Đáp án:
$x=8$
Giải thích các bước giải:
$\sqrt{5x^2+14x+9}-\sqrt{x^2-x-20}=5\sqrt{x+1}$
$ĐKXĐ:x \geq 5$
$pt↔\sqrt{5x^2+14x+9}=5\sqrt{x+1}+\sqrt{x^2-x-20}$
$↔5x^2+14x+9=25(x+1)+x^2-x-20+10\sqrt{(x+1(x^2-x-20)}$
$↔5x^2+14x+9=x^2+24x+5+10\sqrt{(x+1)(x^2-x-20)}$
$↔4x^2-10x+4=10\sqrt{(x+1)(x^2-x-20)}$
$↔2x^2-5x+2=5\sqrt{(x+1)(x^2-x-20)}$
$↔(2x^2-5x+2)^2=25(x+1)(x^2-x-20)$
$↔4x^4+25x^2+4-20x^3-20x+8x^2=25(x^3-x^2-20x+x^2-x-20)$
$↔4x^4-20x^3+33x^2-20x+4=25(x^3-21x-20)$
$↔25x^3-525x-500=4x^4-20x^3+33x^2-20x+4$
$↔4x^4-45x^3+33x^2+505x+504=0$
$↔4x^4-32x^3-13x^3+104x^2-71x^2+568x-63x+504=0$
$↔4x^3(x-8)-13x^2(x-8)+71x(x-8)-63(x-8)=0$
$↔(x-8)(4x^3-13x^2+71x-63)=0$
$TH1:$
$x-8=0$
$↔x=8(TM)$
$TH2:$
$4x^3-13x^2+71x-63=0$
$↔4x.x^2-13x^2+71x-63=0$
$+)x \geq 5$
$↔4x \geq 20$
$↔4x^3 \geq 20x^2$
$↔4x^3-12x^2 \geq 20x^2-8x^2>0$
$71x \geq 355$
$+)71x-63 \geq 292>0$
$↔4x^3-13x^2+71x-63>0$
$↔\text{phương trình này vô nghiệm}$
$\text{vậy phương trình có nghiệm duy nhất x=8}$
