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

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

\(Q=x+1\)

Không thể tìm được GTLN hay GTNN của Q.

b)

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

Để \(\frac{3Q}{\sqrt{x}}\) nguyên thì \(\frac{3}{\sqrt{x}}\)nguyên hay \(\sqrt{x}\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Vì \(\sqrt{x}\)dương nên \(\sqrt{x}\in\left\{1;3\right\}\)

Vậy x=1, x=9 là các giá trị cần tìm

1) Khi x = 36 thì A = \(\frac{\sqrt{36}+4}{\sqrt{36}+2}\Leftrightarrow\frac{5}{4}\)

Vậy khi x = 36 thì A = \(\frac{5}{4}\)

2) B = \((\frac{\sqrt{x}\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}+\frac{4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}):\frac{x+16}{\sqrt{x}+2}\)

= \(\frac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}.\frac{\sqrt{x}+2}{x+16}=\frac{x+16}{x-16}.\frac{\sqrt{x}+2}{x+16}\)

= \(\frac{\sqrt{x}+2}{x-16}\)

Vậy B = \(\frac{\sqrt{x}+2}{x-16}\)

b: \(=\dfrac{\left|x\right|+\left|x-2\right|+1}{2x-1}=\dfrac{x+x-2+1}{2x-1}=\dfrac{2x-1}{2x-1}=1\)

c: \(=\left|x-4\right|+\left|x-6\right|\)

=x-4+6-x=2

ĐKXĐ: ...

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

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

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

b/ Do \(x+\sqrt{x}+1=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

\(\Rightarrow\) Để \(D>0\Leftrightarrow1-\sqrt{x}>0\Leftrightarrow\sqrt{x}< 1\Rightarrow0\le x< 1\)

\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)ĐK x>=0 x khác -1

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

b/ x =\(\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{3+2\sqrt{3}+1}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)

\(\Rightarrow\sqrt{x}=\frac{\sqrt{3}+1}{2}\)

Em thay vào tính nhé!

c) với x>1

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

Áp dụng bất đẳng thức Cosi 

A\(\ge2\sqrt{2}+3\)

Xét dấu bằng xảy ra ....

dấu bằng xảy ra khi nào v ạ ??

ĐKXĐ:...

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

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

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

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

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