\(a,A=2\sqrt{20}-\dfrac{2}{\sqrt{3}+1}-\sqrt{80}+\sqrt{4+2\sqrt{3}}\\ =2.2\sqrt{5}-\dfrac{2\left(\sqrt{3}-1\right)}{\sqrt{3^2}-1}-4\sqrt{5}+\sqrt{\left(\sqrt{3}+1\right)^2}\\ =-\dfrac{2\left(\sqrt{3}-1\right)}{2}+\left|\sqrt{3}+1\right|\\ =-\sqrt{3}+1+\sqrt{3}+1\\ =2\)

\(B=\left(1+\dfrac{x+\sqrt{x}}{1+\sqrt{x}}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\left(dk:x\ge0,x\ne1\right)\\ =\left(1+\dfrac{\sqrt{x}\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\right)\left(1-\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\right)\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\\ =1-x\)

\(b,A=4\sqrt{B}\Leftrightarrow4\sqrt{1-x}=2\\ \Leftrightarrow\sqrt{1-x}=\dfrac{1}{2}\\ \Leftrightarrow\left|1-x\right|=\dfrac{1}{4}\)

\(\Leftrightarrow1-x=\dfrac{1}{4}\\ \Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)

Vậy \(x=\dfrac{3}{4}\) thì \(A=4\sqrt{B}\).

a) \(A=2\sqrt{20}-\dfrac{2}{\sqrt{3}+1}-\sqrt{80}+\sqrt{4+2\sqrt{3}}\)

\(A=2\cdot2\sqrt{5}-\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}-4\sqrt{5}+\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}\cdot1+1^2}\)

\(A=4\sqrt{5}-\dfrac{2\left(\sqrt{3}-1\right)}{2}-4\sqrt{5}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

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

\(A=-\sqrt{3}+1+\sqrt{3}+1\)

\(A=2\)

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

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

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

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

\(B=1-x\)

b) Ta có: \(A=4\sqrt{B}\)

\(\Rightarrow2=4\sqrt{1-x}\)

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

\(\Leftrightarrow1-x=\dfrac{1}{4}\)

\(\Leftrightarrow x=1-\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)

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

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

=\(\dfrac{2\sqrt{x}+1}{\sqrt{x}-1}\) 

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

Để A nguyên thì \(\sqrt{x}-1\)∈Ư₍₂₎

⇒x∈{4;0;9}

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

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

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

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

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

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

b: Để Q là số nguyên thì \(2x⋮x-1\)

=>\(2x-2+2⋮x-1\)

=>\(2⋮x-1\)

=>\(x-1\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{2;0;3;-1\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{0;2;3\right\}\)

Biểu thức gì vậy bạn?

tìm các giá trị nguyên của x để biểu thức P=A.B  nhận giá trị nguyên

a: Thay x=9 vào A, ta được:

\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=\dfrac{-5}{2}\)

\(B=\dfrac{3\sqrt{x}-15+20-2\sqrt{x}}{x-25}=\dfrac{\sqrt{x}+5}{x-25}=\dfrac{1}{\sqrt{x}-5}\)

b: Để \(A=B\cdot\left|x-4\right|\) thì \(\left|x-4\right|=\dfrac{A}{B}=\dfrac{\sqrt{x}+2}{\sqrt{x}-5}:\dfrac{1}{\sqrt{x}-5}=\sqrt{x}+2\)

\(\Leftrightarrow x-4=\sqrt{x}+2\)

\(\Leftrightarrow x-\sqrt{x}-6=0\)

=>x=9