Bài 6:
a: \(B=\frac{x}{x+3}+\frac{2x}{x-3}-\frac{9-3x^2}{9-x^2}\)
\(=\frac{x\left(x-3\right)+2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3x^2-9}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2-3x+2x^2+6x-3x^2+9}{\left(x-3\right)\left(x+3\right)}=\frac{3x+9}{\left(x-3\right)\left(x+3\right)}=\frac{3}{x-3}\)
b: |2x+1|=5
=>2x+1=5 hoặc 2x+1=-5
=>2x=4 hoặc 2x=-6
=>x=2(nhận) hoặc x=-3(loại)
Thay x=2 vào B, ta được:
\(B=\frac{3}{2-3}=\frac{3}{-1}=3\)
c: \(B\left(x+10\right)-6=0\)
=>\(\frac{3\left(x+10\right)}{x-3}-6=0\)
=>3(x+10)-6(x-3)=0
=>x+10-2(x-3)=0
=>x+10-2x+6=0
=>16-x=0
=>x=16(nhận)
Bài 5:
a: Thay x=-6 vào B, ta được;
\(B=\frac{-6+4}{-6+2}=\frac{-2}{-4}=\frac12\)
b: \(A=\frac{4}{x+2}+\frac{3}{x-2}+\frac{-5x-2}{x^2-4}\)
\(=\frac{4\left(x-2\right)+3\left(x+2\right)-5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x-8+3x+6-5x-2}{\left(x-2\right)\left(x+2\right)}=\frac{2x-4}{\left(x-2\right)\left(x+2\right)}=\frac{2}{x+2}\)
c: C=A:B
\(=\frac{2}{x+2}:\frac{x+4}{x+2}=\frac{2}{x+4}\)
\(C\left(x^2-16\right)=5x+7\)
=>\(\frac{2\left(x-4\right)\left(x+4\right)}{x+4}=5x+7\)
=>5x+7=2(x-4)=2x-8
=>3x=-15
=>x=-5(nhận)
Bài 1 :
`A=((x^2-3)/(x^2-9)+1/(x-3)) : x/(x+3)`
`A=((x^2-3)/((x-3)(x+3)) + (x+3)/((x-3)(x+3))) * ( x+3)/x`
`A=(x^2-3+x+3)/((x-3)(x+3)) * (x+3)/x`
`A=(x*(x+1))/((x-3)(x+3)) * (x+3)/x`
`A=(x+1)/(x-3)`
`b)` thay `x=4` ta có :
`A=(4+1)/(4-3) = 5`
`c)`Thay `A=3`
`=> (x+1)/(x-3)=3`
`=> x+1 = 3x-9`
`=> -2x = -10`
`=> x=5`
Bài 3:
` A= ((a+2)/(x+1) - (a-2)/(a-1)) * (a+1)/a`
`A= (((a+2)(a-1))/((a+1)(a-1)) - ((a-2)(a+1))/((a-1)(a+1))) * (a+1)/a`
`A= ((a^2+a-2)/((a-1)(a+1)) - (a^2-a-2)/((a-1)(a+1))) * (a+1)/a`
`A = (a^2+a-2 -a^2 +a +2)/((a-1)(a+1)) * (a+1)/a`
`A= 2a/((a+1)(a-1)) * (a+1)/a = 2/(a-1)`
`b)`để `B=1` thì :
`3/(a^2 - 1) =1`
`=> a^2 - 1 = 3`
`=> a^2 = 4`
`=> a = 2` hoặc `a=-2`
`c)` Có `A=2B`
`=> 2/(a-1) = 2*(3/(a^2-1))`
`=> 2(a-1) = 6/((a-1)(a+1))`
`=> (2a+2)/((a-1)(a+1) = 6/((a-1)(a+1))`
`=> 2x+2=6`
`=> 2x=4`
`=> x=2`
