Đáp án:
Bài 4:
$a) (2x+5)(2x-7)-(2x-3)^2=36$
$⇔ 4x^2-14x+10x-35 - 4x^2+12x-9 =36$
$⇔4x^2-4x^2-14x+10x+12x-35-9-36=0$
$⇔8x-80=0$
$⇔8x=80$
$⇔x=80 : 8$
$⇔x=10$
Vậy $x=10$
$b) 6x.(x-1999)-x+1999=0$
$⇔6x(x-1999)-(x-1999)=0$
$⇔(x-1999)(6x-1)=0$
$⇔$\(\left[ \begin{array}{l}x-1999=0\\6x-1=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=1999\\x=-\dfrac{1}{6}\end{array} \right.\)
$\text{Vậy x ∈ { 1999; $-\dfrac{1}{6}$}}$
$c) x^2-9-4(x+3)=0$
$⇔x^2-9 -4x-12=0$
$⇔x^2-4x-21=0$
$⇔x^2-7x+3x-21=0$
$⇔x(x-7)+3(x-7)=0$
$⇔(x-7)(x+3)=0$
$⇔$\(\left[ \begin{array}{l}x-7=0\\x+3=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=7\\x=-3\end{array} \right.\)
$\text{Vậy x ∈ {7; -3}}$
Bài 6 :
$10,3 . 9,7 - 9,7 . 0,3+10,3^2 -10,3.0,3$
$=9,7.(10,3 -0,3)+106,09 -3,09$
$=9,7 . 10 +103$
$=97 +103$
$=200$
