bài 7:
a: \(G=\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}\)
\(=\frac{3}{x+3}+\frac{1}{x-3}+\frac{18}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3\left(x-3\right)+x+3+18}{\left(x-3\right)\left(x+3\right)}=\frac{3x-9+x+21}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{4x+12}{\left(x-3\right)\left(x+3\right)}=\frac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{4}{x-3}\)
b: G=4
=>\(\frac{4}{x-3}=4\)
=>x-3=1
=>x=4(nhận)
Bài 6:
a: \(F=\frac{1}{x+5}+\frac{2}{x-5}-\frac{2x+10}{\left(x+5\right)\left(x-5\right)}\)
\(=\frac{1}{x+5}+\frac{2}{x-5}-\frac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}\)
\(=\frac{1}{x+5}+\frac{2}{x-5}-\frac{2}{x-5}=\frac{1}{x+5}\)
b: F=-3
=>\(x+5=-\frac13\)
=>\(x=-\frac13-5=-\frac{16}{3}\) (nhận)
\(9x^2-42x+49\)
\(=\left(3x\right)^2-2\cdot3x\cdot7+7^2\)
\(=\left(3x-7\right)^2=\left(3\cdot\frac{-16}{3}-7\right)^2=\left(-16-7\right)^2=\left(-23\right)^2=529\)
Bài 3:
a: ĐKXĐ: \(9x^2-6x+1<>0\)
=>\(\left(3x-1\right)^2<>0\)
=>3x-1<>0
=>3x<>1
=>x<>1/3
b: Thay x=-8 vào C, ta được:
\(C=\frac{3\cdot\left(-8\right)^2-\left(-8\right)}{9\left(-8\right)^2-6\cdot\left(-8\right)+1}=\frac{3\cdot64+8}{9\cdot64+6\cdot8+1}=\frac{200}{625}=\frac{8}{25}\)
c: \(C=\frac{3x^2-x}{9x^2-6x+1}\)
\(=\frac{x\left(3x-1\right)}{\left(3x-1\right)^2}=\frac{x}{3x-1}\)
d: C<0
=>\(\frac{x}{3x-1}<0\)
=>0<x<1/3
