c/ \(C'=\frac{1}{\frac{1}{3-2\sqrt{x}}}.\frac{1}{\frac{1}{\sqrt{3-2\sqrt{x}}}+1}=\frac{\sqrt{\left(3-2\sqrt{x}\right)^3}}{1+\sqrt{\left(3-2\sqrt{x}\right)}}\)

Đặt \(\sqrt{\left(3-2\sqrt{x}\right)}=a\)

\(\Rightarrow C'=\frac{a^3}{a+1}=a^2-a+1-\frac{1}{a+1}\)

Đế C' nguyên thì a + 1 là ước của 1

\(\Rightarrow a=0\)

\(\Rightarrow\sqrt{\left(3-2\sqrt{x}\right)}=0\)

\(\Rightarrow x=\frac{9}{4}\left(l\right)\)

Vậy không có x.

Không biết có nhầm chỗ nào không nữa. Lam biếng kiểm tra lại quá. You kiểm tra lại hộ nhé. Thanks

a/ \(C=\left(\frac{2\sqrt{x}}{2x-5\sqrt{x}+3}-\frac{5}{2\sqrt{x}-3}\right):\left(3+\frac{2}{1-\sqrt{x}}\right)\)

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

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

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

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

Câu b, c tự làm nhé

1) ĐKXĐ: \(x>0;x\ne4;x\ne9\)

(*lười lắm, ko chép lại đề nha :V*)

\(P=\frac{\left(2+\sqrt{x}\right)^2+\sqrt{x}\left(2-\sqrt{x}\right)+4x+2\sqrt{x}-4}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{2\sqrt{x}-\left(\sqrt{x}+3\right)}{\sqrt{x}\left(2-\sqrt{x}\right)}\\ =\frac{4+4\sqrt{x}+x+2\sqrt{x}-x+4x+2\sqrt{x}-4}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\\ =\frac{4x+8\sqrt{x}}{2+\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}-3}\\ =\frac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\cdot\frac{\sqrt{x}}{\sqrt{x}-3}=\frac{4x}{\sqrt{x}-3}\)

2) Để P>0 thì

\(\frac{4x}{\sqrt{x}-3}>0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x>0\\\sqrt{x}-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}4x< 0\\\sqrt{x}-3< 0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\\sqrt{x}>3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\\sqrt{x}< 3\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x>9\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 9\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>9\\x< 0\left(ktm\right)\end{matrix}\right.\)

Vậy với \(x>9\) thì \(P>0\).

Chúc bạn học tốt nha.

Ta có :

\(B=\left(\frac{1}{x-4}-\frac{1}{x+4\sqrt{x}+4}\right).\frac{x+2\sqrt{x}}{\sqrt{x}}\)

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

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

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

\(=\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(P=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\frac{-x+x\sqrt{x}+6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{x-\sqrt{x}-x+x\sqrt{x}+6-x-\sqrt{x}-2\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{x\sqrt{x}-x-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{x\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{\left(x-4\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

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

b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

\(Q=\frac{\left(x+27\right)P}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\frac{\left(x+27\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\frac{x+27}{\sqrt{x}+3}\)

\(Q=\frac{x+27}{\sqrt{x}+3}\ge6\\ \Leftrightarrow\frac{x+27}{\sqrt{x}+3}-6\ge0\\ \Leftrightarrow\frac{x+27-6\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\ge0\\ \Leftrightarrow\frac{x-6\sqrt{x}+45}{\sqrt{x}+3}\ge0\)

Dễ thấy \(x-6\sqrt{x}+45=\left(\sqrt{x}-3\right)^2+36\ge36>0\forall x\ge0\)

\(\sqrt{x}+3\ge3>0\forall x\ge0\)

=> Ko có giá trị nào của x thỏa mãn yêu cầu

P/s: Nếu đề là \(x\sqrt{x}+27\)thì sẽ khác một chút :v

Sửa khúc gần cuối :v

\(Q=\frac{x+27}{\sqrt{x}+3}\ge6\\ \Leftrightarrow\frac{x+27}{\sqrt{x}+3}-6\ge0\\ \Leftrightarrow\frac{x+27-6\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\ge0\\ \Leftrightarrow\frac{x-6\sqrt{x}+9}{\sqrt{x}+3}\ge0\\ \Leftrightarrow\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}+3}\ge0\)

Dễ thấy \(\left\{{}\begin{matrix}\left(\sqrt{x}-3\right)^2\ge0\forall x\ge0\\\sqrt{x}+3>0\forall x\ge0\end{matrix}\right.\)

Vậy với mọi \(x\ge0;x\ne1;x\ne4\)thì thỏa mãn đề bài

(chỗ x khác mấy là kết hợp điều kiện để rút gọn được biểu thức P và Q)

\(ĐK:\left\{{}\begin{matrix}x\ge0\\x\ne9\\x\ne25\end{matrix}\right.\)

\(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{2x}{9-x}-1\right):\left(\frac{\sqrt{x}-1}{x-3\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)-2x-\left(x-9\right)}{x-9}:\frac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\frac{-2x-3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{-\sqrt{x}+5}\)

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

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

ĐKXĐ :\(x\) > 0 , x\(\ne9\)

\(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{2x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}-1\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{2}{\sqrt{x}}\right)=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+2x-\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)=\)\(\frac{3\sqrt{x}-x+2x-9+x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x-3}\right)}{\sqrt{x}-1-2\sqrt{x}+6}=\frac{2x-3\sqrt{x}-9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{5-\sqrt{x}}=\frac{\left(\sqrt{x}+3\right)\left(2\sqrt{x}-3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{5-\sqrt{x}}=\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\sqrt{x}-5}\)

câu 2

\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)

câu 1

\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)

\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

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

\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)

1) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

b) \(P=\frac{\sqrt{x}-1}{\sqrt{x}+2}< 0\)

Dễ thấy \(\sqrt{x}+2\ge2>0\forall x\ge0\)

Nên để \(P< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)

Vậy với \(0\le x< 1\)thì P<0

(Câu trả lời bằng hình ảnh)

Có vẻ hum nào cũng có bào nhề