Giải phương trình m) (x^2 – 16)^2 – (x – 4)^2 = 0 n) (5x^2 – 2x + 10)^2 = (3x^2 + 10x – 8)^2
2 câu trả lời
Đáp án:
m) (x^2-16)^2-(x-4)²=0
⇔ (x²-16-x+4).(x²-16+x-4)=0
⇔ [(x-4).(x+4)-1.(x-4)].[(x-4).(x+4)+1.(x-4)]=0
⇔ (x-4).(x+4-1).(x-4).(x+4+1)=0
⇔ (x-4)².(x+3).(x+5)=0
⇔ \left[ \begin{array}{l}(x-4)^2=0\\x+3=0\\x+5=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=4\\x=-3\\x=-5\end{array} \right.
\text{Vậy S={4;-3;-5}}
n) (5x^2-2x+10)^2= (3x^2+10x-8)^2
⇔ (5x²-2x+10)^2- (3x^2+10x-8)^2=0
⇔ (5x^2-2x+10-3x²-10x+8).(5x²-2x+10+3x²+10x-8)=0
⇔ (2x²-12x+18).(8x²+8x+2)=0
⇔ [2.(x²-6x+9)].[2.(4x²+4x+1)]=0
⇔ 4.(x-3)^2.(2x+1)^2=0
⇔ \left[ \begin{array}{l}(x-3)^2=0\\(2x+1)^2=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=3\\x=-0,5\end{array} \right.
\text{Vậy S={3;-0,5}}
Đáp án:
Giải thích các bước giải:
m, (x^2-16)^2-(x-4)^2=0
[(x-4)(x+4)]^2-(x-4)^2=0
[(x-4)(x+4)-(x-4)][(x-4)(x+4)+(x-4]=0
[(x-4)(x+4-1)][(x-4)(x+4+1)]=0
(x-4)^2(x+3)(x+5)=0
vậy nghiệm S thuộc{4,-3,-5}
n,(5x^2 – 2x + 10)^2 = (3x^2 + 10x – 8)^2
(5x^2 – 2x + 10)^2-(3x^2 + 10x – 8)^2=0
(5x^2 – 2x + 10-3x^2 - 10x + 8)(5x^2 – 2x + 10+3x^2 + 10x – 8)
(2x^2-12x+18)(8x^2+8x+2)=0
(2x(x-3)-2(x-3))(4x(2x+1)+2(2x+1))=0
2(x-1)(x-3)2(2x+1)^2=0
vậy tập nghiệm s thuộc{1,3,-1/2}