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

b: \(P=\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

a: Khi x=64 thì \(P=\dfrac{8+1}{8+2}=\dfrac{9}{10}\)

\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)=\left(5+4\sqrt{2}\right)\left[3^2-\left(2\sqrt{1+\sqrt{2}}\right)^2\right]=\left(5+4\sqrt{2}\right)\left[9-4\left(1+\sqrt{2}\right)\right]=\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)=\left(5+4\sqrt{2}\right)\left(5-4\sqrt{2}\right)=5^2-\left(4\sqrt{2}\right)^2=25-16.2=25-32=-7\)

a: \(A=\dfrac{3+\sqrt{1-a^2}}{\sqrt{1+a}}\cdot\dfrac{3+\sqrt{1-a^2}}{\sqrt{1-a^2}}=\sqrt{1-a}\)

b: \(a=\sqrt{3}\left(2-\sqrt{3}\right)=2\sqrt{3}-3\)

Khi \(a=2\sqrt{3}-3\) thì \(A=\sqrt{1-2\sqrt{3}+3}=\sqrt{3}-1\)

a: \(M=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)

b: Để M=9/2 thì \(\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}=\dfrac{9}{2}\)

=>2x-5 căn x+2=0

=>(căn x-2)(2 căn x-1)=0

=>x=1/4 hoặc x=4

c: \(M-4=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)

=>M>4

3 căn 56 nha

\(3\sqrt{56}-2\sqrt{98}-5\sqrt{18}-\sqrt{63}+2\sqrt{28}=3\sqrt{4.14}-2\sqrt{49.2}-5\sqrt{9.2}-\sqrt{9.7}+2\sqrt{4.7}=3.2\sqrt{14}-2.7\sqrt{2}-5.3\sqrt{2}-3\sqrt{7}+2.2\sqrt{7}=6\sqrt{14}-14\sqrt{2}-15\sqrt{2}-3\sqrt{7}+4\sqrt{7}=6\sqrt{14}-29\sqrt{2}+\sqrt{7}\)

Giải:

\(\sqrt{42-10\sqrt{17}}+\sqrt{33-8\sqrt{17}}\)

\(=\sqrt{\left(5-\sqrt{17}\right)^2}+\sqrt{\left(4-\sqrt{17}\right)^2}\)

\(=\left|5-\sqrt{17}\right|+\left|4-\sqrt{17}\right|\)

\(=5-\sqrt{17}+\sqrt{17}-4\)

\(=1\)

Vậy ...

\(\sqrt{42-10\sqrt{17}}+\sqrt{33-8\sqrt{17}}=\sqrt{25-2.5.\sqrt{17}+17}+\sqrt{16-2.4.\sqrt{17}+17}=\sqrt{\left(5-\sqrt{17}\right)^2}+\sqrt{\left(4-\sqrt{17}\right)^2}=\left|5-\sqrt{17}\right|+\left|4-\sqrt{17}\right|=5-\sqrt{17}+\sqrt{17}-4=1\)

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

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

mà 1+căn 3<2+căn 6

nên 1-căn 3>2-căn 6

b: \(\left(2-\sqrt{2}\right)^2=6-4\sqrt{2}\)

(1/2)^2=1/4

mà 6-4căn 2-1/4>0

nên 2-căn 2>1/2

a) ta có \(2-\sqrt{6}=\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)>1-\sqrt{3}\)

b) ta có : \(2-\sqrt{2}=\dfrac{1}{2}\left(4-2\sqrt{2}\right)\)

mà ta có : \(2\sqrt{2}< 3\) (vì \(8< 9\))

\(\Rightarrow4-2\sqrt{2}>4-3>1\) \(\Rightarrow2-\sqrt{2}>\dfrac{1}{2}\)

c) ta có : \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=4008+2\sqrt{2003.2005}\)

\(\left(2\sqrt{2004}\right)^2=8016=4008+4008=4008+2\sqrt{2004.2004}\)

mà ta có : \(x^2\ge x^2-1\Rightarrow x^2>\left(x-1\right)\left(x+1\right)\)

\(\Rightarrow4008+2\sqrt{2004.2004}>4008+2\sqrt{2003.2005}\)

\(\Rightarrow2\sqrt{2004}>\sqrt{2003}+\sqrt{2005}\)

Mysterious Person giúp mk

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

Mysterious Person giups mk nha

a) điều kiện xác định : \(x>0;x\ne1\)

ta có : \(A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right)\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)

\(\Leftrightarrow A=\left(\dfrac{x}{2\sqrt{x}}-\dfrac{1}{2\sqrt{x}}\right)\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\right)\)

\(\Leftrightarrow A=\left(\dfrac{x-1}{2\sqrt{x}}\right)\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

\(\Leftrightarrow A=\left(\dfrac{x-1}{2\sqrt{x}}\right)\left(\dfrac{-4x}{x-1}\right)=-2\sqrt{x}\)

b) để \(A>-6\Leftrightarrow-2\sqrt{x}>-6\Leftrightarrow\sqrt{x}< 3\Leftrightarrow0< x< 9\) và \(x\ne1\)

vậy ....

Mysterious Person giúp mk

Đk: x >0 ; x khác 1

sau khi rút gọn ra -2

b, 9>x>0