a) \(=2\sqrt{5}-3\sqrt{5}+\sqrt{5}-1=-1\)

b) \(=\left[\sqrt{14}+\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{5}\right)}{\sqrt{2}+\sqrt{5}}\right].\sqrt{\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{3}{2}}\right)^2}\)

\(=\left(\sqrt{14}+\sqrt{6}\right)\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{3}{2}}\right)\)

\(=\sqrt{49}-\sqrt{21}+\sqrt{21}-\sqrt{9}\)

\(=7-3=4\)

a) \(\sqrt{20}-15\sqrt{\dfrac{1}{5}}+\sqrt{\left(1-\sqrt{5}\right)^2}=2\sqrt{5}-3\sqrt{5}+\sqrt{5}-1=-1\)

b) \(\left(\dfrac{14}{\sqrt{14}}+\dfrac{\sqrt{12}+\sqrt{30}}{\sqrt{2}+\sqrt{5}}\right).\sqrt{5-\sqrt{21}}=\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{10-2\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)=4\)

Bài 1:

a) Ta có: \(5\sqrt{12}-\sqrt{45}-3\sqrt{48}+\sqrt{75}\)

\(=5\cdot2\cdot\sqrt{3}-\sqrt{3}\cdot\sqrt{15}-3\cdot\sqrt{3}\cdot4+5\sqrt{3}\)

\(=10\sqrt{3}-3\sqrt{5}-12\sqrt{3}+5\sqrt{3}\)

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

b) Ta có: \(\left(1+\frac{5-\sqrt{5}}{1-\sqrt{5}}\right)\left(\frac{5+\sqrt{5}}{1+\sqrt{5}}+1\right)\)

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

\(=\frac{6-2\sqrt{5}}{1-\sqrt{5}}\cdot\frac{6+2\sqrt{5}}{1+\sqrt{5}}\)

\(=\frac{6^2-\left(2\sqrt{5}\right)^2}{1^2-\left(\sqrt{5}\right)^2}=\frac{36-20}{1-5}=\frac{16}{-4}=-4\)

2)

a) Ta có: \(P=x-\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}}{\sqrt{x}+2}\right)\cdot\frac{x-4}{\sqrt{4x}}\)

\(=x-\left(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\)

\(=x-\frac{x+2\sqrt{x}+x-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\)

\(=x-\frac{2x}{2\sqrt{x}}\)

\(=x-\sqrt{x}\)

b) Ta có: \(x=7-\sqrt{48}\)

\(=\frac{14-2\sqrt{48}}{2}=\frac{8-2\cdot2\sqrt{2}\cdot\sqrt{6}+6}{2}\)

\(=\frac{\left(2\sqrt{2}-\sqrt{6}\right)^2}{2}=\frac{\left[\sqrt{2}\cdot\left(2-\sqrt{3}\right)\right]^2}{2}\)

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

Thay \(x=\left(2-\sqrt{3}\right)^2\) vào biểu thức \(P=x-\sqrt{x}\), ta được:

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

\(=7-4\sqrt{3}-\left|2-\sqrt{3}\right|\)

\(=7-4\sqrt{3}-\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))

\(=7-4\sqrt{3}-2+\sqrt{3}\)

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

c) Ta có: \(P=x-\sqrt{x}\)

\(=x-2\cdot\sqrt{x}\cdot\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\)

\(=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\)

Ta có: \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall x\)

Dấu '=' xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\)

\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\)

hay \(x=\frac{1}{4}\)(nhận)

Vậy: Giá trị nhỏ nhất của biểu thức \(P=x-\sqrt{x}\) là \(-\frac{1}{4}\) khi \(x=\frac{1}{4}\)

Bài 20:

a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)

b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)

\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)

c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)

\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)

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

=2

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

\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)

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

\(=8+4\sqrt{3}-4\sqrt{3}-6\)

=2

`a)A=(3-sqrt5)sqrt{3+sqrt5}+(3+sqrt5)sqrt{3-sqrt5}`

`=sqrt{3-sqrt5}sqrt{3+sqrt5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`

`=sqrt{9-5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`

`=2(sqrt{3+sqrt5}+sqrt{3-sqrt5})`

`=sqrt2(sqrt{6+2sqrt5}+sqrt{6-2sqrt5})`

`=sqrt2(sqrt{(sqrt5+1)^2}+sqrt{(sqrt5+1)^2})`

`=sqrt2(sqrt5+1+sqrt5-1)`

`=sqrt{2}.2sqrt5`

`=2sqrt{10}`

`b)B=(5+sqrt{21})(sqrt{14}-sqrt6)sqrt{5-sqrt{21}}`

`=sqrt{5+sqrt{21}}sqrt{5-sqrt{21}}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`

