tốt !

   #Mong được vote 5* và hay nhất ạ !

   Bình luận

2. a,2x²(x-4)+3=2x²-5

   ⇔2x³-8x²+3=2x²-5

   ⇔2x³-10x²+8=0

   ⇔2x³-2x²-8x²+8=0

   ⇔2x²(x-1)-8(x-1)=0

   ⇔(x-1)(2x²-8)=0

   ⇔2(x-1)(x²-4)=0

   ⇔\(\left[ \begin{array}{l}x-1=0\\x^2-4=0\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x=1\\x=±2\end{array} \right.\) 

   Vậy x=1,x=±2

   b,

    

   Bình luận