Question

 Cho biểu thức :B=\(\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)

 a) Rút gọn B 

b) Tim x de B =\(\frac{-3}{5}\)

c)Tim x de B<0

Answer 1

\(B=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-3\right)\left(x-1\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+3-1}{x+3}\)\(=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}=\frac{3\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}=\frac{3}{x-3}\)Điều kiện \(x\ne3\) \(\Rightarrow\frac{-3}{5}=\frac{3}{x-3}\Leftrightarrow x-3=-5\Leftrightarrow x=-2\)\(B=\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)

Answer 2

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

Answer 3

\(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{x+3}\right):\left(1-\frac{1}{x+3}\right)\)

\(B=\left[\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x-3\right)}{\left(x+1\right)\left(x+3\right)}\right]\)    \(:\left[\frac{x+3-1}{x+3}\right]\)

\(B=\frac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}:\frac{x+2}{x+3}\)

\(B=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}\)

\(B=\frac{3.\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}\)

\(B=\frac{3}{x-3}\)

b)   \(B=\frac{-3}{5}\Leftrightarrow\frac{3}{x-3}=\frac{-3}{5}\)

\(\Leftrightarrow-3x+9=15\)

\(\Leftrightarrow-3x=6\)

\(\Leftrightarrow x=-2\)

vậy....

c) \(B< 0\Leftrightarrow\frac{3}{x-3}< 0\)

\(\Leftrightarrow x-3< 0\)   vì \(3>0\)

\(\Leftrightarrow x< 3\)

vậy....