1: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-8}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-\sqrt{x}-5\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}-4}{\sqrt{x}}\)

2: \(P=A\cdot B=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

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

=>P>2

a) \(P=\dfrac{\sqrt{a}\left[\left(\sqrt{a}\right)^3+1\right]}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)

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

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

\(P=a+\sqrt{a}-2\sqrt{a}-1+1\)

\(P=a-\sqrt{a}\)

b) Với a > 1 thì \(a>\sqrt{a}\) , do đó \(P=a-\sqrt{a}>0\), suy ra \(\left|P\right|=P\)

c) \(A=a-\sqrt{a}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Vậy A nhỏ nhất bằng \(-\dfrac{1}{4}\) khi cà chỉ khi \(\sqrt{a}=\dfrac{1}{2}\) hay \(a=\dfrac{1}{4}\)

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

b: a>1 nên P>0

\(\Leftrightarrow P=\left|P\right|\)

a: Ta có: \(B=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{a\sqrt{a}-a-\sqrt{a}+1}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}\)

\(=\dfrac{6\sqrt{a}-6+10-2\sqrt{a}}{\left(\sqrt{a}-1\right)^2\cdot\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}\)

\(=\dfrac{4\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\cdot\dfrac{1}{4\sqrt{a}}\)

\(=\dfrac{1}{\sqrt{a}}\)

a) \(B=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{a\sqrt{a}-a-\sqrt{a}+1}\right).\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{\left(a-1\right)\left(\sqrt{a}-1\right)}\right).\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{6\left(\sqrt{a}-1\right)+10-2\sqrt{a}}{\left(a-1\right)\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{4\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{1}{\sqrt{a}}\)

b) \(C=B.\left(a-\sqrt{a}+1\right)=\dfrac{a-\sqrt{a}+1}{\sqrt{a}}=\sqrt{a}-1+\dfrac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\dfrac{1}{\sqrt{a}}}-1=1\)(bất đẳng thức Cauchy cho 2 số dương)

a)

ĐK: \(a>0\)

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

b)

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

\(\Rightarrow\left|P\right|=P\)

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

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

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

\(=a-\sqrt{a}\)

c) Để A=2 thì \(a-\sqrt{a}-2=0\)

\(\Leftrightarrow\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)=0\)

\(\Leftrightarrow a=4\)

a: \(B=\dfrac{1}{\sqrt{x}+1}\)

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

=>B>=1

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

\(P\cdot\sqrt{x}+2x-\sqrt{x}=3x-2\sqrt{x-4}+3\)

=>\(x+\sqrt{x}+1+2x-\sqrt{x}=3x+3-2\sqrt{x-4}\)

=>\(-2\sqrt{x-4}+3=1\)

=>x-4=1

=>x=5

a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)

\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)

\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)

=-a-1

b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)

\(=\left|3a-5\right|-2a+4\)

\(=5-3a-2a+4\)

=9-5a

c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)

\(=4a-3-\left|2a-1\right|\)

\(=4a-3-2a+1\)

\(=2a-2\)

d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)

\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)

\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)

\(=-a^2\)

\(a,B=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{x+\sqrt{x}-6}\left(x>0;x\ne6\right)\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x+3\sqrt{x}+\sqrt{x}+3+2\sqrt{x}-4-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\\)

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

`b,` Tớ tính mãi ko ra, xl cậu nha=')