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

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

a, Với x ≥ 0, x ≠1 
P= [ \(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\)] : \(\dfrac{\sqrt{x}-1}{2}\)  = 
\(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)]
: \(\dfrac{\sqrt{x}-1}{2}\)
P= \(\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\):\(\dfrac{\sqrt{x}-1}{2}\)
P= \(\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{2}\)
P= \(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
P= \(\dfrac{2}{x+\sqrt{x}+1}\)
b, Ta có : \(x+\sqrt{x}+1=\left(\sqrt{x}\right)^2+2.\dfrac{1}{2}.\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}\)= (\(\sqrt{x}+\dfrac{1}{2}\))² +\(\dfrac{3}{4}\) >\(0\)  ∀ x
=> \(\dfrac{3}{x+\sqrt{x}+1}>0\) ∀ x

=> P > 0 với mọi x ≥ 0 và x ≠ 1

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

1, vt : \(\left(1-\dfrac{5+\sqrt{2}}{\sqrt{2}+1}\right).\sqrt{3+2\sqrt{2}}\)

=\(\dfrac{\sqrt{2}+1-5-\sqrt{2}}{\sqrt{2}+1}.\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}\)

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

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

=-4

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

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

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

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

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

1. (1−5+√2√2+1)⋅√3+2√2=−4√2+1√(√2+1)2=−4(1−5+22+1)⋅3+22=−42+1(2+1)2=−4.

2. Với x>0;x≠1x>0;x≠1 ta có:
A=(√xx+√x−1√x−1):2x+√x−2A=(xx+x−1x−1):2x+x−2
⇔A=(√x√x(√x+1)−1√x−1):2(√x−1)(√x+2)⇔A=(xx(x+1)−1x−1):2(x−1)(x+2)
⇔A=−2(√x−1)(√x+1)⋅(√x−1)(√x+2)2⇔A=−2(x−1)(x+1)⋅(x−1)(x+2)2
⇔A=−(√x+2)√x+1⇔A=−(x+2)x+1. Vạyy với x>0;x≠1x>0;x≠1, ta có A=−(√x+2)√x+1A=−(x+2)x+1.

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)

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

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

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

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

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

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

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

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

\(=\dfrac{x+2\sqrt{x}+1-2\sqrt{x}}{\sqrt{x}}=\dfrac{x+1}{\sqrt{x}}>0\)

=>P>2

đkxđ : \(x\ge0,x\ne1\)

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

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

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

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

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

\(0< x< 1\)

\(\Rightarrow\sqrt{x}< 1\)

\(\Rightarrow\sqrt{x}-1< 0\)

\(\Rightarrow-\sqrt{x}\left(\sqrt{x}-1\right)>0\)

\(x=3+2\sqrt{2}\)

\(\Rightarrow A=-\sqrt{3+2\sqrt{2}}\left(\sqrt{3+2\sqrt{2}}-1\right)\)\(=-\left|1+\sqrt{2}\right|\cdot\left(\left|1+\sqrt{2}\right|-1\right)\)\(=\left(-1-\sqrt{2}\right)\left(1+\sqrt{2}-1\right)=\left(-1-\sqrt{2}\right)\left(\sqrt{2}\right)=-\sqrt{2}-2\)

A=-3; B=

1) Thay x=1x=1 vào biểu thức: A=√1+2√1−2A=1+21−2
A=−3A=−3
2) Chứng minh B=√x√x+2B=xx+2 với x≥0,x≠4x≥0,x≠4.
B=√x+2(√x−2)(√x+2)+(√x+1)(√x−2)(√x+2)(√x−2)−2√x(√x+2)(√x−2)B=x+2(x−2)(x+2)+(x+1)(x−2)(x+2)(x−2)−2x(x+2)(x−2)
=√x+2+x−√x−2−2√x(√x+2)(√x−2)=x−2√x(√x+2)(√x−2)=x+2+x−x−2−2x(x+2)(x−2)=x−2x(x+2)(x−2)
=√x(√x−2)(√x+2)(√x−2)=√x√x+2=x(x−2)(x+2)(x−2)=xx+2
3) Tìm xx để A⋅B≥0A⋅B≥0
A⋅B=√x+2√x−2⋅√x√x+2=√x√x−2A⋅B=x+2x−2⋅xx+2=xx−2.
TH1: x=0⇒√x=0⇒A⋅B=0x=0⇒x=0⇒A⋅B=0 (TM)
TH2: x>0⇒√x>0⇒√x−2>0⇒x>4x>0⇒x>0⇒x−2>0⇒x>4
Kết hợp điêu kiện: x=0x=0 hoặc x>4x>4 thỏa mãn yêu cầu.

x

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

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

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

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

c: Ta có: \(x+\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ

\(\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}>0\forall x\)

a. \(B=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\\ =\left(\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-1\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}-1}{2}\\ =\dfrac{\left(\sqrt{x}-1\right)^2.2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)

b.Ta có:

\(B=\dfrac{2}{x+\sqrt{x}+1}\). Mà \(\left[{}\begin{matrix}2>0\\x+\sqrt{x}+1=\left[\left(\sqrt{x}\right)^2+2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}\right]+\dfrac{3}{4}>0\end{matrix}\right.\)

Vậy B>0 \(\forall x\)