Câu 1: A
Câu 2: D
Câu 3: B
Câu 4: A
Câu 5: C
Câu 6: C
Câu 7: A
Câu 8: D
Bài 2:
a: \(A=\left(\frac{x}{x+2}+\frac{2}{x-2}+\frac{4x}{4-x^2}\right):\frac{2x+1}{8x+16}\)
\(=\frac{x\left(x-2\right)+2\left(x+2\right)-4x}{\left(x-2\right)\left(x+2\right)}\cdot\frac{8\left(x+2\right)}{2x+1}\)
\(=\frac{x^2-2x+2x+4-4x}{\left(x-2\right)}\cdot\frac{8}{2x+1}=\frac{x^2-4x+4}{\left(x-2\right)}\cdot\frac{8}{2x+1}\)
\(=\frac{8\left(x-2\right)^2}{\left(x-2\right)\left(2x+1\right)}=\frac{8\left(x-2\right)}{2x+1}=\frac{8x-16}{2x+1}\)
b: Khi \(x=-2\frac12=-2,5\) thì \(A=\frac{8\cdot\left(-2.5\right)-16}{2\cdot\left(-2,5\right)+1}=\frac{-20-16}{-5+1}=\frac{-36}{-4}=9\)
c: Để A nguyên thì 8x-16⋮2x+1
=>8x+4-20⋮2x+1
=>-20⋮2x+1
mà 2x+1 lẻ
nen 2x+1∈{1;-1;5;-5}
=>2x∈{0;-2;4;-6}
=>x∈{0;-1;2;-3}
Kết hợp ĐKXĐ, ta được: x∈{0;-1;-3}
