Question

[Nguyễn Minh Bảo Anh]

Bài 1 :Tìm x để căn thức có nghĩa

a) \(\sqrt{-2\text{x}+3}\)

b)\(\sqrt{\frac{-5}{x^2+6}}\)

c)\(\sqrt{\frac{4}{x+3}}\)

Bài 2 : Rút gọn

a)\(\sqrt{\left(4+\sqrt{2}\right)^2}\)

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

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

Bài 3

a) \(\sqrt{9-4\sqrt{5}}\) - \(\sqrt{5}\) = -2

b) \(\sqrt{23+8\sqrt{7}}\) - \(\sqrt{7}\) = 4

c) \(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)

Bài 4 : Rút gọn

a) \(\frac{x^2-5}{x+\sqrt{5}}\) với x khác \(\sqrt{5}\)

b) \(\frac{x^2+2\sqrt{2}+2}{x^2-2}\)

c) x - 4 +\(\sqrt{16-8\text{x}+x^2}\) với x >4

Answer 1

[Nguyễn Minh Bảo Anh]

Bài 1 :

a)\(\sqrt{-2\text{x}+3}\) <=> -2x+3 \(\ge\)0 <=> -2x \(\ge\) -3 <=> x\(\le\) \(\frac{3}{2}\)

b)\(\sqrt{\frac{4}{x+3}}< =>x+3>0< =>x>-3\)

Bài 2 :

a)\(\sqrt{\left(4+\sqrt{2}\right)^2}=\left|4+\sqrt{2}\right|=4+\sqrt{2}\)

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

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

Bài 3 :

a) \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)

VT = \(\sqrt{5-2.2.\sqrt{5}+2^2}-\sqrt{5}\)

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

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

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

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

= -2 = VP

b)\(\sqrt{23+8\sqrt{7}}-\sqrt{7}=4\)

VT = \(\sqrt{7+2.4.\sqrt{7}+4^2}-\sqrt{7}\)

= \(\sqrt{\left(\sqrt{7}+4\right)^2}-\sqrt{7}\)

= |\(\sqrt{7}+4\)| -\(\sqrt{7}\)

=\(\sqrt{7}+4-\sqrt{7}\)

= 4 =VP

c) \(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)

VT = \(16-8\sqrt{7}+7\)

= 23 - \(8\sqrt{7}\) = VP

Bài 4:

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

Tương tự

Bài 5 :

a) \(\sqrt{x^2+6\text{x}+9}=3\text{x}-1\)

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

=> x+3 = 3x-1

+) x+3 =3x-1 => x= 2

+)x+3=-3x-1 => x= \(\frac{-1}{2}\) ( không tmđk)

b)+c) Tương tự