giải hệ pt: \(\left\{{}\begin{matrix}x+y-\sqrt{xy}=3\\\sqrt{x+1}+\sqrt{y+1}=4\end{matrix}\right.\)

b, \(Q=\frac{2\left(1-2\sqrt{x}\right)}{1-1+2\sqrt{x}}=\frac{1-2\sqrt{x}}{\sqrt{x}}\)

Để Q nhận giá trị nguyên thì \(\left(1-2\sqrt{x}\right)⋮\sqrt{x}\)

Ta có : \(\frac{1-2\sqrt{x}}{\sqrt{x}}=-2+\frac{1}{\sqrt{x}}\)\(\Rightarrow\sqrt{x}\inƯ\left(1\right)=\left\{\pm1\right\}\)

Do \(\sqrt{x}\ge0\Rightarrow\sqrt{x}=1\Rightarrow x=1\)

\(a,đkxđ:x\ge0\)

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

đkxđ : x\(\ge\)\(\)0

Để \(\frac{2\sqrt{x}-3}{\sqrt{x}+1}\) là số nguyên thì \(\left(2\sqrt{x}-3\right)⋮\left(\sqrt{x}+1\right)\)

Ta có \(\frac{2\sqrt{x}-3}{\sqrt{x}+1}=2+\frac{-5}{\sqrt{x}+1}\Rightarrow\left(-5\right)⋮\left(\sqrt{x}+1\right)\Rightarrow\left(\sqrt{x}+1\right)\inƯ\left(-5\right)\)

Mà Ư(-5)=\(\left\{1,-1,5,-5\right\}\)

Do \(\sqrt{x}+1\ge1\Rightarrow\left(\sqrt{x}+1\right)\in\left\{1,5\right\}\Rightarrow x\in\left\{0,16\right\}\)

\(a,đkxđ:x>0,x\ne1\)

\(E=\frac{\sqrt{x}.\sqrt{x}}{(\sqrt{x}-1).\sqrt{x}}-\frac{2\sqrt{x}-1}{\sqrt{x}(\sqrt{x}-1)}\)

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

\(b,ĐểP>0\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}>0.do\sqrt{x}>0\Rightarrow\)\(\sqrt{x}-1>0\Rightarrow x>1\)

khoảng cách tư ô tô đến tòa nhà là = \(\tan28\) độ \(\times\) 60=31,9 m