\(ĐKXĐ:x\ge0;x\ne1;0\)

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

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

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

\(A=\frac{2x+2+2\sqrt{x}}{\sqrt{x}}\)

\(A=2\sqrt{x}+\frac{2}{\sqrt{x}}+2\)

a/d bđt cauchy 

\(2\sqrt{x}+\frac{2}{\sqrt{x}}\ge2\sqrt{2.2}=2.2=4\)

\(A\ge4+2=6\)

\(< =>A>5\)

dấu "=" xảy ra khi x=1

a, \(\left(\frac{1}{x+2\sqrt{x}}-\frac{1}{\sqrt{x}+2}\right):\frac{1-\sqrt{x}}{x+4\sqrt{x}+4}\)ĐK : x >= 0 ; \(x\ne1\)

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

b, \(F=\frac{5}{2}\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{5}{2}\Rightarrow2\sqrt{x}+4=5\sqrt{x}\Leftrightarrow3\sqrt{x}=4\Leftrightarrow x=\frac{16}{9}\)

ĐK : x > 0 , x khác 1

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

Để bthuc = 5/2 thì \(\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{5}{2}\Rightarrow2\sqrt{x}+4=5\sqrt{x}\Leftrightarrow3\sqrt{x}=4\Leftrightarrow x=\frac{16}{9}\left(tm\right)\)

Điều kiện Quỳnh mới đúng nhé, quên cái căn x kia :v 

=1+ \(\sqrt[]{x-1}\)

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

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

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

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

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

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

Em không chắc câu c, d đâu nha

a) ĐK: \(1\ne\sqrt{x-1}\text{và }x\ge1\Leftrightarrow x\ne2;x\ge1\)

b) \(ĐK:\left\{{}\begin{matrix}x\ge\frac{5}{2}\\\sqrt{x+3}\ne\sqrt{2x-5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{5}{2}\\x\ne8\end{matrix}\right.\)

c) ĐK: \(\left\{{}\begin{matrix}x^2\ge36\left(1\right)\\x\ge-6\left(2\right)\end{matrix}\right..\text{Giải (1) }\Leftrightarrow\left[{}\begin{matrix}x\ge6\\x\le-6\end{matrix}\right.\)

Kết hợp (2) suy ra \(x=-6\text{ hoặc }x\ge6\)

d) ĐK: \(\frac{3x-2}{x+4}\ge0\). tức là 3x - 2 và x + 4 đồng dấu.

TH1: \(\left\{{}\begin{matrix}3x-2\ge0\\x+4>0\left(\text{do x phải khác -4}\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{2}{3}\\x>-4\end{matrix}\right.\Leftrightarrow x\ge\frac{2}{3}\)

TH2: \(\left\{{}\begin{matrix}3x-2< 0\\x+4< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \frac{2}{3}\\x< -4\end{matrix}\right.\Leftrightarrow x< -4\)

Do vậy ĐKXĐ: x >= 2/3 hoặc x<-4

e) ĐK: \(x\in\mathbb{R}\)

\(ĐK:\sqrt{x-1}\ne1;\sqrt{x-1}\ge0\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ge1\end{matrix}\right.\)

\(ĐK:\sqrt{x+3}\ne\sqrt{2x-5}\Leftrightarrow x+3\ne2x-5\Leftrightarrow x-8\ne0\Leftrightarrow x\ne8;x\ge\frac{5}{2}\) \(ĐK:\left\{{}\begin{matrix}x^2\ge36\\x\ge-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\le-6\\x\ge6\end{matrix}\right.\\x\ge-6\end{matrix}\right.\) \(\Leftrightarrow x=-6;x\ge6\)

a,

\(\frac{5\sqrt{60}\cdot3\sqrt{15}}{15\sqrt{50}\cdot2\sqrt{18}}\\ =\frac{5\cdot\sqrt{2^2\cdot15}\cdot3\sqrt{15}}{15\sqrt{2\cdot5^2}\cdot2\sqrt{2\cdot3^2}}\\ =\frac{5\cdot2\cdot3\cdot15}{15\cdot5\cdot2\cdot3\cdot3}=\frac{1}{3}\)

b,

\(\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}\\ =\frac{3-\sqrt{2}+3+\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}\\ =\frac{6}{3^2-2}=\frac{6}{7}\)

c,

\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\\ =\frac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}\\ =\frac{5-2\sqrt{15}+3+5+2\sqrt{15}+3}{5-3}\\ =\frac{16}{2}=8\)

d, Với \(x,y\ge0;x\ne y\), ta được:

\(\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\\ =\frac{\sqrt{x\cdot x^2}-\sqrt{y\cdot y^2}}{\sqrt{x}-\sqrt{y}}\\ =\frac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}\\ =\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}^3\right)}{\sqrt{x}-\sqrt{y}}\\ =\frac{\left(\sqrt{x}-\sqrt{y}\right)\left[\left(\sqrt{x}\right)^2+\sqrt{x\cdot y}+\left(\sqrt{y}\right)^2\right]}{\sqrt{x}-\sqrt{y}}\\ =x+y+\sqrt{xy}\)

Chúc bạn học tốt nha.

với đoạn câu b \(\frac{3-\sqrt{2}+3+\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}\) vì sao như thế vậy bạn