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a) \(B=\left[\frac{21}{\left(x+3\right)\left(x-3\right)}+\frac{x-4}{x-3}-\frac{\left(x-1\right)}{x+3}\right]:\left(\frac{x+3-1}{x+3}\right)\)
ĐK: \(\hept{\begin{cases}x\ne3\\x\ne-3\end{cases}}\)
\(=\left[\frac{21+x-4-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right]:\left(\frac{x+2}{x+3}\right)\)
\(=\left[\frac{21+x-4-x^2+3x+x-3}{\left(x+3\right)\left(x-3\right)}\right]\times\left(\frac{x+3}{x+2}\right)\)
\(=\left(\frac{-x^2+5x+14}{x-3}\right)\left(\frac{1}{x+2}\right)\)
\(=\frac{-\left(x^2+2x-7x-14\right)}{\left(x-3\right)\left(x+2\right)}\)
\(=\frac{-\left(x+2\right)\left(x-7\right)}{\left(x-3\right)\left(x+2\right)}\)
\(=\frac{7-x}{x-3}\)
b) \(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
Mà \(x\ne-3\)
\(\Rightarrow x=2\)
Thế \(x=2\)vào B ta được:
\(B=\frac{7-2}{2-3}=-5\)
c) \(B=\frac{7-x}{x-3}=\frac{-3}{5}\)
\(\Leftrightarrow5\left(7-x\right)=-3\left(x-3\right)\)
\(\Leftrightarrow35-5x+3x-9=0\)
\(\Leftrightarrow-2x=-26\)
\(\Leftrightarrow x=13\)
Vậy để \(B=\frac{-3}{5}\)thì \(x=13\)
d) B<0\(\Rightarrow\frac{7-x}{x-3}< 0\)
TH1: \(\hept{\begin{cases}7-x< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x>7\\x>3\end{cases}\Rightarrow}x>7}\)
TH2: \(\hept{\begin{cases}7-x>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 7\\x< 3\end{cases}\Rightarrow}x< 3}\)
Để B<0 thì x>7 hoặc x<3
a) \(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\) ĐKXĐ: x khác =-3; x khác -2
\(B=\frac{21+x^2-x-12-x^2+4x-3}{\left(x+3\right)\left(x-3\right)}:\frac{x+2}{x+3}\)
\(B=\frac{3x+6}{\left(x+3\right)\left(x-3\right)}:\frac{x+2}{x+3}\)
\(B=\frac{3\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{x+2}\)
\(B=\frac{3}{x-3}\)
b) bước đầu tiên ta phải tìm x:
\(\left|2x+1\right|=5\)
TH1: 2x+1=5 TH2: 2x+1=-5
2x=4 2x=-6
x=2 (nhận) x=-3 (loại)
thay x=2 vào biểu thức B, ta được:
\(B=\frac{3}{2-3}=\frac{3}{-1}=-3\)
vậy B=-3 tại x=2
c) Để \(B=-\frac{3}{5}\)thì \(\frac{3}{x-3}=-\frac{3}{5}\)
\(\Leftrightarrow-3\left(x-3\right)=15\)
\(\Leftrightarrow x-3=-5\)
\(\Leftrightarrow x=-2\)
vậy \(x=-2\)thì \(B=-\frac{3}{5}\)
d) để B<0 thì \(\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)
vậy để B<0 thì x phải < 3 và x khác -3
Bài 2:
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)
b: Thay x=1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)
Thay x=-1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)
c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)
=>6(x-2)=-1/2
=>x-2=-1/12
hay x=23/12
a/ Ta có \(A=\frac{\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}}{1-\frac{x}{x+2}}\)với \(\hept{\begin{cases}x\ne\pm2\\x\ne0\end{cases}}\)
\(A=\frac{\frac{x}{x^2-4}+\frac{x-2-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}}{\frac{x+2-x}{x+2}}\)
\(A=\frac{\frac{x}{x^2-4}+\frac{x-2-2x-4}{x^2-4}}{\frac{2}{x+2}}\)
\(A=\frac{\frac{x-x-6}{x^2-4}}{\frac{2}{x+2}}\)
\(A=\frac{-6}{x^2-4}.\frac{x+2}{2}\)
\(A=\frac{-3}{x-2}\)
b/ Ta có \(x=-4\)thoả mãn ĐKXĐ
Vậy với \(x=-4\):
\(A=\frac{-3}{x-2}=\frac{-3}{-4-2}=\frac{1}{2}\)
c/ Khi \(A\inℤ\)
=> \(\frac{-3}{x-2}\inℤ\)
=> \(-3⋮\left(x-2\right)\)
=> x - 2 là ước của -3
Ta có bảng sau:
| x-2 | -1 | -2 | -3 | -6 | 1 | 2 | 3 | 6 |
| x | 1 | 0 | -1 | -4 | 3 | 4 | 5 | 8 |
Mà ĐKXĐ \(\hept{\begin{cases}x\ne\pm2\\x\ne0\end{cases}}\)
=> \(x\in\left\{\pm1;\pm4;3;5;8\right\}\)
Vậy khi \(x\in\left\{\pm1;\pm4;3;5;8\right\}\)thì \(A\inℤ\).
A=(1/x-2 - (2x/(2-x)(2+x) - 1/2+x) ) *(2-x)/x
=(1/x-2 - x^2+5x-2/(2-x)(2+x))*2-x/x
=(-x^3-4x^2+12x/(x-2)(2-x)(2+x))*2-x/x
= - x(x-2)(x+6)(2-x)/x(x-2)(2-x)(2+x)
= - x+6/x+2
\(B=\left(\frac{1-x^3}{1-x}-x\right):\frac{1-x^2}{1-x-x^2+x^3}\) \(ĐKXĐ:x\ne\pm1\)
\(B=\left[\frac{\left(1-x\right)\left(x^2+x+1\right)}{\left(1-x\right)}-x\right]:\frac{\left(x+1\right)\left(1-x\right)}{\left(1-x\right)-x^2\left(1-x\right)}\)
\(B=\left(x^2+x+1-x\right):\frac{\left(x+1\right)\left(1-x\right)}{\left(1-x\right)\left(1-x^2\right)}\)
\(B=\left(x^2+1\right):\frac{x+1}{\left(x+1\right)\left(1-x\right)}\)
\(B=\frac{x^2+1}{1-x}\)
vậy \(B=\frac{x^2+1}{1-x}\)
b) \(x=-1\frac{2}{3}\)
\(x=\frac{-5}{3}\)
khi đó \(B=\frac{\left(\frac{-5}{3}\right)^2+1}{1+\frac{5}{3}}\)
\(B=\frac{\frac{25}{9}+1}{\frac{8}{3}}\)
\(B=\frac{34}{9}:\frac{8}{3}\)
\(B=\frac{17}{12}\)
vậy \(B=\frac{17}{12}\) khi \(x=-1\frac{2}{3}\)
c) \(B< 0\Leftrightarrow\frac{x^2+1}{1-x}< 0\)
\(\Leftrightarrow\hept{\begin{cases}x^2+1>0\\1-x< 0\end{cases}}\)hoặc \(\hept{\begin{cases}x^2+1< 0\\1-x>0\end{cases}}\)
đến đây bạn giải tiếp