Đề 1
Bài 1:
a, $(x+3)^2-(x-3)^2=6x+18$
$⇔(x+3-x+3)(x+3+x-3)=6x+18$
$⇔6.2x=6x+18$
$⇔12x=6x+18$
$⇔6x=18$
$⇔x=3$
Vậy...
b, $\dfrac{x+3}{x-2}=\dfrac{5}{x-2}{3-x}$ ĐKXĐ: $x\neq2;3$
$⇒(x+3)(3-x)=5$
$⇔9-x^2=5$
$⇔x^2=4$
$⇔x=±2$ (giá trị $x=2$ không thỏa mãn điều kiện)
Vậy $x=-2$.
c, $\dfrac{12x^2+30x-21}{16x^2-9}-\dfrac{3x-7}{3-4x}=\dfrac{6x+5}{4x+3}$ ĐKXĐ: $x\neq±\dfrac{3}{4}$
$⇒12x^2+30x-21+(3x-7)(4x+3)=(6x+5)(4x-3)$
$⇔12x^2+30x-21+12x^2+9x-28x-21=24x^2-18x+20x-15$
$⇔12x^2+30x-21+12x^2+9x-28x-21-24x^2+18x-20x+15=0$
$⇔9x-27=0$
$⇔x=3$ (tmđk)
Vậy...
d, $\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-x-2}$ ĐKXĐ: $x\neq-1;2$
$⇒4(x-2)-2(x+1)=x+3$
$⇔4x-8-2x-2-x-3=0$
$⇔x=13$ (tmđk)
Vậy...
