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1) Bạn đánh nhầm \(\sqrt{x}+3\rightarrow\sqrt{x+3}\); \(\sqrt{x}-3\rightarrow\sqrt{x-3}\)
Sửa : \(ĐKXĐ:x\ne\pm\sqrt{3}\)
a) \(M=\frac{x-\sqrt{x}}{x-9}+\frac{1}{\sqrt{x}+3}-\frac{1}{\sqrt{x}-3}\)
\(\Leftrightarrow M=\frac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow M=\frac{x-\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow M=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow M=\frac{\sqrt{x}+2}{\sqrt{x}+3}\)
b) Để \(M=\frac{3}{4}\)
\(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}+3}=\frac{3}{4}\)
\(\Leftrightarrow4\sqrt{x}+8=3\sqrt{x}+9\)
\(\Leftrightarrow\sqrt{x}-1=0\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1\)(tm)
Vậy để \(A=\frac{3}{4}\Leftrightarrow x=1\)
c) Khi x = 4
\(\Leftrightarrow M=\frac{\sqrt{4}+2}{\sqrt{4}+3}\)
\(\Leftrightarrow M=\frac{2+2}{2+3}\)
\(\Leftrightarrow M=\frac{4}{5}\)
Vậy khi \(x=4\Leftrightarrow M=\frac{4}{5}\)
Bài 1 :
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
\(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}-2}\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
b) Để \(A< -1\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< -1\)
\(\Leftrightarrow\sqrt{x}-2< -\sqrt{x}-1\)
\(\Leftrightarrow2\sqrt{x}< 1\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{2}\)
\(\Leftrightarrow x< \frac{1}{4}\)
Vậy để \(A< -1\Leftrightarrow x< \frac{1}{4}\)
\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)\(\left(ĐKXĐ:x\ne4\right)\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
b) Với \(x=3\)( thỏa mãn ĐKXĐ ) ta có \(P=\frac{3\sqrt{3}}{\sqrt{3}+2}=-9+6\sqrt{3}\)
c) A ở đâu ???? '-'
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
\(B=\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac{7\sqrt{x}-9}{x-9}\\ =\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{7\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\frac{x+3\sqrt{x}-\sqrt{x}-3-7\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\frac{x-5\sqrt{x}+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+3}\)
b) ĐKXĐ: x>0
\(x=\frac{1}{\sqrt{2}-1}-\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}+1-\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{2}{\left(\sqrt{2}\right)^2-1}=\frac{2}{2-1}=2\)
\(\Rightarrow A=\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{\sqrt{2}-2}{\sqrt{2}}=\frac{\sqrt{2}\left(1-\sqrt{2}\right)}{\sqrt{2}}=1-\sqrt{2}\)
c)
\(P=\frac{A}{B}=\frac{\frac{\sqrt{x}-2}{\sqrt{x}+3}}{\frac{\sqrt{x}-2}{\sqrt{x}}}=\frac{\sqrt{x}-2}{\sqrt{x}+3}\cdot\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{\sqrt{x}}{\sqrt{x}+3}\)
Còn khúc sau ko hiểu cho lắm ._.
giúp mình đi huhu