Question

[vũ tiền châu]

giải phương trình vô tỉ sau 

\(\sqrt{2-x^2}+\sqrt{2-\frac{1}{x^2}}=4-\left(x+\frac{1}{x}\right)\)

Answer 1

[Rio Va]

Đặt \(a=\sqrt{2-x^2};b=\sqrt{2-\frac{1}{x^2}};c=x+\frac{1}{x}\)

xet x<0 vt < 2 căn 2<3, vt >4=>loại=>x>0=>c>=2;

ta có a+b=4-c;

a^2+b^2=4-x^2-1/x^2=6-c^2;

\(=>\hept{\begin{cases}2a+2b=8-2c\left(2\right)\\a^2+b^2=6-c^2\left(1\right)\end{cases}}\)

trừ 1 cho 2=>a^2-2a+b^2-2b=-c^2-2-2c=>a^2-2b+1+b^2-2b+1=-c^2+2c-1+1

=>\(\left(a-1\right)^2+\left(b-1\right)^2=-\left(c-1\right)^2+1\)

\(< =>\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=1\)

ta lại có (a-1)^2>=0;(b-1)^2>=0;(c-1)^2>=(2-1)^2=1=>Vế trái>=1=Vế phải, dấu bằng xảy ra<=>

\(\hept{\begin{cases}a=1\\b=1\\c=2\end{cases}< =>x=1}\)

Answer 2

[Trần Hữu Ngọc Minh]

Bạn tham khảo nhé:Điều kiện bạn tự tìm nhé

pt\(\Leftrightarrow\sqrt{2-x^2}+x-2+\sqrt{2-\frac{1}{x^2}}+\frac{1}{x}-2=0\)

\(\Leftrightarrow\frac{2-x^2-\left(x-2\right)^2}{\sqrt{2-x^2}-x+2}+\frac{2-\frac{1}{x^2}-\left(\frac{1}{x}-2\right)^2}{\sqrt{2-\frac{1}{x^2}}-\frac{1}{x}+2}=0\)

\(\Leftrightarrow\frac{-2\left(x^2-2x+1\right)}{\sqrt{2-x^2}-x+2}+\frac{-2\left(\frac{1}{x^2}-\frac{2}{x}+1\right)}{\sqrt{2-\frac{1}{x^2}}-\frac{1}{x}+2}=0\)

\(\Leftrightarrow\frac{\left(x-1\right)^2}{\sqrt{2-x^2}-x+2}+\frac{\left(\frac{1}{x}-1\right)^2}{\sqrt{2-\frac{1}{x^2}}-\frac{1}{x}+2}=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(\frac{1}{\sqrt{2-x^2}-x+2}+\frac{\frac{1}{x^2}}{\sqrt{2-\frac{1}{x^2}}-\frac{1}{x}+2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\Leftrightarrow x=1\left(N\right)\\\frac{1}{\sqrt{2-x^2}-x+2}+\frac{1}{x\sqrt{2x^2-1}-x+2x^2}=0\left(1\right)\end{cases}}\)

\(\left(1\right)\Leftrightarrow x\sqrt{2x^2-1}-x+2x^2+\sqrt{2-x^2}-x+2=0\)

Nhân 2 vào ta có:

\(\Leftrightarrow2x\sqrt{2x^2-1}-4x+4x^2+4+2\sqrt{2-x^2}=0\)

\(\Leftrightarrow\left(x+\sqrt{2x^2-1}\right)^2+\left(\sqrt{2-x^2}+1\right)^2+2\left(x-1\right)^2=0\left(VN\right)\)

Vậy phương trình có 1 nghiệm duy nhất là \(x=1\)