tìm x biết a, x^2-x=20 b,4x^2+19x+9x=0 c, (4x-3)*(4x+3)-5 x(3x-1)=-7

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1. Đáp án:

    

   Giải thích các bước giải:

   a/ $x^2-x=20$
   => $x^2-x-20=0$
   => $x^2+4x-5x-20=0$
   => $(x+4)(x-5)=0$
   => \(\left[ \begin{array}{l}x+4=0\\x-5=0\end{array} \right.\)
   => \(\left[ \begin{array}{l}x=-4\\x=5\end{array} \right.\)

   b/ $4x^2+19x+9x=0$
   => $4x^2+28x=0$
   => $4x(x+7)=0$
   => \(\left[ \begin{array}{l}x=0\\x+7=0\end{array} \right.\)
   => \(\left[ \begin{array}{l}x=0\\x=-7\end{array} \right.\)

   c/ $(4x-3)(4x+3)-5x(3x-1)=-7$
   => $16x^2-9-15x^2+5x+7=0$
   => $x^2+5x-2=0$
   => $4x^2+20x-8=0$
   => $4x^2+20x+25-33=0$
   => $(2x+5)^2-33=0$
   => $(2x+5+\sqrt{33})(2x+5-\sqrt{33})=0$
   => \(\left[ \begin{array}{l}2x+5+\sqrt{33}=0\\2x+5-\sqrt{33}=0\end{array} \right.\)

   => \(\left[ \begin{array}{l}x=\frac{-5-\sqrt{33}}{2}\\x=\frac{-5+\sqrt{33}}{2}\end{array} \right.\)
   Chúc bạn học tốt !!!

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2. a) `x^2-x = 20`

   `<=> x^2 – 5x + 4x – 20 =0`

   `<=> x(x-5) + 4(x-5) =0`

   `<=> (x+4)(x-5)=0`

   `<=>` \(\left[ \begin{array}{l}x=-4\\x=5\end{array} \right.\) 

   b) `4x^2+19x+9x=0

   `<=> 4x^2+28x =0`

   `<=> x^2+7x=0`

   `<=> x(x+7) =0`

   `<=>` \(\left[ \begin{array}{l}x=0\\x=-7\end{array} \right.\) 

   c) `(4x-3)(4x+3) – 5x(3x-1) = -7`

   `<=> 16x^2 – 9 – 15x^2 + 5x + 7=0`

   `<=> x^2 + 5x – 2 =0`

   Có: `\Delta = 5^2 – 4.1.(-2) = 33 > 0`

   `=>` PT có 2 nghiệm phân biệt: `x_(1,2) = (-5\pm\sqrt33)/2`

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