Question

[ninh binh Fpt]

1. Tinh 

  \(\left(3-\sqrt{2}\right).\sqrt{7+4\sqrt{3}}\)

\(\left(\sqrt{3}+\sqrt{5}\right).\sqrt{7-2\sqrt{10}}\)

\(\left(2+\sqrt{5}\right).\sqrt{9-4\sqrt{5}}\)

\(\left(\sqrt{6}+\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)

\(\sqrt{2}.\left(\sqrt{8}-\sqrt{32}+3\sqrt{18}\right)\)

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

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

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

2. Giai pt

\(\sqrt{4x^2-4x+1}=5\)

\(\sqrt{4x-12}+\frac{1}{3}.\sqrt{9x-27}=4+\sqrt{x-3}\)

\(\sqrt{4x+8}-\sqrt{9x+18}-2\sqrt{x+2}=21\)

\(\left(3-2\sqrt{x}\right).\left(2+3\sqrt{x}\right)=16-6x\)

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

Answer 1

[Ngọc Châu]

bài 1:

a)\(\left(3-\sqrt{2}\right)\sqrt{7+4\sqrt{3}}\)

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

\(=\left(3-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)\(do2>\sqrt{3}\)

\(=6+3\sqrt{3}-2\sqrt{2}-\sqrt{6}\)

b) \(\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}\)

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

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

\(=\sqrt{15}-\sqrt{6}+5-\sqrt{10}\)

c)\(\left(2+\sqrt{5}\right)\sqrt{9-4\sqrt{5}}\)

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

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

\(=5-4\)

\(=1\left(hđt.3\right)\)

d)\(\left(\sqrt{6}+\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)

\(=\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{8-2\sqrt{15}}\)

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

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

\(=5-3\)

\(=2\)

e)\(\sqrt{2}\left(\sqrt{8}-\sqrt{32}+3\sqrt{18}\right)\)

\(=\sqrt{2}\left(2\sqrt{2}-4\sqrt{2}+9\sqrt{2}\right)\)

\(=2\left(2-4+9\right)\)

\(=2.7=14\)

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

\(=2-\sqrt{6-2\sqrt{5}}\)

\(=2-\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=2-\left(\sqrt{5}-1\right)\)

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

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

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

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

\(=\sqrt{3}-\sqrt{6}-2\)

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

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

\(=\left(2-\sqrt{\left(\sqrt{5}+1\right)^2}\right)+2\sqrt{5}\)

\(=2-\left(\sqrt{5}+1\right)+2\sqrt{5}\left(do\sqrt{5}>1\right)\)

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

\(=1-\sqrt{5}\)

bài 2)

a) \(\sqrt{4x^2-4x+1}=5\)

\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5\)

\(\Leftrightarrow2x-1=5\)hoặc \(\Leftrightarrow2x-1=-5\)

\(\Leftrightarrow x=3\)hoặc \(\Leftrightarrow x=-2\)

Vậy x = 3 hoặc x = -2