\(\left(\dfrac{1}{\sqrt{x}}-\sqrt{x}\right):\left(\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\) (ĐK: \(x>0\))

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

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

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

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

c: 

b; 

Sửa đề: \(\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)

Yêu cầu đề bài là gì bạn nên ghi đầy đủ để được hỗ trợ tốt hơn.

\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)

Khi x=căn 2 thì \(A=\dfrac{\sqrt{2}+16}{\sqrt{\sqrt{2}}+3}\)

a: \(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)

b: Để A=4 thì \(x+16-4\sqrt{x}-12=0\)

=>x=4

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

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

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

\(=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\)

b: \(B=\dfrac{x\sqrt{x}+26\sqrt{x}-19}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{2x+6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)

\(P=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)} \)P=\(\dfrac{-x+16\sqrt{x}+x\sqrt{x}-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
P=\(\dfrac{\left(16+x\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{16+x}{\sqrt{x}+3}\)

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

\(ĐK:x\ge0;x\ne1;x\ne-2\)

\(P=\dfrac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\dfrac{3x+3\sqrt{x}-9+x-\sqrt{x}+3\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)

b.\(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2}{\sqrt{x}+2}+\dfrac{\sqrt{x}+2}{\sqrt{x}+2}+\dfrac{\sqrt{x}+2}{\sqrt{x}+2}+\dfrac{2}{\sqrt{x}+2}\)

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

Để P lớn nhất thì \(\sqrt{x}+2\) nhỏ nhất

Mà \(\sqrt{x}+2\ge2\) \(\Rightarrow Min=2\)

\(\Rightarrow P\le1+1+1+\dfrac{2}{2}=1+1+1+1=4\)

Vậy \(P_{max}=4\) khi \(x=0\)

A=\(\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)

=\(\dfrac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

=\(\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

=\(\dfrac{x\sqrt{x}+16\sqrt{x}-x-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(x+16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{x+16}{\sqrt{x}+3}\)