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- A =\(\frac{x^2+3+2x-6-x-3}{x^2-9}\)
- A =\(\frac{x^2+x-6}{x^2-9}\)
- A = \(\frac{\left(x+3\right)\left(x-2\right)}{\left(x+3\right)\left(x-3\right)}\)
- A = \(\frac{x-2}{x-3}\)
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 \(\frac{3}{2}< x< 3\)
\(e ) Để \) \(M\)\(\in\)\(Z \) \(thì\) \(1 \)\(⋮\)\(x +3\)
\(\Leftrightarrow\)\(x + 3 \)\(\in\)\(Ư\)\((1)\)\(= \) { \(\pm\)\(1 \) }
\(Lập\) \(bảng :\)
| \(x +3\) | \(1\) | \(- 1\) |
| \(x\) | \(-2\) | \(- 4\) |
\(Vậy : Để \) \(M\)\(\in\)\(Z\) \(thì\) \(x\)\(\in\){ \(- 4 ; - 2\) }
e) Để M \(\in\)Z <=> \(\frac{1}{x+3}\in Z\)
<=> 1 \(⋮\)x + 3 <=> x + 3 \(\in\)Ư(1) = {1; -1}
Lập bảng:
| x + 3 | 1 | -1 |
| x | -2 | -4 |
Vậy ....
f) Ta có: M > 0
=> \(\frac{1}{x+3}\) > 0
Do 1 > 0 => x + 3 > 0
=> x > -3
Vậy để M > 0 khi x > -3 ; x \(\ne\)3 và x \(\ne\)-3/2
mk làm luôn
a)\(A=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right).\)
=\(\frac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)
=\(\frac{\left(3x+3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right).3}\)
=\(\frac{3x+3\sqrt{x}-1}{9\sqrt{x}-3}\)
=
Ta có : A=20+21+22+23+...+22010A=20+21+22+23+...+22010
3A=2+22+23+24+...+220113A=2+22+23+24+...+22011
=> 2A=3A−A=(21+22+...+22011)−(20+21+...+22010)
=>2A=22011−12A=22011−1
=>A=22011−12A=22011−12
=> A < B ( vì 22011−12<2201122011−12<22011 )
Bexiu bạn đang làm cái j thế!!?