Bài 4: Quy đồng mẫu thức nhiều phân thức

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

Quy đồng mẫu các phân thức:
1) 7x-1/2x^2+6x; 3-2x/x^2-9
2) 2x-1/x-x^2; x+1/2-4x+2x^2
3) x-1/x^3+1; 2x/x^2-x+1; 2/x+1
4) 7/5x; 4/x-2y; x-y/8y^2-2x^2
5) x/x^3-1; x+1/x^2-x; x-1/x^2+x+1
6) x/x^2-2ax+a^2; x+a/x^2-ax

Nguyễn Nam
21 tháng 11 2017 lúc 18:34

1)

\(\dfrac{7x-1}{2x^2+6x}=\dfrac{7x-12}{x\left(x+3\right)}\)

\(\dfrac{3-2x}{x^2-9}=\dfrac{3-2x}{\left(x-3\right)\left(x+3\right)}\)

MTC: \(x\left(x-3\right)\left(x+3\right)\)

\(\dfrac{7x-1}{2x^2+6x}=\dfrac{7x-12}{x\left(x+3\right)}=\dfrac{\left(x-3\right)\left(7x-12\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{7x^2-12x-21x+36}{x\left(x-3\right)\left(x+3\right)}=\dfrac{7x^2-33x+36}{x\left(x-3\right)\left(x+3\right)}\)

\(\dfrac{3-2x}{x^2-9}=\dfrac{3-2x}{\left(x-3\right)\left(x+3\right)}=\dfrac{ x\left(3-2x\right)}{x\left(x-3\right)\left(x+3\right)}\dfrac{3x-2x^2}{x\left(x-3\right)\left(x+3\right)}\)

2)

\(\dfrac{2x-1}{x-x^2}=\dfrac{2x-1}{x\left(1-x\right)}\)

\(\dfrac{x+1}{2-4x+2x^2}=\dfrac{x+1}{2\left(1-2x+x^2\right)}=\dfrac{x+1}{2\left(1-x\right)^2}\)

MTC: \(2x\left(1-x\right)^2\)

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

\(\dfrac{x+1}{2-4x+2x^2}=\dfrac{x+1}{2\left(1-2x+x^2\right)}=\dfrac{x+1}{2\left(1-x\right)^2}=\dfrac{ x\left(x+1\right)}{2x\left(1-x\right)^2}=\dfrac{x^2+x}{2x\left(1-x\right)^2}\)

Phương Trâm
21 tháng 11 2017 lúc 19:22

Phần còn lại nhé :v

3.

\(x^3+1=\left(x+1\right)\left(x^2-x+1\right)\)

\(x^2-x+1=x^2-x+1\)

\(x+1=x+1\)

MTC: \(\left(x+1\right)\left(x^2-x+1\right)\)

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

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

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

4.

\(5x\)

\(x-2y=x-2y=-\left(2y-x\right)\)

\(8y^2-2x^2=2\left(4y^2-x^2\right)=2\left(2y-x\right)\left(2y+x\right)\)

MTC: \(-10x\left(2y-x\right)\left(2y+x\right)\)

\(\dfrac{7}{5x}=\dfrac{7\left(2y-x\right)\left(2y+x\right)-2}{5x\left(2y-x\right)\left(2y+x\right)-2}=\dfrac{-14\left(2y-x\right)\left(2y+x\right)}{-10x\left(2y-x\right)\left(2y+x\right)}\)

\(\dfrac{4}{x-2y}=\dfrac{4\left(2y-x\right)\left(2y+x\right)10x}{-\left(2y-x\right)\left(2y+x\right)10x}=\dfrac{40x\left(2y-x\right)\left(2y+x\right)}{-10x\left(2y-x\right)\left(2y+x\right)}\)

\(\dfrac{x-y}{8y^2-2x^2}=\dfrac{\left(x-y\right)-5x}{2\left(2y-x\right)\left(2y+x\right)-5x}=\dfrac{-5x\left(x-y\right)}{-10x\left(2y-x\right)\left(2y+x\right)}\)

5.

\(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)

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

\(x^2+x+1\)

MTC: \(x\left(x-1\right)\left(x^2+x+1\right)\)

\(\dfrac{x}{x^3-1}=\dfrac{x.x}{\left(x-1\right)\left(x^2+x+1\right)x}=\dfrac{x^2}{x\left(x-1\right)\left(x^2+x+1\right)}\)

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

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

6.

\(x^2-2ax+a^2=\left(x-a\right)^2\)

\(x^2-ax=x\left(x-a\right)\)

MTC: \(x\left(x-a\right)^2\)

\(\dfrac{x}{x^2-2ax+a^2}=\dfrac{x.x}{\left(x-a\right)^2x}=\dfrac{x^2}{x\left(x-a\right)^2}\)

\(\dfrac{x+a}{x^2-ax}=\dfrac{\left(x+a\right)\left(x-a\right)}{x\left(x-a\right)\left(x-a\right)}=\dfrac{x^2-a^2}{x\left(x-a\right)^2}\)


Các câu hỏi tương tự
Sách Giáo Khoa
Xem chi tiết
Đỗ Loan
Xem chi tiết
Thee Bao
Xem chi tiết
Xem chi tiết
Mie Mao
Xem chi tiết
Tuyết Ly
Xem chi tiết
Dương Trần Thùy
Xem chi tiết
Sách Giáo Khoa
Xem chi tiết
Sách Giáo Khoa
Xem chi tiết