a: \(\Leftrightarrow\sqrt{6}\left(x+1\right)=5\sqrt{6}\)

=>x+1=5

=>x=4

b: =>x^2/10=1,1

=>x^2=11

=>x=căn 11 hoặc x=-căn 11

c: =>(4x+3)/(x+1)=9 và (4x+3)/(x+1)>=0

=>4x+3=9x+9

=>-5x=6

=>x=-6/5

d: =>(2x-3)/(x-1)=4 và x-1>0 và 2x-3>=0

=>2x-3=4x-4 và x>=3/2

=->-2x=-1 và x>=3/2

=>x=1/2 và x>=3/2

=>Ko có x thỏa mãn

e: Đặt căn x=a(a>=0)

PT sẽ là a^2-a-5=0

=>\(\left[{}\begin{matrix}a=\dfrac{1+\sqrt{21}}{2}\left(nhận\right)\\a=\dfrac{1-\sqrt{21}}{2}\left(loại\right)\end{matrix}\right.\)

=>x=(1+căn 21)^2/4=(11+căn 21)/2

\(P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+7}{4-x}\left(x>0;x\ne4\right)\\ P=\dfrac{\left(3-\sqrt{x}\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\\ P=\dfrac{\sqrt{x}+6-x-x-3\sqrt{x}-2+2\sqrt{x}+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\sqrt{x}}\\ P=\dfrac{-2x+11}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\sqrt{x}}\\ P=\dfrac{-2x\sqrt{x}+11\sqrt{x}+\left(\sqrt{x}+2\right)\left(x-4\right)}{\sqrt{x}\left(x-4\right)}\)

\(P=\dfrac{-2x\sqrt{x}+11\sqrt{x}+x\sqrt{x}-4\sqrt{x}+2x-8}{\sqrt{x}\left(x-4\right)}\\ P=\dfrac{-x\sqrt{x}+8\sqrt{x}+2x-8}{\sqrt{x}\left(x-4\right)}\)

a: ĐKXĐ: x>0; x<>1

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

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

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

b: Q>2

=>Q-2>0

=>\(\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)

=>căn x-1>0

=>x>1

a) ĐK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

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

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

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

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

b) Q>2 <=> \(\dfrac{x}{\sqrt{x}-1}>2\Leftrightarrow x>2\sqrt{x}-2\)

\(\Leftrightarrow x-2\sqrt{x}+2>0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+1>0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2\ge0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1\le0\\\sqrt{x}-1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le1\end{matrix}\right.\)

KL:.....

Yêu cầu đề?

m mem đề đâu 

g, \(y=\sqrt{-x^2+4x-5}=\sqrt{-\left(x-2\right)^2-1}\)

\(\Rightarrow\) Hàm số này không xác định với mọi x.

h, \(y=\sqrt{x^2+2x+2}=\sqrt{\left(x+1\right)^2+1}>0\forall x\)

\(\Rightarrow\) Hàm số này xác định với mọi x.

i, \(y=\sqrt{\left(x-1\right)\left(x-3\right)}\) xác định khi:

\(\left(x-1\right)\left(x-3\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3\ge0\\x-1\le0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le1\end{matrix}\right.\)

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

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

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

b: khi x=6-2căn 5 thì \(P=\dfrac{6-2\sqrt{5}-3\sqrt{5}+3-5}{\left(\sqrt{5}-3\right)\left(\sqrt{5}-4\right)\cdot\sqrt{5}}\)

\(=\dfrac{-5\sqrt{5}+4}{\sqrt{5}\left(\sqrt{5}-3\right)\left(\sqrt{5}-4\right)}\)

ĐKXĐ: \(x\ge0;x\ne1\)

\(\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{3}{\sqrt{x}+3}\)

\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{15\sqrt{x}-11+3x+7\sqrt{x}-6-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{3x+19\sqrt{x}-14}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}+7\right)\left(3\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(D=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{3\sqrt{x}+1}{x-1}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\left(x\ge0;x\ne1\right)\\ D=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\\ D=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\cdot\dfrac{1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

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

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

các biểu thức trong căn pt hết về HĐT rồi phá ra là done