tìm x 6x^2-(2x+5)(3x-2)=7 (3x-5)(7-5x)-(5x+2)(2-3x)=4

0 bình luận về “tìm x 6x^2-(2x+5)(3x-2)=7 (3x-5)(7-5x)-(5x+2)(2-3x)=4”

1. 6x^2-(2x+5)(3x-2)=7

   ⇔6x^2-(6x^2-4x+15x-10)=7

   ⇔6x^2-6x^2+4x-15x+10=7

   ⇔-9x+10=7

   ⇔-9x=7-10

   ⇔-11x=-3

   ⇔x=-3:(-9)

   ⇔x=$\frac{3}{11}$ .

   Vậy x=$\frac{3}{11}$ .

   (3x-5)(7-5x)-(5x+2)(2-3x)=4

   ⇔21x-15x^2-35+25x-(10x-15x^2+4-6x)=4

   ⇔21x-15x^2-35+25x-10x+15x^2-4+6x=4

   ⇔42x-39=4

   ⇔42x=4+39

   ⇔42x=43

   ⇔x=43:42

   ⇔x=$\frac{43}{42}$ .

   Vậy x=$\frac{43}{42}$ .

    

   Trả lời

2. Đáp án:

   a)x=$\frac{43}{42}$ 

   b)x=$\frac{3}{11}$ 

   Giải thích các bước giải:

    a)6x²-(2x+5)(3x-2)=7

   ⇒21x-15x²-35+25x-(10x-15x²+4-6x)=4

   ⇒21x-15x²-35+25x-10x+15x²-4+6x=4

   ⇒42x-39=4

   ⇒42x=4+39=43

   ⇒x=$\frac{43}{42}$ 

   b)(3x-5)(7-5x)-(5x+2)(2-3x)=4

   ⇒6x²-6x²+4x-15x+10=7

   ⇒-11x=-3

   ⇒x=$\frac{3}{11}$ 

   Trả lời