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

Những câu hỏi liên quan
Tuấn Nguyễn
Xem chi tiết
Nguyễn Lê Phước Thịnh
20 tháng 12 2022 lúc 23:09

\(=\dfrac{x+\sqrt{x}-3\sqrt{x}+1}{x-1}:\dfrac{x+2\sqrt{x}+1-4\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

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

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)

Nguyễn Thị Thu Phương
Xem chi tiết
Akai Haruma
6 tháng 8 2021 lúc 17:03

1.

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

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

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

Akai Haruma
6 tháng 8 2021 lúc 17:06

2.

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

\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)

Akai Haruma
6 tháng 8 2021 lúc 17:09

3.

\(A=\left[\frac{\sqrt{x}(\sqrt{x}-3)+2\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\frac{3x+9}{(\sqrt{x}-3)(\sqrt{x}+3)}\right]:\frac{1}{(\sqrt{x}+3)^2}\)

\(=\frac{3\sqrt{x}-9}{(\sqrt{x}-3)(\sqrt{x}+3)}.(\sqrt{x}+3)^2=\frac{3(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}(\sqrt{x}+3)^2=3(\sqrt{x}+3)\)

TÊN HỌ VÀ
Xem chi tiết
Hquynh
10 tháng 4 2023 lúc 21:37

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

Ngô Chí Thành
Xem chi tiết
Nguyễn Lê Phước Thịnh
10 tháng 7 2022 lúc 20:35

\(A=\left(\dfrac{3\sqrt{x}}{\sqrt{x}+3}+\dfrac{3\sqrt{x}}{x-9}\right):\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\dfrac{3x-9\sqrt{x}+3\sqrt{x}}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{3x-6\sqrt{x}}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)

Ly Ly
Xem chi tiết
An Thy
5 tháng 7 2021 lúc 15:58

\(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)

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

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

Nguyên Phan
Xem chi tiết
Nguyễn Lê Phước Thịnh
24 tháng 7 2021 lúc 21:43

Ta có: \(\left(\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}-1}\right):\left(\sqrt{x}+\dfrac{\sqrt{x}}{x-1}\right)\)

\(=\dfrac{x\sqrt{x}+1-\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x\sqrt{x}-\sqrt{x}+\sqrt{x}}{x-1}\)

\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}+\sqrt{x}}{x\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+1}{x\sqrt{x}}\)

Nấm Chanel
Xem chi tiết
ĐỖ NV1
Xem chi tiết
Nguyễn Lê Phước Thịnh
23 tháng 4 2023 lúc 14:24

\(=\left(\dfrac{1-x}{\sqrt{x}}\right):\dfrac{\sqrt{x}-1+1-x}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\dfrac{1-x}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}\)

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

Ngọc Vũ
Xem chi tiết
Nguyễn Hoàng Minh
21 tháng 10 2021 lúc 11:48

\(A_1=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

\(A_2=\left[\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]:\dfrac{x-\sqrt{x}+1}{x+1}\\ A_2=\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x-\sqrt{x}+1}\\ A_2=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+1}{x-\sqrt{x}+1}\)

nguyen ngoc son
Xem chi tiết
Nguyễn Lê Phước Thịnh
5 tháng 2 2022 lúc 22:22

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

b: \(=\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\)

Nguyễn Huy Tú
5 tháng 2 2022 lúc 22:42

a, \(=\left(\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+1\right)\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\)

b, với x > 0 

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

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