Đề câu c co bị sai ko vậy bạn? (y - 2\(\sqrt{x}\) +1)

a: \(=\sqrt{3}+1-\sqrt{3}=1\)

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

c: Sửa đề:\(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)

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

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

= \(\dfrac{x+3\sqrt{x}+2}{\sqrt{x}+1}.\dfrac{-x+3\sqrt{x}-2}{\sqrt{x}-1}\)

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

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

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

= \(\dfrac{x+3\sqrt{x}+2}{\sqrt{x}+1}:\dfrac{-x+3\sqrt{x}-2}{\sqrt{x}-1}\)

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

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

\(a.A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)

\(\left(x\ge0;x\ne1\right)\)

\(b.A=\dfrac{1}{2}\Leftrightarrow\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}-\dfrac{1}{2}=0\)

\(\Leftrightarrow\dfrac{4-10\sqrt{x}-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}=0\)

\(\Leftrightarrow-11\sqrt{x}+1=0\)

\(\Leftrightarrow x=\dfrac{1}{121}\left(TM\right)\)

KL...........

Cảm ơn nhiều nha :)

ĐKXĐ: \(x\ge0,x\ne1\)

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

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

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

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

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

a. \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}-x-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\)

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

b. \(\:B=\dfrac{4\sqrt{3+2\sqrt{2}}}{3+2\sqrt{2}+4}=\dfrac{4+4\sqrt{2}}{7+2\sqrt{2}}=\dfrac{\left(4+4\sqrt{2}\right).\left(7-2\sqrt{2}\right)}{\left(7+2\sqrt{2}\right).\left(7-2\sqrt{2}\right)}=\dfrac{12+20\sqrt{2}}{41}\)

a: \(Q=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

b: Để Q=1/2 thì \(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{1}{2}\)

=>-10căn x+4=căn x+3

=>-11 căn x=-1

=>x=1/121

Bài 1: 

a: \(B=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)

b: Để B=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)

=>3 căn x=3

=>căn x=1

hay x=1(loại)