Question

[thiyy]

a)A=\(\dfrac{1}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\) với a>\(\dfrac{1}{2}\)     

b)A=\(\dfrac{\sqrt{x-2\sqrt{x-1}}}{\sqrt{x-1}-1}\)+\(\dfrac{\sqrt{x+2\sqrt{x-1}}}{\sqrt{x-1+1}}\) với x>2 

c)\(\dfrac{a+b}{b^2}\)\(\sqrt{\dfrac{a^2b^4}{a^2+2ab+b^2}}\) với a+b>0; b≠0

d)A=\(\left(\sqrt{\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\) với a≥0; a≠1

e)A=\(\dfrac{x-1}{\sqrt{y}-1}\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)}{\left(x-1\right)^4}}\) với x≠1; y≠1; y>o

f)A=\(\sqrt{\dfrac{m}{1-2x+x^2}}\)\(\sqrt{\dfrac{4m-8mx+4mx^2}{81}}\) với m>0; x≠4

g)A=\(\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right)\)\(\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\) với x>0; x≠4

h)\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\)\(\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\) với a≥0; a≠1

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(A=\dfrac{1}{2a-1}\cdot\sqrt{5a^2}\cdot\left|2a-1\right|\)

\(=\dfrac{2a-1}{2a-1}\cdot a\sqrt{5}=a\sqrt{5}\)(do a>1/2)

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

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

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

c:

\(=\dfrac{a+b}{b^2}\cdot\dfrac{ab^2}{a+b}=a\)

d: Sửa đề: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)

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

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

e:

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

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

f:

\(A=\sqrt{\dfrac{m}{\left(1-x\right)^2}\cdot\dfrac{4m\left(1-2x+x^2\right)}{81}}\)

\(=\sqrt{\dfrac{m}{\left(x-1\right)^2}\cdot\dfrac{4m\left(x-1\right)^2}{81}}\)

\(=\sqrt{\dfrac{4m^2}{81}}=\dfrac{2m}{9}\)