Tìm x 7) (5x – 1) (5x + 1) = 25x^2 – 7x + 15 8) 25x (x – 1) – (5x – 1)^2 = 14 9) (x + 3) (4 – x) + (x + 1) (x – 1) = 10 10) (x – 2)^2 – (x + 1) (x +

0 bình luận về “Tìm x 7) (5x – 1) (5x + 1) = 25x^2 – 7x + 15 8) 25x (x – 1) – (5x – 1)^2 = 14 9) (x + 3) (4 – x) + (x + 1) (x – 1) = 10 10) (x – 2)^2 – (x + 1) (x +”

1. Đáp án:

    

   Giải thích các bước giải:

   $7.(5x-1)(5x+1)=25x^2-7x+15$

   $⇔25x^2-1=25x^2-7x+15$

   $⇔25x^2-25x^2+7x=15+1$

   $⇔7x=16$

   $⇔x=\dfrac{16}{7}$

   $8,25x(x-1)-(5x-1)^2=14$

   $⇔25x^2-25x-(25x^2-10x+1)=14$

   $⇔25x^2-25x-25x^2+10x+1=14$

   $⇔-(25-10x)=14-1$

   $⇔-15x=13$

   $⇔x=-\dfrac{13}{15}$

   $9,(x+3)(4-x)+(x+1)(x-1)=10$

   $⇔x(4-x)+3(4-x)+x^2-1=10$

   $⇔4x-x^2+12-3x+x^2=10+1$

   $⇔(4x-3x)-(x^2-x^2)=11-12$

   $⇔x=-11$

   $10.(x-2)^2-(x+1)(x+3)=-7$

   $⇔x^2-4x+4-(x^2+3x+x+3)=-7$

   $⇔x^2-4x-x^2-4x-3=-11$

   $⇔-8x=-8$

   $⇔x=1$

   $11,(x+1)^2-(x+2)(x-2)=0$

   $⇔x^2+2x+1-x^2+4=0$

   $⇔2x=0-1-4$

   $⇔2x=-5$

   $⇔x=-\dfrac{5}{2}$

   $12. (2x+3)^2-4(x-1)^2=16$

   $⇔4x^2+9x+12-4(x^2-2x+1)=16$

   $⇔4x^2+9x-4x^2+8x-4=16-12$

   $⇔(4x^2-4x^2)+(9x+8x)=4+4$

   $⇔17x=8$

   $⇔x=\dfrac{8}{17}$

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