Những câu hỏi liên quan
Đỗ Phương Thảo
Xem chi tiết
Chu Công Đức
20 tháng 2 2020 lúc 9:16

\(ĐKXĐ:x\ne\pm2\)

\(\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)

\(=\left[\frac{2}{x+2}-\frac{4}{\left(x+2\right)^2}\right]:\left[\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-1}{x-2}\right]\)

\(=\left[\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right]:\left[\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right]\)

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

\(=\frac{2x}{\left(x+2\right)^2}:\frac{-x}{\left(x-2\right)\left(x+2\right)}=\frac{2x}{\left(x+2\right)^2}.\frac{-\left(x-2\right)\left(x+2\right)}{x}\)

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

Khách vãng lai đã xóa
Bùi Anh Tuấn
20 tháng 2 2020 lúc 9:17

\(\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)

\(\Leftrightarrow\left(\frac{2}{x+2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{1}{2-x}\right)\)

\(\Leftrightarrow\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2+x+2}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\frac{2x}{\left(x+2\right)^2}\cdot\frac{\left(x-2\right)\left(x+2\right)}{x+4}\)

\(\Leftrightarrow\frac{2x^2-4x}{\left(x+2\right)\left(x+4\right)}\)

Khách vãng lai đã xóa
Kim Trân Ni
Xem chi tiết
KAl(SO4)2·12H2O
9 tháng 12 2019 lúc 14:47

\(\frac{3\left(x+1\right)}{x+2}-\frac{3x-6}{x^2-4}\)

\(=\frac{3\left(x+1\right)}{x+2}-\left(\frac{3x-6}{x^2-4}\right)\)

\(=\frac{3x^2-6x^2-12x+24}{x^3+2x^2-4x-8}\)

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

\(=\frac{3x-6}{x+2}\)

Khách vãng lai đã xóa
KAl(SO4)2·12H2O
9 tháng 12 2019 lúc 14:51

\(\frac{x^2+4x+4}{1-x}.\frac{\left(1-x\right)^2}{3\left(x+2\right)^3}\)

\(=\frac{x^2+4x+4}{1-x}.\left[\frac{\left(1-x\right)^2}{3\left(x+2\right)^3}\right]\)

\(=\frac{x^4+2x^3-3x^2-4x+4}{-3x^4-15x^3-18x^2+12x+24}\)

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

\(=\frac{-x+1}{3x+6}\)

Khách vãng lai đã xóa
Yukihira Souma
Xem chi tiết
Phạm Hồ Thanh Quang
15 tháng 7 2017 lúc 7:46

a) ĐKXĐ: \(x;y\ne0,x\ne\frac{y}{2},y\ne\frac{x}{2}\)
\(\frac{y}{2x^2-xy}+\frac{4x}{y^2-2xy}=\frac{y}{x\left(2x-y\right)}-\frac{4x}{y\left(2x-y\right)}\)\(=\frac{y^2-4x^2}{xy\left(2x-y\right)}=\frac{\left(y-2x\right)\left(y+2x\right)}{xy\left(2x-y\right)}\)
\(=\frac{-\left(y+2x\right)}{xy}\)

b) ĐKXĐ: \(x\ne2;x\ne-2\)
\(\frac{1}{x+2}+\frac{3}{x^2-4}+\frac{x-14}{\left(x^2+4x+4\right)\left(x-2\right)}\)\(=\frac{1}{x+2}+\frac{3}{\left(x-2\right)\left(x+2\right)}+\frac{x-14}{\left(x+2\right)^2\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x+2\right)+3\left(x+2\right)+x-14}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\frac{\left(x^2+4x+4\right)-16}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{\left(x+2\right)^2-16}{\left(x+2\right)^2\left(x-2\right)}=\frac{\left(x+2-4\right)\left(x+2+4\right)}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\frac{x+6}{\left(x+2\right)^2}\)

nguyenthithuhien
Xem chi tiết
Nguyễn Thị Quỳnh Anh
Xem chi tiết
Nguyễn Thanh Thảo
Xem chi tiết
công chúa nhỏ
Xem chi tiết
Nguyễn Thị Thu Hiền
Xem chi tiết
Mathematics❤Trần Trung H...
4 tháng 4 2020 lúc 20:34

Phân thức đại số.Phân thức đại số.

Khách vãng lai đã xóa
Huy Công Tử
Xem chi tiết
KAl(SO4)2·12H2O
3 tháng 4 2020 lúc 16:59

a) \(\frac{x^3+4x^2+x-2}{x+1}=\frac{\left(x+1\right)\left(x^2+3x-2\right)}{x+1}=x^2+3x-2\)

b) \(\frac{x-3}{2x-2}+\frac{1}{x-1}=\frac{x^2-2x+1}{2x^2-4x+2}=\frac{\left(x-1\right)\left(x-1\right)}{2\left(x-1\right)\left(x-1\right)}=\frac{1}{2}\)

Khách vãng lai đã xóa