[Toán 10] Giải các phương trình

L

leminhnghia1

Giải:

ĐK: xx \geq 1-1

2(x2+2)=5x3+12(x^2+2)=5\sqrt{x^3+1}

    2[(x2x+1)+(x+1)]=5(x+1)(x2x+1)\iff 2[(x^2-x+1)+(x+1)]=5\sqrt{(x+1)(x^2-x+1)}

Đặt x+1=a;x2x+1=b;(a,b\sqrt{x+1}=a; \sqrt{x^2-x+1}=b ;(a,b \geq 0)0). Thay vào ta có:

Pt     2(a2+b2)5ab=0\iff 2(a^2+b^2)-5ab=0

    (2ab)(a2b)=0\iff (2a-b)(a-2b)=0

    2a=b\iff 2a=b v a=2ba=2b

...............
 
L

leminhnghia1

Giải:

ĐK: xx \geq 11

PT     x2x1(x1)x+x2x=0\iff x-2\sqrt{x-1}-(x-1)\sqrt{x}+\sqrt{x^2-x}=0

    (x1)2x1+1x(x1)(x11)=0\iff (x-1)-2\sqrt{x-1}+1-\sqrt{x(x-1)}(\sqrt{x-1}-1)=0

    (x11)2x(x1)(x11)=0\iff (\sqrt{x-1}-1)^2-\sqrt{x(x-1)}(\sqrt{x-1}-1)=0

    (x11)(x11x(x1))=0\iff (\sqrt{x-1}-1)(\sqrt{x-1}-1-\sqrt{x(x-1)})=0

    x1=1\iff \sqrt{x-1}=1 v x11x(x1)=0\sqrt{x-1}-1-\sqrt{x(x-1)}=0

Với x11x(x1)=0\sqrt{x-1}-1-\sqrt{x(x-1)}=0 (vô nghiệm)
............
 
Top Bottom