Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề và hỗ trợ bạn tốt hơn nhé.

a: Thay x=2 vào B, ta được:

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

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

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

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

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

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

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

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

b) Mà với \(x\ge0\) và \(x\ne1\) thì 

\(x+\sqrt{x}+1\ge0\) và \(2>0\) nên \(P>0\)

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

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

b: x+căn x+1+1>=1>0

2>0

=>P>0 với mọi x thỏa mãn x>=0 và x<>1

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

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

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

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

\(a,P=\frac{x+2}{x-2}+\frac{x}{x+2}-\frac{4}{x^2-4}\)

\(P=\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{4}{\left(x-2\right)\left(x+2\right)}\)

\(P=\frac{x^2+4x+4+x^2-2x-4}{x^2-4}\)

\(P=\frac{2x^2+2x}{x^2-4}\)

\(P=\frac{2x^2+2x}{x^2-4}\)               (1)

\(b,x^2-3x=0\)

\(\Leftrightarrow x\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=3\left(tm\right)\end{cases}}\)

thay vào (1) ta có : 

\(P=\frac{2\cdot3^2+2\cdot3}{3^2-4}=\frac{24}{5}\)

Đề bị lỗi công thức rồi bạn.

Với  x ≥ 0 ,   x ≠ 1 ,   x ≠ 4 ta có:

Q = x + 27 . P x + 3 x − 2 = x + 27 x + 3 = x − 9 + 36 x + 3 = x − 3 + 36 x + 3 = − 6 + x + 3 + 36 x + 3 ≥ − 6 + 12 = 6

\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)