K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

8 tháng 8 2021

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

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

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

\(=\dfrac{4}{4-3}\)

\(=4\)

b) \(Q=\left(1+\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\right).\dfrac{1}{\sqrt{x}}vớix>0,x\ne4\)

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

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

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

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

1 tháng 2 2022

a, \(A=\dfrac{4\left(3-\sqrt{7}\right)}{2}+2\sqrt{7}=\dfrac{12}{2}=6\)

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

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

1 tháng 2 2022

nhờ bạn làm rõ vì sao \(\dfrac{\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{2-\sqrt{x}}{x-1}\) lại bằng \(\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

mình xin cảm ơn

7 tháng 12 2021

\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)

8 tháng 12 2021

\(P=-\dfrac{3}{5}\) sao suy ra đc \(\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\) thế

 

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 8 2021

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

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

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

Tick hộ nha

8 tháng 8 2021

ok

14 tháng 10 2021

\(a,A=4\sqrt{3}-5\sqrt{3}+2-\sqrt{3}=2-2\sqrt{3}\\ B=\dfrac{x+2\sqrt{x}+8+2\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-4}\\ b,B-\dfrac{1}{2}A=\dfrac{\sqrt{x}}{\sqrt{x}-4}-\dfrac{1}{2}\left(2-2\sqrt{3}\right)=0\\ \Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-4}=1+\sqrt{3}\\ \Leftrightarrow\sqrt{x}=\left(1+\sqrt{3}\right)\left(\sqrt{x}-4\right)\Leftrightarrow\sqrt{x}=\sqrt{x}-4\sqrt{3}+\sqrt{3x}-4\\ \Leftrightarrow\sqrt{3x}=4\sqrt{3}+4\\ \Leftrightarrow\sqrt{x}=\dfrac{4\sqrt{3}+4}{\sqrt{3}}\\ \Leftrightarrow\sqrt{x}=\dfrac{12+4\sqrt{3}}{3}\\ \Leftrightarrow x=\dfrac{192+96\sqrt{3}}{9}=\dfrac{64+32\sqrt{3}}{3}\)

14 tháng 10 2021

\(\dfrac{\sqrt{x}}{\sqrt{x}-4}=1-\sqrt{3}\)
Nhỉ???

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

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

5 tháng 2 2022

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

b, với x > 0 

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

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

15 tháng 4 2022

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

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

\(=\dfrac{4}{1}=4\)

Vậy \(A=4\)

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

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

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

Vậy \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}\) với \(x>0,x\ne1\)

a: \(=2+\sqrt{3}+2-\sqrt{3}=4\)

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

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

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

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

b: Để A<=3/căn x thì \(\dfrac{x-2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)^2}< =\dfrac{3}{\sqrt{x}}\)

=>\(\dfrac{x-2\sqrt{x}-1-3x+6\sqrt{x}-3}{\left(\sqrt{x}-1\right)^2}< =0\)

=>\(-2x+4\sqrt{x}-4< =0\)

=>\(x-2\sqrt{x}+2>=0\)(luôn đúng)