a) Ta có: \(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)

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

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

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

Ta có: \(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=\frac52\)

=>\(\frac{x}{\sqrt{xy}}+\frac{y}{\sqrt{xy}}=\frac52\)

=>\(2\left(x+y\right)=5\sqrt{xy}\)

=>\(2x-5\sqrt{xy}+2y=0\)

=>\(2x-4\sqrt{xy}-\sqrt{xy}+2y=0\)

=>\(2\sqrt{x}\left(\sqrt{x}-2\sqrt{y}\right)-\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)=0\)

=>\(\left(\sqrt{x}-2\sqrt{y}\right)\left(2\sqrt{x}-\sqrt{y}\right)=0\)

TH1: \(\sqrt{x}-2\sqrt{y}=0\)

=>\(\sqrt{x}=2\sqrt{y}\)

=>x=4y

\(A=\frac{2x+3\cdot\sqrt{xy}}{2x-3\sqrt{xy}}\)

\(=\frac{\sqrt{x}\left(2\sqrt{x}+3\sqrt{y}\right)}{\sqrt{x}\left(2\sqrt{x}-3\sqrt{y}\right)}=\frac{2\sqrt{x}+3\sqrt{y}}{2\sqrt{x}-3\sqrt{y}}\)

\(=\frac{2\cdot\sqrt{4y}+3\sqrt{y}}{2\cdot\sqrt{4y}-3\sqrt{y}}=\frac{4\sqrt{y}+3\sqrt{y}}{4\sqrt{y}-3\sqrt{y}}=\frac{4+3}{4-3}=7\)

TH2: \(2\sqrt{x}-\sqrt{y}=0\)

=>\(\sqrt{y}=2\sqrt{x}\)

=>y=4x

\(A=\frac{2x+3\cdot\sqrt{xy}}{2x-3\sqrt{xy}}\)

\(=\frac{\sqrt{x}\left(2\sqrt{x}+3\sqrt{y}\right)}{\sqrt{x}\left(2\sqrt{x}-3\sqrt{y}\right)}=\frac{2\sqrt{x}+3\sqrt{y}}{2\sqrt{x}-3\sqrt{y}}\)

\(=\frac{2\sqrt{x}+3\cdot\sqrt{4x}}{2\sqrt{x}-3\cdot\sqrt{4x}}=\frac{2\sqrt{x}+6\sqrt{x}}{2\sqrt{x}-6\sqrt{x}}=\frac{8}{-4}=-2\)

Điều kiện \(x\ge0\)

\(\sqrt{x}=x\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

TL:

x=0

x=1

-HT-

Câu trai có vấn đề

Cái này mà kêu lp 4 hỏ

|2x - 5| = x-1

TH1: 2x - 5 = x - 1

2x + ( -5 ) = x + ( -1 )

x + x + ( -5 ) = x + ( -1)

=> x + ( -5) = -1

x = ( -1 ) - ( -5)

x = 4

TH2: 2x + ( -5) = -( x - 1)

2x + (-5) = -x + 1

3x + ( -x) + ( -5) = (-x) + 1

3x + (-5) = 1

3x = 1 - ( -5 ) = 6

x = 6 : 3 = 2

Vậy x = 4 và x = 2

Tick tui 1 tick hen

• x⁴-2x³+10x²-20x=0 =>x³(x-2)+10x(x-2)=0 =>(x-2)(x³+10x)=0 =>x(x-2)(x²+10)=0

 =>x=0 hoặc x=2 hoặc x= - căn 10