Bài 6: Biến đối đơn giản biểu thức chứa căn bậc hai

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

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

\(=12\cdot\sqrt{\dfrac{5}{6}}-12\cdot\sqrt{\dfrac{6}{5}}\)

\(=2\sqrt{30}-\dfrac{12}{5}\sqrt{30}=-\dfrac{2}{5}\sqrt{30}\)

\(A=\dfrac{x-1-9}{\sqrt{x-1}-3}=\dfrac{\left(\sqrt{x-1}-3\right)\left(\sqrt{x-1}+3\right)}{\sqrt{x-1}-3}\)

\(=\sqrt{x-1}+3>=3\)

Dấu '=' xảy ra khi x=1

a: \(=-xy\cdot\dfrac{\sqrt{y}}{\sqrt{x}}=-\sqrt{x}\cdot y\sqrt{y}\)

b: \(=\sqrt{\dfrac{-105x^3}{35\cdot35}}=\dfrac{\sqrt{-105x^3}}{35}\)

c: \(=\sqrt{5a}\cdot\dfrac{a}{7\sqrt{b}}\)

d: \(=-7\sqrt{3}\sqrt{xy}\)

`a, = 12 sqrt 3 - 2 sqrt 5 - 6 sqrt 3 + 5 sqrt 5 - 6 sqrt 3`

`= 3 sqrt 5`

`b, = 12 sqrt 2 + 2 sqrt 2 - 10 sqrt 2 + 4 sqrt 2`

`= 8 sqrt 2`

`c, = 10 sqrt 3 - 4 sqrt 3 + 4sqrt 3 - 10 sqrt 5`

`= 10(sqrt 3 - sqrt 5)`.

\(\left(\dfrac{\sqrt{216}}{3}-\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}\right).\dfrac{1}{\sqrt{6}}.\\ =\left(\dfrac{6\sqrt{6}}{3}-\dfrac{\left(2\sqrt{3}-\sqrt{6}\right)\left(\sqrt{8}+2\right)}{\left(\sqrt{8}-2\right)\left(\sqrt{8}+2\right)}\right).\dfrac{1}{\sqrt{6}}.\\ =\left(2\sqrt{6}-\dfrac{4\sqrt{6}+4\sqrt{3}-4\sqrt{3}-2\sqrt{6}}{4}\right).\dfrac{1}{\sqrt{6}}.\\ =\left(2\sqrt{6}-\dfrac{2\sqrt{6}}{4}\right).\dfrac{1}{\sqrt{6}}.\\ =\left(2\sqrt{6}-\dfrac{\sqrt{6}}{2}\right).\dfrac{1}{\sqrt{6}}.\\ =2\sqrt{6}.\dfrac{1}{\sqrt{6}}-\dfrac{\sqrt{6}}{2}.\dfrac{1}{\sqrt{6}}.\\ =2-\dfrac{1}{2}=\dfrac{3}{2}.\)

\(=\left(\sqrt{5}-1-5\right)\left(\sqrt{5}+6\right)\)

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

=5-121=-116

Lời giải:
a.

$\sqrt{75}+\sqrt{48}-\sqrt{300}=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}=(5+4-10)\sqrt{3}=-\sqrt{3}$

b.

$\sqrt{98}-\sqrt{72}+0,5\sqrt{8}=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=(7-6+1)\sqrt{2}=2\sqrt{2}$

c.

$\sqrt{99}-\sqrt{11}+2\sqrt{44}=3\sqrt{11}-\sqrt{11}+4\sqrt{11}$

$=(3-1+4)\sqrt{11}=6\sqrt{11}$

\(a\sqrt{75}+\sqrt{48}-\sqrt{300}\)

=\(5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\)

=\(\left(4+5-10\right)\sqrt{3}\)

=-\(\sqrt{3}\)

b\(\sqrt{95}-\sqrt{72}+0,5\sqrt{8}\)

=\(\sqrt{95}-6\sqrt{2}+\sqrt{2}\)

=\(\sqrt{95}-\left(6+1\right)\sqrt{2}\)

=\(\sqrt{95}-7\sqrt{2}\)

c\(\sqrt{99}-\sqrt{11}+2\sqrt{44}\)

=\(3\sqrt{11}-\sqrt{11}+4\sqrt{11}\)

=\(\left(3-1+4\right)\sqrt{11}\)

=\(6\sqrt{11}\)