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15 tháng 7 2019

\(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)

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

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

\(=\left|\sqrt{x-1}-1\right|+\sqrt{x-1}+1\)(*)

\(x\ge2\Rightarrow x-1\ge1\Rightarrow\sqrt{x-1}\ge1\Rightarrow\sqrt{x-1}-1\ge0\)

Khi đó (*)\(=\sqrt{x-1}-1+\sqrt{x-1}+1=2\sqrt{x-1}\)(đpcm)

4 tháng 7 2021

\(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)

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

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

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

\(=\sqrt{x-1}-1+\sqrt{x-1}+1\left(x\ge2\right)=2\sqrt{x-1}\)

a) \(\dfrac{1}{\sqrt{5}+\sqrt{7}}=\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)}=\dfrac{\sqrt{7}-\sqrt{5}}{2}\)

c) \(\dfrac{7}{\sqrt{5}-\sqrt{3}+\sqrt{5}}=\dfrac{7}{2\sqrt{5}-\sqrt{3}}=\dfrac{7\left(2\sqrt{5}+\sqrt{3}\right)}{\left(2\sqrt{5}+\sqrt{3}\right)\left(2\sqrt{5}-\sqrt{3}\right)}\)

\(=\dfrac{14\sqrt{5}+7\sqrt{3}}{17}\)

 

 

2 tháng 6 2020

ý a con phân số mk rút gọn ấy nhé tử và mẫu \(\sqrt{5}-1\)

2 tháng 6 2020

a)

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

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

1 tháng 5 2022

1, vt : \(\left(1-\dfrac{5+\sqrt{2}}{\sqrt{2}+1}\right).\sqrt{3+2\sqrt{2}}\)

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

=\(\dfrac{-4}{\sqrt{2}+1}.\sqrt{\left(\sqrt{2}+1\right)^2}\)

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

=-4

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

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

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

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

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

10 tháng 6 2019

a)\(\sqrt{x}+1>\sqrt{x+1}\) (x>0)

Có:\(\left(\sqrt{x}+1\right)^2=x+2\sqrt{x}+1\left(1\right)\) (x>0)

\(\sqrt{\left(x+1\right)^2}=x+1\) (2) (x>0)

từ (1) và (2) =>(đpcm)

b)\(\sqrt{x^2+1}>x\)

Có:\(\sqrt{\left(x^2+1\right)^2}=x^2+1\left(1\right)\)

x2=x2 (2)

Từ (1) và (2) =>(đpcm)

c)\(\frac{1}{2}+a+b\ge\sqrt{a}+\sqrt{b}\left(a,b\ge0\right)\)

Vì a,b >or= 0

=>\(a+b\ge\sqrt{a}+\sqrt{b}\)

\(\Rightarrow\frac{1}{2}+a+b\ge\sqrt{a}+\sqrt{b}\) (đáng lẽ 1/2+a+b> mới phải)

18 tháng 6 2019

Tải app giải toán và kết bạn trao đổi nào cả nhà: https://www.facebook.com/watch/?v=485078328966618

19 tháng 11 2021

Bài 1:

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

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

2) \(VT=\dfrac{\left(6a+1\right)\left(a+6\right)+\left(6a-1\right)\left(a-6\right)}{a\left(a-6\right)\left(a+6\right)}.\dfrac{\left(a-6\right)\left(a+6\right)}{a^2+1}\)

\(=\dfrac{12a^2+12}{a\left(a^2+1\right)}=\dfrac{12\left(a^2+1\right)}{a\left(a^2+1\right)}=\dfrac{12}{a}=VP\)

 

8 tháng 5 2022

a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)

b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

8 tháng 4 2021

a, \(\sqrt{\left(\sqrt{5}-4\right)^2}-\sqrt{5}+\sqrt{20}=4\)

\(VT=\sqrt{\left(4-\sqrt{5}\right)^2}-\sqrt{5}+\sqrt{20}=\left|4-\sqrt{5}\right|-\sqrt{5}+\sqrt{20}\)

\(=4-\sqrt{5}-\sqrt{5}+2\sqrt{5}=4\) hay \(VT=VP\)

Vậy ta có đpcm 

b, Với \(x>0,x\ne4\)

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

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

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

25 tháng 4 2021

1.

Giả sử điều trên là đúng ta có:

\( \left | \sqrt{5}-4 \right |-\sqrt{5}+\sqrt{20}=4\)

Ta có: \(4>\sqrt{5}\)

\(\Rightarrow 4-\sqrt{5}- \sqrt{5}+\sqrt{20}=4\)

\(\Leftrightarrow 4-\sqrt{20}+\sqrt{20}=4\)

\(\Rightarrow đpcm\)

2.