Cho biểu thức N = \(\left(\frac{x+3}{x-3}+\frac{18}{9-x^2}+\frac{x-3}{x+3}\right):\left(1-\frac{x+1}{x+3}\right)\)
a) Rút gọn N
b) Tìm x để N = \(\frac{-1}{2}\)
c) Tìm x để N < 0
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17 tháng 10 2018
\(a)\)\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt{x-3}}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}-3}{\sqrt{x}-3}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\frac{3\sqrt{x}+3}{\sqrt{x}+3}.\frac{\sqrt{x}-3}{\sqrt{x+1}}\)
\(R=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
\(b)\) Ta có : \(R< -1\)
\(\Leftrightarrow\)\(\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}< -1\)
\(\Leftrightarrow\)\(\frac{\sqrt{x}-3}{\sqrt{x}+3}< \frac{-1}{3}\)
\(\Leftrightarrow\)\(3\sqrt{x}-9< -\sqrt{x}-3\)
\(\Leftrightarrow\)\(4\sqrt{x}< 6\)
\(\Leftrightarrow\)\(\sqrt{x}< \frac{3}{2}\)
\(\Leftrightarrow\)\(x< \frac{9}{4}\)
Chúc bạn học tốt ~
VH
22 tháng 8 2020
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 .................
VH
22 tháng 8 2020
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) Ta có: \(N=\left(\frac{x+3}{x-3}+\frac{18}{9-x^2}+\frac{x-3}{x+3}\right):\left(1-\frac{x+1}{x+3}\right)\)
\(=\left(\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\frac{18}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}\right):\left(\frac{x+3}{x+3}-\frac{x+1}{x+3}\right)\)
\(=\frac{x^2+6x+9-18-\left(x^2-6x+9\right)}{\left(x-3\right)\left(x+3\right)}:\frac{2}{x+3}\)
\(=\frac{x^2+6x-9-x^2+6x-9}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{2}\)
\(=\frac{12x-18}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{2}\)
\(=\frac{12x-18}{x-3}\cdot\frac{1}{2}\)
\(=\frac{12x-18}{2x-6}\)
b)
ĐKXĐ: \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
Đặt \(N=-\frac{1}{2}\)
\(\Leftrightarrow\frac{12x-18}{2x-6}=-\frac{1}{2}\)
\(\Leftrightarrow12x-18=\frac{6-2x}{2}\)
\(\Leftrightarrow12x-18=3-x\)
\(\Leftrightarrow12x-18-3+x=0\)
\(\Leftrightarrow13x-21=0\)
\(\Leftrightarrow13x=21\)
hay \(x=\frac{21}{13}\)(tm)
Vậy: Khi \(N=-\frac{1}{2}\) thì \(x=\frac{21}{13}\)
c) Để N<0 thì 12x-18 và 2x-6 khác dấu
*Trường hợp 1:
\(\left\{{}\begin{matrix}12x-18>0\\2x-6< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12x>18\\2x< 6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\frac{3}{2}\\x< 3\end{matrix}\right.\)\(\Leftrightarrow\frac{3}{2}< x< 3\)
*Trường hợp 2:
\(\left\{{}\begin{matrix}12x-18< 0\\2x-6>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12x< 18\\2x>6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \frac{3}{2}\\x>3\end{matrix}\right.\)(vô lý)
Vậy: Khi N<0 thì \(\frac{3}{2}< x< 3\)