Question

[Trần Tuấn Kiệt]

bài 1: thực hiện phép tính

a, (\(\frac{x+1}{x-1}-\frac{x-1}{x+1}\)) : (\(\frac{1}{x+1}-\frac{x}{1-x}+\frac{2}{x^2-1}\))

b, \(\frac{2+x}{2-x}:\frac{4x^2}{4-4x+x^2}\) . (\(\frac{2}{2-x}-\frac{4}{8+x^3}.\frac{4-2x+x^2}{2-x}\))

c, ((\(\frac{3}{x-y}+\frac{3x}{x^2+y^2}\)) : \(\frac{2x+y}{x^2+2xy+y^2}\)) . \(\frac{x-y}{3}\)

bài 2: cho biểu thức M = \(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

a, tìm ĐKXĐ, rút gọn M

b, tìm x để M có giá trị nguyên

Answer 1

[Hoàng Thúy An]

2, a,đkxđ \(x\ne-3;x\ne2\)

mình giải luôn nhé k ghi lại đề nữa

\(=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}\)

\(=\frac{\left(x+2\right)\left(x-2\right)-5-1\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

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

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

\(=\frac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x-4}{x-2}\)

b,\(M=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\)

để M nguyên thì \(\frac{2}{x-2}\) nguyên=>x - 2 là ước của 2,\(Ư_{\left(2\right)}=\left\{-2;-1;1;2\right\}\)

x - 2 = -2 <=> x = 0

x - 2 = -1 <=> x = 1

x - 2 = 1 <=> x = 3

x - 2 =2 <=> x = 4

vậy x = {0;1;3;4}

Answer 2

[Nguyễn Thị Ngọc Hân]

a) \(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{x^2-1}:\frac{x-1+x^2+x+2}{x^2-1}\)

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