a,  \(P=\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{\sqrt{x}+3}{5-\sqrt{x}}-\frac{3x+4\sqrt{x}-5}{x-4\sqrt{x}-5}\)

\(P=\frac{\sqrt{x}+2}{\sqrt{x}+1}+\frac{\sqrt{x}+3}{\sqrt{x}-5}-\frac{3x+4\sqrt{x}-5}{x-4\sqrt{x}-5}\)

\(P=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-5\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-5\right)}-\frac{3x+4\sqrt{x}-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-5\right)}\)

\(P=\frac{x-3\sqrt{x}-10+x+4\sqrt{x}+3-3x-4\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-5\right)}\)

\(P=\frac{-x-3\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-5\right)}\)

\(P=\frac{\left(\sqrt{x}+1\right)\left(-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-5\right)}=\frac{-\sqrt{x}-2}{\sqrt{x}-5}\)

để P > -2 

\(\Rightarrow\frac{-\sqrt{x}-2}{\sqrt{x}-5}>-2\) đoạn này đang chưa nghĩ ra

c, \(P=\frac{-\sqrt{x}-2}{\sqrt{x}-5}\in Z\)  \(\Rightarrow-\sqrt{x}-2⋮\sqrt{x}-5\)

=> -căn x + 5 - 7 ⋮ căn x - 5

=> -(căn x - 5) - 7 ⋮ căn x - 5 

=> 7 ⋮ x - 5 đoạn này dễ

a, Với \(x\ge0;x\ne25\)thì \(P=\frac{\sqrt{x}+2}{5-\sqrt{x}}\)  đoạn này đúng rồi 

\(P>-2\)\(\Leftrightarrow\frac{\sqrt{x}+2}{5-\sqrt{x}}>-2\)

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

\(\Leftrightarrow\frac{12-\sqrt{x}}{5-\sqrt{x}}>0\)

Xét 2 trường hợp cùng âm, cùng dương hoặc "trong trái ngoài cùng"

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}>12\\0\le\sqrt{x}< 5\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x>144\\0\le x< 25\end{cases}}\)

Làm luôn cho đầy đủ =)

\(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\left(đk:x\ge1\right)\)

\(< =>\sqrt{x-2\sqrt{x-1}}^2=\left(\sqrt{x-1}-1\right)^2\)

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

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

Vậy phương trình trên thỏa mãn với mọi \(x\ge1\)

ĐKXĐ : \(x\ge1\)

Bình phương 2 vế lên ta có :

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

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

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

\(\Leftrightarrow0x=0\)( luôn đúng với mọi \(x\ge1\))

Vậy ...............

\(\sqrt{\frac{10}{17}}va\frac{3}{4}\)

Ta có \(\frac{10}{17}>\frac{9}{16}\)

\(\Rightarrow\sqrt{\frac{10}{17}}>\sqrt{\frac{9}{16}}\)

\(\Rightarrow\sqrt{\frac{10}{17}}>\frac{3}{4}\)

Học tốt

Câu 1 là 5 lần

Làm rõ ra nha bạn :))

Dùng lệnh tex ( đề bài )

\(A=(4\sqrt{2}+3).\sqrt{41-24\sqrt{2}}\)

\(A=\left(4\sqrt{2}+3\right).\sqrt{9+2.3.4\sqrt{2}+32}\)

\(A=\left(4\sqrt{2}+3\right).\sqrt{\left(3+4\sqrt{2}\right)^2}\)

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

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

Bạn vào TKHĐ mình xem cho tiện nha

@huybip5cc . Không có bạn ơi !! 

Ta có: \(A=\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right)^2\times\frac{x^2-1}{2}-\sqrt{1-x^2}\)    \(\left(ĐK:x\ge1\right)\)

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

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

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

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

   \(\Leftrightarrow A=x\)

Học tốt

ĐKXĐ : ...............

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

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

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

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

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

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

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

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

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

<=> \(xy+\sqrt{x^2+1}\sqrt{y^2+1}-1=-x\sqrt{x^2+1}-y\sqrt{y^2+1}\)--->Bình phương 2 vế:

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

                                                                                                     \(x^2\left(x^2+1\right)+y^2\left(y^2+1\right)+2xy\sqrt{x^2+1}\sqrt{y^2+1}\)

<=>\(2\left(1-xy-\sqrt{x^2+1}\sqrt{y^2+1}\right)=\left(x^2-y^2\right)^2\ge0\)=>\(1-xy-\sqrt{x^2+1}\sqrt{y^2+1}\ge0\)

<=>\(1-xy\ge\sqrt{x^2+1}\sqrt{y^2+1}>0\)---> Bình phương 2 vế:

\(1+x^2y^2-2xy\ge\left(x^2+1\right)\left(y^2+1\right)\)<=>\(0\ge\left(x+y\right)^2\ge0\)<=>\(x+y=0\Leftrightarrow x=-y\Rightarrow x^2=y^2\)

--> Thay vào A---> \(A=\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=\left(x+\sqrt{y^2+1}\right)\left(y+\sqrt{x^2+1}\right)=1\)