\(A=\frac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{\sqrt{x}-1}\)  \(\left(x\ge0;x\ne1\right)\)

\(A=\frac{x+2\sqrt{x}+1-4\sqrt{x}}{\sqrt{x}-1}=\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)

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

\(B=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)

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

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

\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)+1}{\sqrt{3}-\sqrt{2}}+2\)

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

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

để A = B thì \(\sqrt{x}-1\)= \(\frac{2}{\sqrt{3}-\sqrt{2}}+2\)

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

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

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

tới bước này tui bí :(( mong các bạn giỏi khác giúp bạn :D

Câu 1:

\(\frac{A}{B}\ge\frac{x}{4}+5\Leftrightarrow\frac{\sqrt{x}+4}{\sqrt{x}-1}:\frac{1}{\sqrt{x}-1}\ge\frac{x}{4}+5\)

\(\Rightarrow\sqrt{x}+4\ge\frac{x}{4}+5\Rightarrow x-4\sqrt{x}+4\le0\)

\(\Rightarrow\left(\sqrt{x}-2\right)^2\le0\Rightarrow\sqrt{x}-2=0\Rightarrow x=4\)

Câu 2:

Bạn coi lại đề, biểu thức B không hợp lý

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

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

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

b) \(\frac{A}{B}=\frac{\sqrt{x}+4}{\sqrt{x-1}}:\frac{1}{\sqrt{x}-1}=\sqrt{x}+4\)

Để \(\frac{A}{B}\ge\frac{x}{4}+5\)

\(\Leftrightarrow\sqrt{x}+4\ge\frac{x}{4}+5\)

\(\Leftrightarrow4\sqrt{x}+16\ge x+20\)

\(\Leftrightarrow x-4\sqrt{x}+4\le0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2\le0\)

Mà \(\left(\sqrt{x}-2\right)^2\ge0;\forall x\ge0\)

\(\Rightarrow\left(\sqrt{x}-2\right)^2=0\)

\(\Leftrightarrow x=4\)

Vậy ...

\(B=\frac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}.\frac{x-1}{x-2\sqrt{x}}\)

\(=\frac{x-3\sqrt{x}}{x-2\sqrt{x}}\)

\(=\frac{\sqrt{x}-3}{\sqrt{x}-2}\)

a.Ta co:

\(\frac{\sqrt{x}-3}{\sqrt{x}-2}< 1\left(x\ge0,x\ne4\right)\) 

\(\Leftrightarrow\sqrt{x}-3< \sqrt{x}-2\)

\(\Leftrightarrow3>2\)

Vay \(B< 1\left(\forall x\ge0,x\ne4\right)\)

Lát mình giải 2 câu kia,di ăn com cái

b.Ta co:

\(\frac{\sqrt{x}-3}{\sqrt{x}-2}< \frac{3}{2}\)

\(\Leftrightarrow2\sqrt{x}-6< 3\sqrt{x}-6\)

\(\Leftrightarrow x>0\)

Vay \(B< \frac{3}{2}\left(\forall x>0,x\ne4\right)\)

c.Ta co:

\(\frac{\sqrt{x}-3}{\sqrt{x}-2}>\sqrt{x}-1\)

\(\Leftrightarrow\sqrt{x}-3>x-3\sqrt{x}+2\)

\(\Leftrightarrow x-4\sqrt{x}+5< 0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+1< 0\) (vo ly)

Vay khong co gia tri nao cua x thoa man \(B>\sqrt{x}-1\)

a) ĐK: \(x\ge0,x\ne1,x\ne\frac{1}{4}\)

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

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

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

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

\(A=\frac{x+1}{x+\sqrt{x}+1}\)

Để \(A=\frac{6-\sqrt{6}}{5}\Rightarrow\frac{x+1}{x+\sqrt{x}+1}=\frac{6-\sqrt{6}}{5}\)

\(\Rightarrow5x+5=\left(6-\sqrt{6}\right)x+\left(6-\sqrt{6}\right)\sqrt{x}+6-\sqrt{6}\)

\(\Rightarrow\left(1-\sqrt{6}\right)x+\left(6-\sqrt{6}\right)\sqrt{x}+1-\sqrt{6}=0\)

\(\Rightarrow x-\sqrt{6}.\sqrt{x}+1=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{\sqrt{2}+\sqrt{6}}{2}\\\sqrt{x}=\frac{-\sqrt{2}+\sqrt{6}}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{cases}}\left(tmđk\right)\)

b) Xét \(A-\frac{2}{3}=\frac{x+1}{x+\sqrt{x}+1}-\frac{2}{3}=\frac{3x+3-2x-2\sqrt{x}-2}{3\left(x+\sqrt{x}+1\right)}\)

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

Do \(x\ge0,x\ne1,x\ne\frac{1}{4}\Rightarrow\left(\sqrt{x}-1\right)^2>0\)

Lại có \(x+\sqrt{x}+1=\left(\sqrt{x}+\frac{1}{2}\right)+\frac{3}{4}>0\)

Nên \(A-\frac{2}{3}>0\Rightarrow A>\frac{2}{3}\).

a/ Ta có: \(x+2\sqrt{x}+1=\left(\sqrt{x}+1\right)^2\)

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

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

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

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

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

b/ Thay \(x=\frac{\sqrt{3}}{2+\sqrt{3}}\)  => \(P=\frac{2}{\frac{\sqrt{3}}{2+\sqrt{3}}-1}=\frac{2\left(2+\sqrt{3}\right)}{\sqrt{3}-2-\sqrt{3}}\)

=> \(P=-\left(2+\sqrt{3}\right)\)

c/ \(P=\frac{2}{x-1}=-\frac{4}{\sqrt{x}+1}\) <=> \(\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\frac{2}{\sqrt{x}+1}\)

<=> \(\frac{1}{\sqrt{x}-1}=-2\)

<=> \(1=-2\sqrt{x}+2\)

<=> \(2\sqrt{x}=1=>\sqrt{x}=\frac{1}{2}=>x=\frac{1}{4}\)