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P/s : sửa đề
ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
a) \(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(P=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(P=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(P=\frac{-3\sqrt{x}-3x}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(P=\frac{-3\sqrt{x}\left(1+\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{-3\sqrt{x}}{\sqrt{x}+3}\)
b) \(P< -\frac{1}{2}\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}+\frac{1}{2}< 0\)
\(\Leftrightarrow\frac{-6\sqrt{x}+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)
\(\Leftrightarrow\frac{-5\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)
Mà \(2\left(\sqrt{x}+3\right)>0\)
\(\Rightarrow-5\sqrt{x}+3< 0\)
\(\Leftrightarrow-5\sqrt{x}< -3\)
\(\Leftrightarrow\sqrt{x}>\frac{3}{5}\)
\(\Leftrightarrow x>\frac{9}{25}\)
Vấy .................
c) \(P.\left(\sqrt{x}+3\right)+2\sqrt{x}-2+x=2\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}\left(\sqrt{x}+3\right)+2\sqrt{x}-2+x=2\)
\(\Leftrightarrow-3\sqrt{x}+2\sqrt{x}-2-2+x=0\)
\(\Leftrightarrow-\sqrt{x}-4+x=0\)
\(\Leftrightarrow-\sqrt{x}\left(1-\sqrt{x}\right)=4\)
Còn lại lập bảng tự tìm giá trị của x là ra .( Chú ý : đối chiếu ĐKXĐ )
d)
\(P.\left(\sqrt{x}+3\right)+x\left(\sqrt{x}-m\right)=x-\sqrt{x}\left(3+m\right)\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}\left(\sqrt{x}+3\right)+x\sqrt{x}-xm=x-3\sqrt{x}-m\sqrt{x}\)
\(\Leftrightarrow-3\sqrt{x}+x\sqrt{x}-xm-x+3\sqrt{x}+m\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(x+m\right)-x\left(m+1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left[x+m-m\sqrt{x}-\sqrt{x}\right]=0\)
\(\Leftrightarrow\sqrt{x}\left[m\left(1-\sqrt{x}\right)-\sqrt{x}\left(1-\sqrt{x}\right)\right]=0\)
\(\Leftrightarrow\sqrt{x}=0;m-\sqrt{x}=0;1-\sqrt{x}=0\)
+) \(\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\)
+) \(1-\sqrt{x}=0\)
\(\Leftrightarrow x=1\left(TM\right)\)
+) \(m-\sqrt{x}=0\)
\(\Leftrightarrow\orbr{\begin{cases}m-\sqrt{0}=0\\m-\sqrt{1}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}m=0\\m=1\end{cases}}}\)
Vậy ..................
a) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\\x\ne9\end{cases}}\)
b) \(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x-3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+3}{2\left(\sqrt{x}-1\right)}=\frac{-3\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=-\frac{3}{2\left(\sqrt{x}-3\right)}\)c) Để P nguyên thì \(2\left(\sqrt{x}-3\right)\in\left\{-3;-1;1;3\right\}\)=> x thuộc rỗng.
\(P=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(P=\left(\frac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(P=\left(\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(P=\frac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-5x\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(P=\frac{4\sqrt{x}\left(2+5x\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(P=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(P=\frac{-4x}{3-\sqrt{x}}\)
\(P=\frac{4x}{\sqrt{x}-3}\)
Có:
\(m\left(\sqrt{x}-3\right)P>x+1\)
\(\Leftrightarrow m\left(\sqrt{x}-3\right).\frac{4x}{\sqrt{x}-3}>x+1\)
\(\Leftrightarrow4mx>x+1\)
\(\Leftrightarrow4mx-x>1\)
\(\Leftrightarrow\left(4m-1\right)x>1\)
\(\Leftrightarrow x>\frac{1}{4m-1}\)
Lại có:
\(x>9\)
\(\Rightarrow\frac{1}{4m-1}< 9\)
\(\Leftrightarrow1< 9\left(4m-1\right)\)
\(\Leftrightarrow1< 36m-1\)
\(\Leftrightarrow10< 36m\)
\(\Leftrightarrow m< \frac{5}{18}\)
Ta có: \(M=\frac{\sqrt{x}\left(3-\sqrt{x}\right)+2x}{9-x}:\frac{\sqrt{x}-2-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{x+3\sqrt{x}}{9-x}:\frac{4-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{9-x}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{4-\sqrt{x}}=\frac{x}{\sqrt{x}-4}\)
Khi x > 16 thì \(\sqrt{x}-4>0\), như vậy \(M>y\Leftrightarrow x>m-3x+1\Leftrightarrow4x-1>m\) với mọi x > 16. Vậy m < 15 thì \(M>y\) với mọi x > 16.
Chúc em học tốt ^^
em cám ơn ạk