Bạn chưa đăng nhập. Vui lòng đăng nhập để hỏi bài

Những câu hỏi liên quan
Nguyễn Hoài Linh
Xem chi tiết
Phạm Thị Thúy An
Xem chi tiết
Quyết Tâm Chiến Thắng
Xem chi tiết
KAl(SO4)2·12H2O
1 tháng 2 2019 lúc 12:53

làm a thôi nha :D 

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

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

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

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

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

\(C=\frac{x+1}{x^2-2x+1}.\frac{x^2-1+x+2-x^2}{x-1}\)

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

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

\(C=\frac{x^2+2x+1}{x^3-3x^2+3x-1}\)

Nguyệt
1 tháng 2 2019 lúc 13:20

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

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

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

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

b)\(\text{Để B nguyên }\Rightarrow x^2⋮x-1\)

\(x^2=x^2-1+1=\left(x-1\right).\left(x+1\right)+1\)

\(\Rightarrow\text{Để }x^2⋮x-1\Rightarrow1⋮x-1\Rightarrow x-1\inƯ\left(1\right)=\left\{\pm1\right\}\Rightarrow x\in\left\{2;0\right\}\)

Nguyễn Trần Thanh Loan
Xem chi tiết
Thuy Duong Nguyen
Xem chi tiết
Phạm Gia kiệt
Xem chi tiết
Thuy Duong Nguyen
Xem chi tiết
Trần Thanh Phương
9 tháng 12 2018 lúc 19:41

a) Phân thức M xác định khi và chỉ khi :

+) \(2x-2\ne0\Leftrightarrow x\ne1\)

+) \(2x+2\ne0\Leftrightarrow x\ne-1\)

+) \(1-\frac{x-3}{x+1}\ne0\)

\(\Leftrightarrow x-3\ne x+1\)

\(\Leftrightarrow0x\ne4\left(\text{luôn đúng}\right)\)

Vậy \(x\ne\left\{1;-1\right\}\)

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

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

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

\(M=\frac{8}{2\left(x-1\right)2\left(x+1\right)}\cdot\frac{x+1}{4}\)

\(M=\frac{8\left(x+1\right)}{4\left(x-1\right)\left(x+1\right)\cdot4}\)

\(M=\frac{8\left(x+1\right)}{8\left(x+1\right)\left(x-1\right)}\)

\(M=\frac{1}{x-1}\)

shitbo
9 tháng 12 2018 lúc 20:06

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

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

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

\(=\left[\frac{6x+8-6x+6}{4x^2-4}\right]:\left(\frac{4}{x+1}\right)\)

\(=\frac{14}{4x^2-4}:\left(\frac{4}{x+1}\right)=\frac{14x+14}{16x^2-16}=\frac{7x+7}{8x^2-8}\)

leducminh
Xem chi tiết
Minh
Xem chi tiết
Lê Tài Bảo Châu
31 tháng 10 2019 lúc 22:51

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

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

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

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

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

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

Khách vãng lai đã xóa
Lê Tài Bảo Châu
31 tháng 10 2019 lúc 23:01

b) \(Q=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{5x-5x}{2x\left(x+5\right)}\)

\(=\frac{x\left(x^2+2x\right)}{2x\left(x+5\right)}+\frac{2\left(x-5\right)\left(x+5\right)}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\)

\(=\frac{x^3+2x^2+2\left(x^2-25\right)+50-5x}{2x\left(x+5\right)}\)

\(=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)

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

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

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

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

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

\(=\frac{x-1}{2}\)

Khách vãng lai đã xóa
Lê Tài Bảo Châu
31 tháng 10 2019 lúc 23:02

xin lỗi phần a làm sai mình làm lại

Khách vãng lai đã xóa