Question

Giải phương trình sau: 

1, \(\sqrt{5x+3}\) = \(\sqrt{3-\sqrt{2}}\)

2, \(\sqrt{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}\) = 2

3,\(\sqrt{-4x^2+25}=x\)

Answer 1

1. ĐKXĐ: $x\geq \frac{-3}{5}$

PT $\Leftrightarrow 5x+3=3-\sqrt{2}$

$\Leftrightarrow x=\frac{-\sqrt{2}}{5}$

Answer 2

2. ĐKXĐ: $x\geq \sqrt{7}$ 

PT $\Leftrightarrow (\sqrt{x}-7)(\sqrt{x}+7)=4$

$\Leftrightarrow x-49=4$

$\Leftrightarrow x=53$ (thỏa mãn)

 

Answer 3

Answer 4

3. ĐKXĐ: $\frac{5}{2}\geq x\geq \frac{-5}{2}$

PT \(\Rightarrow \left\{\begin{matrix} x\geq 0\\ -4x^2+25=x^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ x^2=5\end{matrix}\right.\)

\(\Rightarrow x=\sqrt{5}\) (thỏa mãn)

 

Answer 5

1)

ĐKXĐ: \(x\ge-\dfrac{3}{5}\)

Ta có: \(\sqrt{5x+3}=\sqrt{3-\sqrt{2}}\)

\(\Leftrightarrow5x+3=3-\sqrt{2}\)

\(\Leftrightarrow5x=-\sqrt{2}\)

hay \(x=\dfrac{-\sqrt{2}}{5}\)(nhận)

Vậy: \(S=\left\{-\dfrac{\sqrt{2}}{5}\right\}\)

Answer 6

2)

ĐKXĐ: \(x\ge0\)

Ta có: \(\sqrt{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}=2\)

\(\Leftrightarrow\sqrt{x-49}=2\)

\(\Leftrightarrow x-49=4\)

hay x=53(thỏa ĐK)

Answer 7

3) Ta có: \(\sqrt{-4x^2+25}=x\)

\(\Leftrightarrow-4x^2+25=x^2\)

\(\Leftrightarrow-4x^2-x^2=-25\)

\(\Leftrightarrow x^2=5\)

hay \(x\in\left\{\sqrt{5};-\sqrt{5}\right\}\)

Vậy: \(S=\left\{\sqrt{5};-\sqrt{5}\right\}\)