Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right):\dfrac{2\sqrt{x}}{x-4}\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\left(\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\)
\(=\sqrt{x}\)
b) Để P>4 thì \(\sqrt{x}>4\)
hay x>16
Kết hợp ĐKXĐ, ta được: x>16
Vậy: Khi x>16 thì P>4
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Ta có: \(P=\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{2}{\sqrt{x}-2}-\dfrac{4\sqrt{x}}{x-4}\)
\(=\dfrac{x-2\sqrt{x}+2\sqrt{x}+4-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(B=\dfrac{6-7x}{x^2-4}+\dfrac{3}{x+2}-\dfrac{2}{2-x}\)
\(=\dfrac{6-7x+3x-6+2x+4}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{-2x+4}{\left(x+2\right)\left(x-2\right)}\)
\(=-\dfrac{2}{x+2}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
b: Ta có: \(D=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{5\sqrt{x}+5}{x-4}\right)\cdot\dfrac{x-4}{\sqrt{x}}\)
\(=\dfrac{x+4\sqrt{x}+4-x+4\sqrt{x}-5\sqrt{x}-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{x-4}{\sqrt{x}}\)
\(=\dfrac{3\sqrt{x}-1}{\sqrt{x}}\)
\(a,ĐK:x\ne\pm1;x\ne0\\ M=\dfrac{1-x+2x}{\left(1+x\right)\left(1-x\right)}:\dfrac{1-x}{x}\\ M=\dfrac{x+1}{\left(x+1\right)\left(1-x\right)}\cdot\dfrac{x}{1-x}=\dfrac{x}{\left(1-x\right)^2}\\ b,ĐK:x\ge0;x\ne4\\ N=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ N=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
Tất cả đều phải tìm điều kiện
a: ĐKXĐ: \(x\notin\left\{-1;3\right\}\)
Ta có: \(A=\dfrac{x^3-3}{x^2-2x-3}+\dfrac{6-2x}{x+1}+\dfrac{x+3}{3-x}\)
\(=\dfrac{x^3-3-2\left(x-3\right)^2-\left(x+3\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}\)
\(=\dfrac{x^3-3-2x^2+12x-18-x^2-4x-3}{\left(x-3\right)\left(x+1\right)}\)
\(=\dfrac{x^4-3x^2+8x-24}{\left(x-3\right)\left(x+1\right)}\)
\(=\dfrac{x^2\left(x-3\right)+8\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}\)
\(=\dfrac{x^2+8}{x+1}\)
b: Ta có: A=x-2
\(\Leftrightarrow x^2+8=x^2-x-2\)
\(\Leftrightarrow8+x+2=0\)
hay x=-10
a) ĐKXĐ:
\(\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}+2>0\\\sqrt{4x}>0\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\sqrt{x}>2\\\sqrt{x}>-2\\2\sqrt{x}>0\end{matrix}\right.\\\rightarrow \left\{{}\begin{matrix}x>\sqrt{2}\\x>-\sqrt{2}\\x>0\end{matrix}\right.\\ \rightarrow x>\sqrt{2}\)
Vậy \(x>\sqrt{2}\)
b)
\(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right).\dfrac{x-4}{\sqrt{4x}}\\ =\left[\dfrac{\sqrt{x}.\left(\sqrt{x}+2\right)+\sqrt{x}.\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right].\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\\ =\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\\ =\dfrac{2x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\\ =\dfrac{2x}{2\sqrt{x}}=\dfrac{x}{\sqrt{x}}=\dfrac{\sqrt{x}.\sqrt{x}}{\sqrt{x}}=\sqrt{x}\)
Vậy \(M=\sqrt{x}\)
a) ĐKXĐ:
\(\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}+2>0\\\sqrt{4x}>0\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\sqrt{x}>2\\\sqrt{x}>-2\\2\sqrt{x}>0\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}x>4\\x>-4\\x>0\end{matrix}\right.\\ \rightarrow x>4\)
Vậy \(x>4\)
a) Để M xác định thì \(\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
b) \(M=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-1\right)\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-1\right)\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=x-1\)
mơn nha