\(\sqrt{25-x^2}-\sqrt{10-x^2}=3\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(ĐK:x\in R\)
2) \(ĐK:x< 0\)
3) \(ĐK:x\in\varnothing\)
4) \(=\sqrt{\left(x+1\right)^2+2}\)
\(ĐK:x\in R\)
5) \(=\sqrt{-\left(a-4\right)^2}\)
\(ĐK:x\in\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(x+3\right)^2}=12\)
=>\(\left|x+3\right|=12\)
=>\(\left[{}\begin{matrix}x+3=12\\x+3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-15\end{matrix}\right.\)
b: ĐKXĐ: x>=1
\(\sqrt{25x-25}-\sqrt{9x-9}=10\)
=>\(5\sqrt{x-1}-3\sqrt{x-1}=10\)
=>\(2\sqrt{x-1}=10\)
=>x-1=25
=>x=26(nhận)
![](https://rs.olm.vn/images/avt/0.png?1311)
c1 \(\sqrt{25-x^2}=3+\sqrt{10-x^2}\) (dk \(-5\le x\le5\) )
bp 2 ve \(25-x^2=9+6\sqrt{10-x^2}+10-x^2\)
\(\Leftrightarrow6\sqrt{10-x^2}=6\)
\(\Leftrightarrow\sqrt{10-x^2}=1\Leftrightarrow10-1=x^2\)
\(\Leftrightarrow x=+-3\)
c2 \(\left(\sqrt{25-x^2}-4\right)-\left(\sqrt{10-x^2}-1\right)=0\)
\(\Leftrightarrow\frac{\left(25-x^2-16\right)}{\left(\sqrt{25-x^2}+4\right)\left(\sqrt{25-x^2}-4\right)}-\frac{10-x^2-1}{\left(\sqrt{10-x^2}-1\right)\left(\sqrt{10-x^2}+1\right)}=0\)
\(\Leftrightarrow\left(9-x^2\right)\left(....\right)=0\)
\(\Leftrightarrow x=+-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a - b =3
a2 - b2 = 15
=> a+b = 5
=> a =4 ; b = 1
=> 25 - x2 = 16 => x = + -3 thỏa mãn 10 -x2 =1
ĐKXĐ: \(-\sqrt{10}\le x\le\sqrt{10}\)
Ta có: \(\sqrt{25-x^2}-\sqrt{10-x^2}=3\)
\(\Leftrightarrow\sqrt{25-x^2}=3+\sqrt{10-x^2}\)
\(\Leftrightarrow\sqrt{\left(25-x^2\right)^2}=\left(3+\sqrt{10-x^2}\right)^2\)
\(\Leftrightarrow25-x^2=9+6\cdot\sqrt{10-x^2}+10-x^2\)
\(\Leftrightarrow25-x^2=6\sqrt{10-x^2}-x^2+19\)
\(\Leftrightarrow25-x^2-6\sqrt{10-x^2}+x^2-19=0\)
\(\Leftrightarrow-6\sqrt{10-x^2}+6=0\)
\(\Leftrightarrow-6\sqrt{10-x^2}=-6\)
\(\Leftrightarrow10-x^2=1\)
\(\Leftrightarrow x^2=9\)
hay \(\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-3\left(nhận\right)\end{matrix}\right.\)
Vậy: S={3;-3}