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Bài Làm:
a, \(P=\frac{x+3}{\sqrt{x}-2}:\left(\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{x-4}\right)\)
\(=\frac{x+3}{\sqrt{x}-2}:\left(\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(=\frac{x+3}{\sqrt{x}-2}:\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x-3}{\sqrt{x}-2}:\frac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x-3}{\sqrt{x}-2}:\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+3}{\sqrt{x}-2}:\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{x+3}{\sqrt{x}-2}:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(=\frac{x+3}{\sqrt{x}-2}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{x+3}{\sqrt{x}}\)
a/ ĐKXĐ:...
\(E=\left(\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\left(x-1\right)}{x-1}\right):\left(\frac{x-1}{\sqrt{x}}\right)\)
\(E=\left(\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+4x\sqrt{x}-4\sqrt{x}}{x-1}\right).\frac{\sqrt{x}}{x-1}\)
\(E=\frac{4x^2}{\left(x-1\right)^2}\)
Bn ơi! Kia là chia \(\sqrt{x}-\frac{1}{\sqrt{x}}\) hay nhân z? Bn xem lại đề bài nhé! Theo mk là nhân thì nó sẽ ra kết quả ngắn gọn hơn nhìu :D
Bài 1:
a/ ĐKXĐ: \(x\ge2;x\ne11\)
b/ \(P=\frac{\left(x-5\right)\left(\sqrt{x-2}+\sqrt{3}\right)}{x-2-3}=\sqrt{x-2}+\sqrt{3}\)
c/ \(\sqrt{x-2}\ge0\forall x\in R\Rightarrow P=\sqrt{x-2}+\sqrt{3}\ge\sqrt{3}\forall x\in R\)
"="\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
\(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a/ \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt[]{x-3}\right)}\right]:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt[]{x-3}}\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}+\sqrt{x}-3}{\sqrt{x}-3}\right].\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
=> \(R=\frac{3\sqrt{x}-3}{\sqrt{x}-3}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
b/ Để R<-1 => \(\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< -1\)
<=> \(3\sqrt{x}-3< -\sqrt{x}-1\)
<=> \(4\sqrt{x}< 2\)=> \(\sqrt{x}< \frac{1}{2}\) => \(-\frac{1}{4}< x< \frac{1}{4}\)
Chỗ => R = \(\left(\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\) là sao vậy ạ?
em viết nhầm đề nha.M = \(\frac{y}{\sqrt{xy}-x}+\frac{x}{\sqrt{xy}+y}-\frac{x+y}{\sqrt{xy}}\)mới đúng
\(M=\frac{-5x-5}{\sqrt{x}-2}\)
\(\Leftrightarrow M\left(\sqrt{x}-2\right)=-5x-5\)
\(\Leftrightarrow5x+5+M\sqrt{x}-2M=0\)
\(\Leftrightarrow5x+\sqrt{x}\cdot M+\left(5-2M\right)=0\)
\(\Delta=M^2-4\cdot5\cdot\left(5-2M\right)\)
\(=M^2+40M-100\)
Pt có nghiệm khi \(\Delta\ge0\)
\(\Leftrightarrow M^2+40M+400-500\ge0\)
\(\Leftrightarrow\left(M+20\right)^2\ge500\)
\(\Leftrightarrow M+20\ge\sqrt{500}\)
\(\Leftrightarrow M\ge10\sqrt{5}-20\)
Dấu "=" khi tự tìm.