Tim x\(3< \left|x-1\right|< 5\)
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a) +) Nếu 6,8 - x > 0 => 6,8 > x thì |6,8 - x| = 6,8 - x
=> 6,8 - x > \(\frac{3}{5}\) <=> 6,8 - \(\frac{3}{5}\) > x => 6,2 > x. Kết hợp => x < 6,2
+) Nếu 6,8 - x < 0 => 6,8 < x thì |6,8 - x| = -(6,8 - x) = x - 6,8
=> x - 6,8 > \(\frac{3}{5}\) => x > 0,6 + 6,8 => x > 7,4. Kết hợp => x > 7,4
Vậy ....

- \(B=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-3\right)\left(x-1\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+3-1}{x+3}\)\(=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}=\frac{3\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}=\frac{3}{x-3}\)
- Điều kiện \(x\ne3\) \(\Rightarrow\frac{-3}{5}=\frac{3}{x-3}\Leftrightarrow x-3=-5\Leftrightarrow x=-2\)
- \(B=\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)
a) B=(\(\frac{21}{x^2-9}\)-\(\frac{x-4}{3-x}\)-\(\frac{x-1}{3+x}\)) : (1-\(\frac{1}{x+3}\)) (ĐK: x khác +-3)
=(\(\frac{21}{\left(x-3\right).\left(x+3\right)}\)+\(\frac{x-4}{x-3}\)-\(\frac{x-1}{x+3}\)) : (1-\(\frac{1}{x+3}\))
=(\(\frac{21+\left(x+4\right).\left(x+3\right)-\left(x-1\right).\left(x-3\right)}{\left(x-3\right).\left(x+3\right)}\):(\(\frac{x+3-1}{x+3}\))
=(\(\frac{3x+6}{\left(x-3\right).\left(x+3\right)}\)) . (\(\frac{x+3}{x+2}\))
=(\(\frac{3.\left(x+2\right)}{\left(x-3\right).\left(x+3\right)}\). \(\frac{x+3}{x+2}\)
=\(\frac{3}{x-3}\)
b) B=\(\frac{3}{x-3}\)=\(\frac{-3}{5}\)
(=) \(\frac{3.5}{x-3}\)=-3
(=) -3.(x-3) = 15
(=) -3x=6
(=) x=-2
vậy x=2 thì B=\(\frac{-3}{5}\)
c) B=\(\frac{3}{x-3}\)<0
(=) 3 < x - 3
(=) -x < - 3 - 3
(=) x > 6
Vậy với x > 6 thì B < 0
Ta có : 3 < | x - 1| < 5
\(\Rightarrow\) |x - 1| = 4
\(\Rightarrow\) x - 1 = 4 \(\Rightarrow\) x = 5
x - 1 = - 4 \(\Rightarrow\) x = - 3
Vậy x \(\in\){5; -3}
\(3< \left|x-1\right|< 5\)
\(\Leftrightarrow\orbr{\begin{cases}3< x-1< 5\\-3>x-1>-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}4< x< 6\\-2>x>-4\end{cases}}\)