\(\sqrt{x^2+a^2}=t>=IaI\); t^2=x^2+a^2
\(t-\frac{5a}{t}=x\) TH1. (x>=0 (*)
\(t^2-10a+\frac{25a^2}{t^2}=x^2=t^2-a^2\)
\(25a^2\cdot\left(\frac{1}{t}\right)^2+a^2-10a=0\)
\(t^2=\frac{25a^2}{10a-a^2}=0\)
\(x^2=\frac{25a}{\left(10-a\right)}-a^2\)
sau do Bien luan theo dk ton tai nghiem
x>=0; t>=IaI
TH2. x<0 (*) doi dau lai
Dai qua moi roi
nhan cheo len van con can
noi de lam the nao? kho day khong tin thu tim ra x xem
cach khac: co ve nhe nhang hon giai bien luan luon
\(x^2+a^2=x.\sqrt{x^2+a^2}+5a\)a
x=0 =>a^2=5a => \(\orbr{\begin{cases}a=0\\a=5\end{cases}}\)
x<>0 => \(\orbr{\begin{cases}a\ne0\\a\ne5\end{cases}}\)
chia 2 ve cho x^2 (dat 1/x^2=y^2)
\(1+a^2y^2=\sqrt{1+a^2y^2}+5ay^2\)
(a^2-5a)y^2=\(\sqrt{1+a^2y^2}-1\)
lien hop[..]
(a^2-5a)y^2[...]=a^2y^2
chia y^2 <>0
(a^2-5a).[..]=a^2
\(\sqrt{1+a^2y^2}+1=\frac{a^2}{a^2-5a}=\frac{a}{a-5}\)
\(\sqrt{1+a^2y^2}=\frac{a}{a-5}-1=\frac{5}{a-5}\)
a<=5 phuong trinh vo nghiem
a>5
1+a^2y^2=\(\frac{25}{\left(a-5\right)^2}\)
\(y^2=\frac{25-\left(a-5\right)^2}{\left(a-5\right)^2.a^2}=\frac{10-a}{\left(a-5\right)^2.a}\)
a>10 => vo nghiem
5<a<=10
x^2=\(\frac{a\left(5-a\right)^2}{10-a}\)
x=+-\(\left(a-5\right)\sqrt{\frac{a}{10-a}}\)
ban tu kiem tra xem co khi bien doi sai dau do
