Question

[Nguyễn Hoàng Linh]

Giai phương trình :

a) \(\sqrt{x+3}=1+\sqrt{2}\)

b) \(\sqrt{10+\sqrt{5x}}=\sqrt{6}+2\)

c) \(\sqrt{x^2-16}-\sqrt{x-4}=0\)

d) \(x-6\sqrt{x}+5=0\)

e) \(\sqrt{x-3}>hoặc=7\)

g) \(\sqrt{x+1}< hoặc=3\)

Answer 1

[Akai Haruma]

a) ĐKXĐ: \(x\geq -3\)

Ta có: \(\sqrt{x+3}=1+\sqrt{2}\)

\(\Rightarrow x+3=(1+\sqrt{2})^2\)

\(\Leftrightarrow x+3=1+2+2\sqrt{2}=3+2\sqrt{2}\)

\(\Leftrightarrow x=2\sqrt{2}\) (thỏa mãn)

Vậy \(x=2\sqrt{2}\)

b) ĐK: \(x\geq 0\)

Có: \(\sqrt{10+\sqrt{5x}}=\sqrt{6}+2\)

\(\Rightarrow 10+\sqrt{5x}=(\sqrt{6}+2)^2=6+4+4\sqrt{6}\)

\(\Leftrightarrow \sqrt{5x}=4\sqrt{6}=\sqrt{96}\)

\(\Leftrightarrow x=\frac{96}{5}\) (thỏa mãn)

Vậy.....

c) ĐK: \(x\geq 4\)

Ta có: \(\sqrt{x^2-16}-\sqrt{x-4}=0\)

\(\Leftrightarrow \sqrt{(x-4)(x+4)}-\sqrt{x-4}=0\)

\(\Leftrightarrow \sqrt{x-4}(\sqrt{x+4}-1)=0\)

\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-4}=0\\ \sqrt{x+4}=1\end{matrix}\right. \Leftrightarrow \left[\begin{matrix} x=4\\ x=-3\end{matrix}\right.\) (loại $x=-3$ vì $x\geq 4$)

Vậy \(x=4\)

Answer 2

[Akai Haruma]

d) ĐK: \(x\ge 0\)

Ta có: \(x-6\sqrt{x}+5=0\)

\(\Leftrightarrow (x-\sqrt{x})-5(\sqrt{x}-1)=0\)

\(\Leftrightarrow \sqrt{x}(\sqrt{x}-1)-5(\sqrt{x}-1)=0\)

\(\Leftrightarrow (\sqrt{x}-5)(\sqrt{x}-1)=0\)

\(\Leftrightarrow \left[\begin{matrix} \sqrt{x}-5=0\\ \sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=25\\ x=1\end{matrix}\right.\) (đều t/m)

e) ĐK: \(x\geq 3\)

\(\sqrt{x-3}\geq 7\)

\(\Leftrightarrow x-3\geq 49\)

\(\Leftrightarrow x\geq 52\). Kết hợp với ĐK suy ra \(x\geq 52\)

f) ĐK: \(x\geq -1\)

Ta có: \(\sqrt{x+1}\leq 3\)

\(\Leftrightarrow x+1\leq 9\)

\(\Leftrightarrow x\leq 8\)

Kết hợp với ĐK suy ra \(-1\leq x\leq 8\)