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

Những câu hỏi liên quan
êfe
Xem chi tiết
Hoàng Quang Kỳ
Xem chi tiết
Nguyễn Thị Quỳnh Trang
Xem chi tiết
Nguyễn Lê Phước Thịnh
1 tháng 3 2022 lúc 15:00

b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)

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

\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)

\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)

Minh Thư
Xem chi tiết
do phuong nam
22 tháng 6 2018 lúc 17:05

A = \(\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right).\left(x+1\right)}-\frac{x+3}{2\left(x+2\right)}\right).\frac{4x^2-4}{5}\)

A = \(\left(\frac{\left(x+1\right)^2+3.2-\left(x+3\right).\left(x-1\right)}{2\left(x-1\right).\left(x+1\right)}\right).\frac{4x^2-4}{5}\)

A = \(\left(\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right).\left(x+1\right)}\right).\frac{4\left(x^2-1\right)}{5}\)

A = \(\frac{10}{2\left(x-1\right).\left(x+1\right)}.\frac{4\left(x-1\right).\left(x+1\right)}{5}\)

A = 4

Phạm Da Đen
Xem chi tiết
Nguyễn Trần Thanh Loan
Xem chi tiết
Funny Suuu
Xem chi tiết
Nguyễn Phương
19 tháng 3 2020 lúc 12:01

=[x(x-2)/2(x2+4)-2x2/(4+x2)(2-x)][x(x-2)(x+1)/x3]

={[x(x-2)(2-x)-4x2 ]/2(2-x)(4+x2)} .[x(x-2)(x+1)/x3 ]

=[-x(x2+4)/2(2-x)(4+x2)].[x(x-2)(x+1)/x3 ]

=-x.x(x-2)(x+1)/2(2-x)x3

=(x+1)/2x

Khách vãng lai đã xóa
Bae Sooji
Xem chi tiết
Nguyễn Công Tỉnh
2 tháng 7 2019 lúc 14:35

Chép đề đúng chưa bạn? 2 phân số đầu có ngoặc không vậy?

Bae Sooji
2 tháng 7 2019 lúc 14:42

Nguyễn Công Tỉnh đúng r bạn, mình sửa lại r

Nguyễn Công Tỉnh
2 tháng 7 2019 lúc 14:56

Bạn tự tìm ĐKXĐ nhé!

\(B=\left(\frac{x}{x^2-x-6}-\frac{x-1}{3x^2-4x-15}\right):\frac{x^4-2x^2+1}{3x^2+11x+10}.\left(x^2-2x+1\right)\)

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

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

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

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

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

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

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

Vậy...

Minh
Xem chi tiết
Lê Tài Bảo Châu
1 tháng 11 2019 lúc 20:58

a) \(P=\frac{2}{2x+3}+\frac{3}{2x+1}-\frac{6x+5}{\left(2x+3\right)\left(2x-3\right)}\)

\(=\frac{2\left(2x+1\right)\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}+\frac{3\left(2x+3\right)\left(2x-3\right)}{\left(2x+1\right)\left(2x+3\right)\left(2x-3\right)}-\frac{\left(6x+5\right)\left(2x+1\right)}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)

\(=\frac{\left(4x+2\right)\left(2x-3\right)+3\left(4x^2-9\right)-12x^2-16x-5}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)

\(=\frac{8x^2-8x-6+12x^2-27-12x^2-16x-5}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)

\(=\frac{8x^2-24x-38}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)

Check hộ mình xem nghi nghi sai sai

Khách vãng lai đã xóa
Lê Tài Bảo Châu
1 tháng 11 2019 lúc 21:06

b) \(Q=\left(\frac{x+1}{2x-1}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)

\(=\left(\frac{x+1}{2x-1}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right).\frac{4x^2-4}{5}\)

\(=\left(\frac{2\left(x+1\right)\left(x-1\right)\left(x+1\right)}{2\left(2x-1\right)\left(x-1\right)\left(x+1\right)}+\frac{2.3\left(2x-1\right)}{2\left(x-1\right)\left(x+1\right)\left(2x-1\right)}-\frac{\left(x+3\right)\left(2x-1\right)\left(x-1\right)}{2\left(x+1\right)\left(2x-1\right)\left(x-1\right)}\right).\frac{4x^2-4}{5}\)

\(=\frac{2\left(x+1\right)\left(x^2-1\right)+12x-6-\left(2x^2+5x-3\right)\left(x-1\right)}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)

\(=\frac{2\left(x^3+x^2-x-1\right)+12x-6-2x^3-5x^2+3x+2x^2+5x-3}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)

\(=\frac{2x^3+2x^2-2x-2+20x-2x^3-3x^2-9}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)

\(=\frac{-x^2+18x-11}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(=\frac{-x^2+18x-11}{\left(2x-1\right)}.\frac{2}{5}\)

\(=\frac{-2x^2+36x-22}{5\left(2x-1\right)}\)

Khách vãng lai đã xóa