Question

[Quách Thi Tuyen]

Giúp mình :>

Answer 1

[Quỳnh Anh]

Trả lời:

Bài 1:

a, \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}\)

\(=3\sqrt{2}-4\sqrt{3^2.2}+2\sqrt{4^2.2}\)

\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}\)

\(=-\sqrt{2}\)

b, \(\sqrt{\left(2+\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)

\(=\left|2+\sqrt{5}\right|+\sqrt{5-6\sqrt{5}+9}\)

\(=2+\sqrt{5}+\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{5}.3+3^2}\)

\(=2+\sqrt{5}+\sqrt{\left(\sqrt{5}-3\right)^2}\)

\(=2+\sqrt{5}+\left|\sqrt{5}-3\right|\)

\(=2+\sqrt{5}+\left(3-\sqrt{5}\right)\)

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

c, \(\frac{2}{\sqrt{3}-1}+\frac{2}{\sqrt{3}+1}\)

\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}+\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)

\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}\right)^2-1^2}+\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}\right)^2-1^2}\)

\(=\frac{2\left(\sqrt{3}+1\right)}{3-1}+\frac{2\left(\sqrt{3}-1\right)}{3-1}\)

\(=\frac{2\left(\sqrt{3}+1\right)}{2}+\frac{2\left(\sqrt{3}-1\right)}{2}\)

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

Bài 2:

a, \(\sqrt{2x+1}-3=1\left(ĐK:x\ge-\frac{1}{2}\right)\)

\(\Leftrightarrow\sqrt{2x+1}=4\)

\(\Leftrightarrow2x+1=16\)

\(\Leftrightarrow2x=15\)

\(\Leftrightarrow x=\frac{15}{2}\left(tm\right)\)

Vậy x = 15/2