`=sqrt{25-21}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`

`=2sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`

`=2sqrt2sqrt{5+sqrt{21}}(sqrt{7}-sqrt3)`

`=2sqrt{10+2sqrt{21}}(sqrt{7}-sqrt3)`

`=2sqrt{(sqrt3+sqrt7)^2}(sqrt{7}-sqrt3)`

`=2(sqrt3+sqrt7)(sqrt{7}-sqrt3)`

`=2(7-3)`

`=8`

`c)C=sqrt{4+sqrt7}-sqrt{4-sqrt7}`

`=sqrt{(8+2sqrt7)/2}-sqrt{(8-2sqrt7)/2}`

`=sqrt{(sqrt7+1)^2/2}-sqrt{(sqrt7+1)^2/2}`

`=(sqrt7+1)/sqrt2-(sqrt7-1)/2`

`=2/sqrt2=sqrt2`

`d)D=\sqrt{2+sqrt3}+sqrt{14-5sqrt3}+sqrt2`

`=>sqrt2D=sqrt{4+2sqrt3}+sqrt{28-10sqrt3}+2`

`=>sqrt2D=sqrt{(sqrt3+1)^2}+sqrt{(5-sqrt3)^2}+2`

`=>sqrt2D=8`

`=>D=4sqrt2`

a) Ta có: \(\frac{6}{\sqrt{2}-\sqrt{3}+3}\)

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

\(=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{5-2\sqrt{6}-9}\)

\(=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{-4-2\sqrt{6}}\)

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

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

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

b) Ta có: \(\left(\frac{4}{\sqrt{5}+1}-\frac{4}{\sqrt{5}-1}\right):\sqrt{3+2\sqrt{2}}\)

\(=\left(\frac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}-\frac{4\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\right):\sqrt{2+2\cdot\sqrt{2}\cdot1+1}\)

\(=\left(\frac{4\left(\sqrt{5}-1\right)}{4}-\frac{4\left(\sqrt{5}+1\right)}{4}\right):\sqrt{\left(\sqrt{2}+1\right)^2}\)

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

\(=-\frac{2}{\sqrt{2}+1}\)(Vì \(\sqrt{2}+1>0\))

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

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

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

b,\(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)  \(=\sqrt{8\sqrt{3}}-2\sqrt{50\sqrt{3}}+4\sqrt{8\sqrt{3}}\)

\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}\)

\(=0\)

d,\(A=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)

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

\(\sqrt2A=\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\)

\(\sqrt2A=\sqrt{(\sqrt5-1)^2}\) \(+\sqrt{(\sqrt5+1)^2}\)    \(=\sqrt5-1 +\sqrt5+1=2\sqrt5\)

\(\Rightarrow A=\dfrac{2\sqrt5}{\sqrt2}\) \(=\sqrt{10}\)

a. \(\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}=\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\)

\(=\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\) 

\(=\frac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{3\sqrt{5}-3+5-\sqrt{5}}{2\left(\sqrt{5}+1\right)}\)  

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

\(\sqrt{242}.\sqrt{26}.\sqrt{130}.\sqrt{0,9}-\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)\)

\(=\sqrt{121}.\sqrt{2}.\sqrt{2}.\sqrt{13}.\sqrt{13}.\sqrt{10}.\sqrt{0,9}-\left(2-1\right)\)

\(=11.2.13.\sqrt{9}-1=286.3-1=857\)

\(\frac{3-\sqrt{6}}{\sqrt{12}-\sqrt{8}}-\frac{\sqrt{15}-\sqrt{5}}{2\sqrt{12}-4}+\frac{\sqrt{17-4\sqrt{15}}}{4}\)

\(=\frac{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\left(\sqrt{3}-\sqrt{2}\right)}-\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{4\left(\sqrt{3}-1\right)}+\frac{\sqrt{\left(2\sqrt{3}-\sqrt{5}\right)^2}}{4}\)

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

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

giúp nốt mình câu e đi bạn !

Lời giải:

$A=\sqrt{7}-(4\sqrt{3}+2\sqrt{3})+\sqrt{4}$

$=\sqrt{7}-6\sqrt{3}+2$

$B=\frac{7-\sqrt{5}}{\sqrt{5}-1}.\frac{-3+\sqrt{5}}{\sqrt{5}+1}$

$=\frac{(7-\sqrt{5})(-3+\sqrt{5})}{(\sqrt{5}-1)(\sqrt{5}+1)}$

$=\frac{-26+10\sqrt{5}}{4}=\frac{-13+5\sqrt{5}}{2}$

$C=(\sqrt{3}+1).\frac{\sqrt{14}-6\sqrt{3}}{\sqrt{5}+1}$

$=\frac{-18-6\sqrt{3}+\sqrt{14}+\sqrt{42}}{\sqrt{5}+1}$

Với biểu thức của bạn thì càng cố rút gọn thì càng dài. Bạn cần gấp nhưng đề bài lại viết sai?