Question

Cho biểu thức A=\(\left(\frac{1}{x+2}-\frac{2}{x-2}-\frac{x}{4-x^2}\right):\frac{6.\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)

a) Rút gọn biểu thức A

b) Tìm x để A > 0

c) Tìm x biết \(x^2+3x+2=0\)

d) Tìm x để A đạt GTLN, tìm GTLN đó.

Answer 1

a)Có A=\(\left(\frac{1}{x+2}-\frac{2}{x-2}-\frac{x}{4-x^2}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)(ĐKXĐ \(x\ne2,-2,-1\))

=\(\left(\frac{2-x}{\left(2-x\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(2-x\right)\left(x+2\right)}-\frac{x}{\left(2-x\right)\left(2+x\right)}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)

=\(\frac{2-x+2x+4-x}{\left(2-x\right)\left(x+2\right)}.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)

=\(\frac{6\left(2-x\right)\left(x+1\right)}{6\left(2-x\right)\left(x+2\right)^2}\)

=\(\frac{x+1}{\left(x+2\right)^2}\)

b)Có A=\(\frac{x+1}{\left(x+2\right)^2}\)

Để A>0 <=> x+1>0 <=>x>-1

c) Có x²+3x+2=0

<=> x²+2x+x+2=0

<=> x(x+2)+(x+2)=0

<=>(x+1)(x+2)=0

<=> x=-1 hoặc x=-2