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

\(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)
\(=5\sqrt{10}+2.5-\sqrt{\left(25.10\right)}\)
\(=5\sqrt{10}+10-5\sqrt{10}\)
\(=10\)

Giải:

\(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)

\(=5\sqrt{10}+10-\sqrt{250}\)

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

\(=10\)

Vậy ...

\(A=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\sqrt{\dfrac{9x^3}{64}}\)

\(A=4\left|\dfrac{5\sqrt{x}}{2}\right|-\dfrac{8}{3}\left|\dfrac{3\sqrt{x}}{2}\right|-\dfrac{4}{3x}\left|\dfrac{3x\sqrt{x}}{8}\right|\)

Vì \(x>0\) nên:

\(A=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}=\dfrac{11\sqrt{x}}{2}\)

a) \(A=\dfrac{2}{\sqrt{3}+1}+\dfrac{6}{\sqrt{3}-1}+1\)

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

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

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

\(=\dfrac{8\sqrt{3}+6}{2}\)

\(=\dfrac{2\left(4\sqrt{3}+3\right)}{2}\)

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

b: \(B=\dfrac{\sqrt{\dfrac{7+2\sqrt{6}}{2}\cdot2}\cdot\left(\sqrt{6}-1\right)}{2\sqrt{5}}\)

\(=\dfrac{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}{2\sqrt{5}}=\dfrac{5}{2\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)

a,=\(4\sqrt{3}+3\)

b,=\(\dfrac{\sqrt{5}}{2}\)

a: \(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)

\(=\sqrt{\sqrt{3}}\left(2\sqrt{80}-2\sqrt{5}-3\sqrt{20}\right)\)

\(=0\)

b: \(=\sqrt{\sqrt{3}}\left(2\sqrt{8}-2\sqrt{5}-6\sqrt{5}\right)\)

\(=\sqrt{\sqrt{3}}\left(4\sqrt{2}-8\sqrt{5}\right)\)

\(\sqrt{4-2\sqrt{3}}-\sqrt{27}=\sqrt{\left(\sqrt{3}-1\right)^2}-3\sqrt{3}=\sqrt{3}-1-3\sqrt{3}=-1-2\sqrt{3}\)

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

Giải:

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

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

\(=\left|3+\sqrt{5}\right|-2\sqrt{9-5}+\left|3-\sqrt{5}\right|\)

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

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

\(=2\)

Vậy ...

Đề là gì vậy bạn

\(\dfrac{5\sqrt{7}+7\sqrt{5}}{\sqrt{35}}=\dfrac{\sqrt{5}\left(\sqrt{5}.\sqrt{7}+7\right)}{\sqrt{35}}=\dfrac{\sqrt{7}\left(\sqrt{5}.\sqrt{7}+7\right)}{7}=\dfrac{7\sqrt{5}+7\sqrt{7}}{7}=\sqrt{5}+\sqrt{7}\)

\(a.P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)

Để : \(P\in Z\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\in Z\Leftrightarrow\left(\sqrt{x}+1\right)\in\left\{\pm1;\pm2\right\}\)

+) \(\sqrt{x}+1=1\Leftrightarrow x=0\left(TM\right)\)

+) \(\sqrt{x}+1=-1\Leftrightarrow vô-n^o\)

+) \(\sqrt{x}+1=2\Leftrightarrow x=1\left(KTM\right)\)

+) \(\sqrt{x}+1=-2\Leftrightarrow vô-n^o\)

KL.............

\(b.Q=\dfrac{\sqrt{a}+1}{\sqrt{a}+2}=\dfrac{\sqrt{a}+2-1}{\sqrt{a}+2}=1-\dfrac{1}{\sqrt{a}+2}\)

Để : \(Q\in Z\Leftrightarrow\dfrac{1}{\sqrt{a}+2}\in Z\Leftrightarrow\left(\sqrt{a}+2\right)\in\left\{\pm1\right\}\)

+) \(\sqrt{a}+2=1\Leftrightarrow vô-n^o\)

+) \(\sqrt{a}+2=-1\Leftrightarrow vô-n^o\)

KL............

\(c.A=\dfrac{\sqrt{a}-1}{\sqrt{a}-4}=\dfrac{\sqrt{a}-4+3}{\sqrt{a}-4}=1+\dfrac{3}{\sqrt{a}-4}\)

Để : \(A\in Z\Leftrightarrow\dfrac{3}{\sqrt{a}-4}\in Z\Leftrightarrow\left(\sqrt{a}-4\right)\in\left\{\pm1;\pm3\right\}\)

+) \(\sqrt{a}-4=1\Leftrightarrow a=25\left(TM\right)\)

+) \(\sqrt{a}-4=-1\Leftrightarrow a=9\left(TM\right)\)

+) \(\sqrt{a}-4=3\Leftrightarrow a=49\left(TM\right)\)

+) \(\sqrt{a}-4=-3\Leftrightarrow a=1\left(TM\right)\)

KL............

P/s : Mình thấy đề bài b sai nhé , mẫu phải là \(\sqrt{a}-2\) thì mới phù hợp ĐK đã cho .