a) ĐKXĐ: \(x>0\)

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

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

\(A=x-\sqrt{x}=2\)

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

\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)(do \(\sqrt{x}+1\ge1>0\))

b) \(A=x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)>0\)(do \(x>1\))

\(\Leftrightarrow A=x-\sqrt{x}=\left|A\right|\)

c) \(A=x-\sqrt{x}=\left(x-\sqrt{x}+\dfrac{1}{4}\right)-\dfrac{1}{4}\)

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

\(minA=-\dfrac{1}{4}\Leftrightarrow\sqrt[]{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)

\(a,A=\dfrac{x\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1\left(x>0\right)\\ A=\dfrac{x\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-2\sqrt{x}-1+1\\ A=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\\ A=2\Leftrightarrow x-\sqrt{x}-2=0\\ \Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}=2\left(\sqrt{x}>0\right)\\ \Leftrightarrow x=4\left(tm\right)\)

\(b,x>1\Leftrightarrow\sqrt{x}-1>0\\ \Leftrightarrow\left|A\right|=\left|x-\sqrt{x}\right|=\left|\sqrt{x}\left(\sqrt{x}-1\right)\right|=\sqrt{x}\left(\sqrt{x}-1\right)=A\left(\sqrt{x}>0\right)\)

\(c,A=x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\\ A_{min}=-\dfrac{1}{4}\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)

*Rút gọn

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

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

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

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

Ta có: \(C=x-\sqrt{x}+1\)

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

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

Dấu '=' xảy ra khi \(\sqrt{x}=\dfrac{1}{2}\)

hay \(x=\dfrac{1}{4}\)

\(C=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\left(x>0;x\ne1\right)\)

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

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

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

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

Dấu "=" xảy ra khi \(\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{4}\)

Vậy  \(C_{min}=\dfrac{3}{4}\)

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

Áp dụng AM-GM có: \(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\)

Dấu "=" xảy ra khi x=1 (ktm đk)

Suy ra dấu bằng ko xảy ra \(\Rightarrow\sqrt{x}+\dfrac{1}{\sqrt{x}}-1>2-1=1\)

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

\(\Rightarrow N< 2\) mà \(N>0\),\(N\) nguyên

\(\Rightarrow N=1\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=1\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{3+\sqrt{5}}{2}\\\sqrt{x}=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7+3\sqrt{5}}{2}\\x=\dfrac{7-3\sqrt{5}}{2}\end{matrix}\right.\) (tm)

Vậy...

\(\Rightarrow C=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\) * \(\Rightarrow C=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\) Dấu = xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)

* Ta có \(N=\dfrac{2\sqrt{x}}{C}=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}>0\left(1\right)\) 

Xét \(N-2=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}-2=\dfrac{2\sqrt{x}-2x+2\sqrt{x}-2}{x-\sqrt{x}+1}=\dfrac{-2x+4\sqrt{x}-2}{x-\sqrt{x}+1}=\dfrac{-2\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}< 0\left(dox\ne1\right)\Rightarrow N< 2\left(2\right)\) Từ (1) và (2) \(\Rightarrow0< N< 2\). Mà N nguyên nên N=1  \(\Rightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=1\Rightarrow2\sqrt{x}=x-\sqrt{x}+1\Leftrightarrow x-3\sqrt{x}+1=0\)

\(\Delta=9-4=5\Rightarrow\) pt có 2 nghiệm phân biệt: \(x_1=\dfrac{\sqrt{5}+3}{2}\left(TM\right);x_2=\dfrac{3-\sqrt{5}}{2}\left(TM\right)\)

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

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

\(=\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b: B=2/7

=>\(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=\dfrac{2}{7}\)

=>\(2\left(x+\sqrt{x}+1\right)=7\sqrt{x}\)

=>\(2x+2\sqrt{x}-7\sqrt{x}+2=0\)

=>\(2x-5\sqrt{x}+2=0\)

=>\(\left(2\sqrt{x}-1\right)\cdot\left(\sqrt{x}-2\right)=0\)

=>\(\left[{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)

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

B nhỏ nhất khi \(\dfrac{3}{\sqrt{x}+2}\) lớn nhất \(\Rightarrow\sqrt{x}+2\) nhỏ nhất \(\Rightarrow\sqrt{x}\) nhỏ nhất

Mà \(x\ge0\Rightarrow\sqrt{x}\ge0\Rightarrow\sqrt{x}_{min}=0\)

\(\Rightarrow B_{min}=-\dfrac{1}{2}\) khi \(x=0\)

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

\(\sqrt{x}+2>=2\forall x\) thỏa mãn ĐKXĐ

=>\(\dfrac{3}{\sqrt{x}+2}< =\dfrac{3}{2}\forall x\) thỏa mãn ĐKXĐ

=>\(-\dfrac{3}{\sqrt{x}+2}>=-\dfrac{3}{2}\forall x\) thỏa mãn ĐKXĐ

=>\(-\dfrac{3}{\sqrt{x}+2}+1>=-\dfrac{3}{2}+1=-\dfrac{1}{2}\forall x\) thỏa mãn ĐKXĐ

=>\(B>=-\dfrac{1}{2}\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x=0

2+ 6/ căn x -1

Câu 1:

\(a-\sqrt{a}+1=a-2.\sqrt{a}.\frac{1}{2}+\frac{1}{2^2}+\frac{3}{4}\)

\(=(\sqrt{a}-\frac{1}{2})^2+\frac{3}{4}\)

Ta thấy \((\sqrt{a}-\frac{1}{2})^2\geq 0, \forall a\) không âm

\(\Rightarrow a-\sqrt{a}+1=(\sqrt{a}-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}\)

Vậy GTNN của biểu thức là $\frac{3}{4}$. Dấu "=" xảy ra khi \((\sqrt{a}-\frac{1}{2})^2=0\Leftrightarrow a=\frac{1}{4}\)

Câu 2:

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

Ta thấy \((a-1)^2\geq 0, \forall a\) thuộc tập xác định

\(\Rightarrow 2-(a-1)^2\leq 2\)

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

Vậy GTLN của biểu thức là $\sqrt{2}$ khi \((a-1)^2=0\Leftrightarrow a=1\)

Câu 3:

ĐK: $x\geq 1$

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

\(=(\sqrt{x-1}-1)^2\geq 0, \forall x\geq 1\)

Vậy GTNN của biểu thức là $0$

Dấu "=" xảy ra khi \((\sqrt{x-1}-1)^2=0\Leftrightarrow x=2\)

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

Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)

Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+\sqrt{1+x^2}=a>0\\y+\sqrt{1+y^2}=b>0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}1+x^2=a^2+x^2-2ax\\1+y^2=b^2+y^2-2by\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{a^2-1}{2a}\\y=\dfrac{b^2-1}{2b}\end{matrix}\right.\)

Giả thiết trở thành: \(ab=2018\)

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

\(P=\dfrac{1}{2}\left(a+b\right)\left(1-\dfrac{1}{ab}\right)=\dfrac{1}{2}\left(a+b\right).\dfrac{2017}{2018}\ge\sqrt{ab}.\dfrac{2017}{2018}=\dfrac{2017}{\sqrt{2018}}\)

\(P_{min}=\dfrac{2017}{\sqrt{2018}}\)

Dấu "=" xảy ra khi \(x=y=\dfrac{2017}{2\sqrt{2018}}\)