Số 1 bỏ ik nha.

Giá trị của x thì A có nghĩa là : \(x\ne0\)

dễ mà bạn

\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

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

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

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

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

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

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

\(D=\sqrt{2}-1-\sqrt{2}-1\)

\(D=-2\)

CÂU THỨ 2 NHA !!!!!!

XÉT:        \(2VT=2a\sqrt{b-1}+2b\sqrt{a-1}\)

=>    \(2VT=a.2.\sqrt{1}.\sqrt{b-1}+b.2.\sqrt{1}.\sqrt{a-1}\)

TA ÁP DỤNG BĐT CAUCHY 2 SỐ SẼ ĐƯỢC: 

=>    \(2VT\le a\left(1+b-1\right)+b\left(1+a-1\right)\)

=>   \(2VT\le ab+ab\)

=>   \(2VT\le2ab\)

=>   \(VT\le ab\)

=> TA CÓ ĐIỀU PHẢI CHỨNG MINH.

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

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

=>   \(A=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{4+2\sqrt{3}}\)

=>   \(A=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)

=>   \(A=\left(\sqrt{3}+1\right)^2\left(\sqrt{3}-2\right)\)

=>   \(A=\left(4+2\sqrt{3}\right)\left(\sqrt{3}-2\right)\)

=>   \(A=4\sqrt{3}-8+6-4\sqrt{3}\)

=>   \(A=-8+6=-2\)

VẬY \(A=-2\)

\(B=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{2}.\sqrt{4-\sqrt{15}}\)

=>   \(B=\sqrt{8-2\sqrt{15}}\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

=> \(B=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\left(4+\sqrt{15}\right)\)

=>  \(B=\left(\sqrt{5}-\sqrt{3}\right)^2\left(4+\sqrt{15}\right)\)

=>   \(B=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)

=>   \(B=32+8\sqrt{15}-8\sqrt{15}-30\)

=>   \(B=2\)

VẬY    \(B=2\)

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

\(\sqrt{x}-3⋮\sqrt{x}-2\Leftrightarrow\sqrt{x}-2-1⋮\sqrt{x}-2\)

\(\Leftrightarrow-1⋮\sqrt{x}-2\Leftrightarrow\sqrt{x}-2\inƯ\left(-2\right)=\left\{\pm1;\pm2\right\}\)

\(B=\frac{\sqrt{x}-3}{\sqrt{x}-2}\left(x\ge0\right)\)

để B đạt giá trị âm thì \(\sqrt{x}-3\)và \(\sqrt{x}-2\)phải trái dấu nhau 

ta thấy \(\sqrt{x}-3< \sqrt{x}-2\)\(\Rightarrow\hept{\begin{cases}\sqrt{x}-3< 0\\\sqrt{x}-2>0\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{x}< 3\\\sqrt{x}>2\end{cases}\Leftrightarrow}\hept{\begin{cases}x< 9\\x>4\end{cases}\Leftrightarrow}4< x< 9}\)

vậy 4<x<9 thì B đạt giá trị âm

Ta có :

\(\left(5xy^2+9xy-x^2y^2\right):\left(-xy\right)\)

\(=\left(-xy\right).\left(-5y-9+xy\right):\left(-xy\right)\)

\(=-5y-9+xy\)

Thay \(x=1,y=2\) vào ta có :

\(-5y-9+xy=\left(-5\right).2-9+1.2=-17\)