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19 tháng 10 2021

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

 

19 tháng 10 2021

\(15,A=\dfrac{x-1-4\sqrt{x}+4+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ A=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\\ 16,B=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{2\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\\ B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\left(\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\\ B=x-\sqrt{x}-2\sqrt{x}-2+2\sqrt{x}+2=x-\sqrt{x}\)

NV
9 tháng 5 2021

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

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

\(=\dfrac{\sqrt{a}}{\sqrt{a}+1}\)

5 tháng 11 2023

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

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

1: \(A=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-5-\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x-4-\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\dfrac{x-\sqrt{x}-12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)

7 tháng 10 2021

a) \(\sqrt{0,64.a^2}\left(a>0\right)=0,8.\left|a\right|=0,8a\)

b) \(\sqrt{a^2\left(a-2\right)^2}\left(a>2\right)=\left|a\left(a-2\right)\right|=a\left(a-2\right)=a^2-2a\)

c) \(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}\left(a\ge0,a\ne1\right)=\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}=1+\sqrt{a}+a\)

22 tháng 7 2023

bn gửi lại cho rõ hơn đi. chơ ảnh này mờ quá

NV
2 tháng 3 2022

ĐKXĐ: \(x>2\)

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

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

- Với \(x\ge6\Rightarrow A=\dfrac{\sqrt{x-2}-2+\sqrt{x-2}+2}{\dfrac{x-2}{x}}=\dfrac{2x\sqrt{x-2}}{x-2}=\dfrac{2x}{\sqrt{x-2}}\)

- Với \(2< x< 6\Rightarrow A=\dfrac{2-\sqrt{x-2}+\sqrt{x-2}+2}{\dfrac{x-2}{x}}=\dfrac{4x}{x-2}\)

3 tháng 3 2022

em cảm ơn ạ

 

24 tháng 10 2021

d: \(\dfrac{-\left(\sqrt{3}-\sqrt{6}\right)}{1-\sqrt{2}}+\dfrac{6\sqrt{3}+3}{\sqrt{3}}-\dfrac{13}{4+\sqrt{3}}\)

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

\(=2+\sqrt{3}\)