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

Những câu hỏi liên quan
Vangull
Xem chi tiết
Yeutoanhoc
22 tháng 5 2021 lúc 9:04

`P=(2-2/(1-sqrtx)):((x+2)/(x+sqrtx-2)-sqrtx/(sqrtx+2))`
`đk:x>=0,x ne 1`
`P=(2+2/(sqrtx-1)):((sqrtx+2)/((sqrtx+2)(sqrtx-1))-sqrtx/(sqrtx+2))`
`=2(1+1/(sqrtx-1)):((x+2-x+sqrtx)/((sqrtxx+2)(sqrtx-1)))`
`=2(1+1/(sqrtx-1)):(1/(sqrx-1))`
`=2((sqrtx-1+1)/(sqrtx-1)).(sqrtx-1)`
`=2sqrtx`

Nguyễn Châu Mỹ Linh
Xem chi tiết
Nguyễn Lê Phước Thịnh
5 tháng 5 2021 lúc 13:44

Câu 1:

Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)

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

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

Nguyễn Lê Phước Thịnh
5 tháng 5 2021 lúc 13:46

Câu 3: 

Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)

\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)

\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)

\(=\sqrt{a}\left(\sqrt{a}-2\right)\)

\(=a-2\sqrt{a}\)

....
Xem chi tiết
An Thy
22 tháng 6 2021 lúc 18:54

\(\dfrac{8-x}{2+\sqrt[3]{x}}:\left(2+\dfrac{\sqrt[3]{x^2}}{2+\sqrt[3]{x}}\right)+\left(\sqrt[3]{x}+\dfrac{2\sqrt[3]{x}}{\sqrt[3]{x}-2}\right).\left(\dfrac{\sqrt[3]{x^2}-4}{\sqrt[3]{x^2}+2\sqrt[3]{x}}\right)\)

\(\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}{2+\sqrt[3]{x}}:\left(\dfrac{4+2\sqrt[3]{x}+\sqrt[3]{x^2}}{2+\sqrt[3]{x}}\right)+\dfrac{\sqrt[3]{x^2}}{\sqrt[3]{x}-2}.\left(\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x}+2\right)}{\sqrt[3]{x}\left(\sqrt[3]{x}+2\right)}\right)\)

\(\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}{2+\sqrt[3]{x}}.\dfrac{2+\sqrt[3]{x}}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}+\dfrac{\sqrt[3]{x^2}}{\sqrt[3]{x}-2}.\dfrac{\sqrt[3]{x}-2}{\sqrt[3]{x}}\)

\(=\sqrt[3]{x}-2+\sqrt[3]{x}=2\sqrt[3]{x}-2\)

Nguyên Hoàng
Xem chi tiết
Nguyễn Việt Lâm
29 tháng 1 lúc 22:01

ĐKXĐ:...

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

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

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

Trần Minh Thư
Xem chi tiết
Phạm Trần Bỏ Phương
Xem chi tiết
Lê Ng Hải Anh
24 tháng 5 2021 lúc 16:48

\(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\) (ĐK: x ≥ 0, x ≠ 4)

\(=\left[\dfrac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\left[\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+10-x}{\sqrt{x}+2}\right]\)

\(=\left(\dfrac{-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\dfrac{6}{\sqrt{x}+2}\)

\(=\dfrac{\left(-6\right)\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

Vậy...

 

Minh Anh Vũ
Xem chi tiết
Anh Quynh
Xem chi tiết
Hoàng Kiều Quỳnh Anh
Xem chi tiết
Nguyễn Việt Lâm
22 tháng 3 2022 lúc 16:33

\(A=2\left|2-\sqrt{5}\right|-\dfrac{8\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)

\(=2\left(\sqrt{5}-2\right)-\dfrac{8\left(3+\sqrt{5}\right)}{4}=2\sqrt{5}-4-2\left(3+\sqrt{5}\right)\)

\(=2\sqrt{5}-4-6-2\sqrt{5}=-10\)

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

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

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

Phạm Trần Phát
Xem chi tiết
Nguyễn Lê Phước Thịnh
5 tháng 9 2023 lúc 20:35

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

ĐKXĐ: x>0; x<>4

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

\(=\dfrac{x-2\sqrt{x}+3\sqrt{x}+6+3\sqrt{x}-2}{x-4}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x+\sqrt{x}}\)

\(=\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{x+4\sqrt{x}+4}{x+3\sqrt{x}+2}\)

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