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

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

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

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

\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=4\end{matrix}\right.\) \(\Rightarrow a+b=7\)

Bài 5: 

a: Thay \(x=4+2\sqrt{3}\) vào E, ta được:

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

b: Để E<1 thì E-1<0

\(\Leftrightarrow\dfrac{\sqrt{x}-1-\sqrt{x}+3}{\sqrt{x}-3}< 0\)

\(\Leftrightarrow\sqrt{x}-3< 0\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)

c: Để E nguyên thì \(4⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{-2;1;2;4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{4;5;7\right\}\)

hay \(x\in\left\{16;25;49\right\}\)

Câu 2:
a) Ta có \(x=4-2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-2\right)^2}=\sqrt{3}-2\)

Thay \(x=\sqrt{3}-1\) vào \(B\), ta được

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

b) Để \(B\) âm thì \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\) mà \(\sqrt{x}+1\ge1>0\forall x\) \(\Rightarrow\sqrt{x}-2< 0\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)

c) Ta có \(B=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=1-\dfrac{3}{\sqrt{x}+1}\)

Với mọi \(x\ge0\) thì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\dfrac{3}{\sqrt{x}+1}\le3\Rightarrow B=1-\dfrac{3}{\sqrt{x}+1}\ge-2\)

Dấu "=" xảy ra khi \(\sqrt{x}+1=1\Leftrightarrow x=0\)

Vậy \(B_{min}=-2\) khi \(x=0\)

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

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

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

thay vào A=\(\dfrac{-2}{3}\)

b)

A=-1+\(\dfrac{1}{\sqrt{x}+1}\) \(\ge\) -1+\(\dfrac{1}{1}\)=1(vì \(\sqrt{x}\)\(\ge\) 0)

Dấu bằng xẩy ra\(\Leftrightarrow\) x=0

chỗ đó cho thêm x-1 nha

đấu >= thay thành <= rùi nhân thêm x-1>=-1 nữa là lớn nhất bằng 0

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

Mà \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow B=\left(\sqrt{3}-1-1\right)^2=\left(\sqrt{3}-2\right)^2=7-4\sqrt{3}\)

\(b,P=AB=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\\ P=\dfrac{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}=\sqrt{x}-1\\ c,Q=\sqrt{x}+\dfrac{1}{P}=\sqrt{x}+\dfrac{1}{\sqrt{x}-1}\\ Q=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+1\ge2\sqrt{1}+1=3\\ Q_{min}=3\Leftrightarrow\left(\sqrt{x}-1\right)^2=1\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=1\\1-\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\sqrt{x}=2\left(x>1\Leftrightarrow\right)x=4\left(tm\right)\)

a: \(B=\left(\sqrt{x}-1\right)^2=\left(\sqrt{3}-2\right)^2=7-4\sqrt{3}\)

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

a, Ta có : \(x=4\Rightarrow\sqrt{x}=2\)

\(\Rightarrow A=\frac{2+1}{2+2}=\frac{3}{4}\)

Vậy với x = 4 thì A = 3/4 

b, \(B=\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}+5}{x-1}=\frac{3\left(\sqrt{x}+1\right)-\sqrt{x}-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

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

với x=4(t/m DK)

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

thay\(\sqrt{x}\)=2 vào biểu thức A ta được

A=(2+1)/(2+2)

A=3/4

Lời giải:
a.

\(A=\frac{\sqrt{x}(\sqrt{x^3}-1)}{x+\sqrt{x}+1}-\frac{\sqrt{x}(2\sqrt{x}+1)}{\sqrt{x}}+\frac{2(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}-1}\)

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

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

b.

$A=x-\sqrt{x}+1=(x-\sqrt{x}+\frac{1}{4})+\frac{3}{4}$

$=(\sqrt{x}-\frac{1}{2})^2+\frac{3}{4}\geq 0+\frac{3}{4}=\frac{3}{4}$

$\Rightarrow A_{\min}=\frac{3}{4}$

Giá trị này đạt tại $\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}$

a, Để A nhận giá trị dương thì \(A>0\)hay \(x-1>0\Leftrightarrow x>1\)

b, \(B=2\sqrt{2^2.5}-3\sqrt{3^2.5}+4\sqrt{4^2.5}\)

\(=4\sqrt{5}-9\sqrt{5}+16\sqrt{5}=\left(4-9+16\right)\sqrt{5}=11\sqrt{5}\)

( theo công thức \(A\sqrt{B}=\sqrt{A^2B}\))

c, Với \(a\ge0;a\ne1\)

\(C=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\)

\(=\left(\sqrt{a}+1\right)^2.\frac{1}{\left(\sqrt{a}+1\right)^2}=1\)