\(P\left(\sqrt{x}+1\right)=4\sqrt{x}+7\)

\(P\sqrt{x}+P=4\sqrt{x}+7\)

\(P\sqrt{x}+P-4\sqrt{x}-7=0\)

\(P\sqrt{x}-4\sqrt{x}+P-7=0\)

\(\sqrt{x}\left(P-4\right)=7-P\)

\(\sqrt{x}=\frac{7-P}{P-4}\)\(\left(P\ne4\right)\)

Vì \(\sqrt{x}\ge0=>\frac{7-P}{P-4}\ge0\)

TH1 :\(\hept{\begin{cases}7-P\ge0\\P-4>0\end{cases}}\)\(\hept{\begin{cases}7\ge P\\P>4\end{cases}}\)

\(7\ge P>4\)

Vì P nguyên dương nên \(P\in\left(7,6,5\right)\)

\(\sqrt{x}\in\left(0,\frac{1}{2},2\right)\)

\(x\in\left(0,\frac{1}{4},4\right)\)

TH2:\(\hept{\begin{cases}7-P\le0\\P-4< 0\end{cases}}\)

\(\hept{\begin{cases}7\le P\\P< 4\end{cases}}\)

Vậy \(x\in\left(0,\frac{1}{4},2\right)\)

x= 4,0000000000000000

\(a,\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}}{x-2\sqrt{x}}\right):\frac{\sqrt{x}+1}{3}\)

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

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

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

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

\(b,\frac{3}{\sqrt{x}-2}=Q=1\)

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

\(5=\sqrt{x}\)

\(\sqrt{25}=\sqrt{x}\)

\(< =>x=5\)

a) điều kiện xác định \(x>=0,x\ne1\)

b)\(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\)

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

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

\(\frac{\left|x\right|-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

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

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

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

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

\(c.1+\frac{1}{\sqrt{x}}\)

để phương trình nghiệm nguyên \(1⋮\sqrt{x}\)

\(\sqrt{1}⋮\sqrt{x}\)

\(1⋮x\)

\(< =>x=1\)

a, ĐK : \(x>0;x\ne1\)

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

c, Ta có : \(\sqrt{x}+1⋮\sqrt{x}\Rightarrow1⋮\sqrt{x}\)do \(\sqrt{x}⋮\sqrt{x}\)

\(\Rightarrow\sqrt{x}\inƯ\left(1\right)=\left\{\pm1\right\}\)

rồi lập bảng, xét đk nhưng hình như ktm hết, ý c em ko chắc lắm vì cùng chưa gặp bài nào như này, ai biết bày em :> 

\(\hept{\begin{cases}x-1=0\\x+y=3\end{cases}}\)   

\(\hept{\begin{cases}x=0+1\\y=3-x\end{cases}}\)   

\(\hept{\begin{cases}x=1\\y=3-1\end{cases}}\)   

\(\hept{\begin{cases}x=1\\y=2\end{cases}}\)

\(\sqrt{x^2-4x+5}\le3x-17\)

bình phương 2 vế \(\Leftrightarrow x^2-4x+5\le9x^2-102x+289\)

\(\Leftrightarrow-8x^2+98x-294\le0\Leftrightarrow4x^2-21x-28x+147\ge0\)

\(\Leftrightarrow\left(4x-21\right)\left(x-7\right)\ge0\)

TH1 : \(\hept{\begin{cases}4x-21\ge0\\x-7\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge\frac{21}{4}\\x\ge7\end{cases}\Leftrightarrow}x\ge7}\)

TH2 : \(\hept{\begin{cases}4x-21\le0\\x-7\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le\frac{21}{4}\\x\le7\end{cases}\Leftrightarrow}x\le\frac{21}{4}}\)

Vậy tập nghiệm của bpt là S = { x | x >= 7 ; x =< 21/4 }