K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
15 tháng 5 2022
\(A=\dfrac{1}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)
AD
2
T
10 tháng 9 2023
\(A=\dfrac{2x+4}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}+\dfrac{2}{1-\sqrt{x}}\)
\(=\dfrac{2x+4}{\sqrt{x^3}-1}+\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}-\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{2x+4}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{2\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{2x+4+x+\sqrt{x}-2-2x-2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
#Toru
10 tháng 9 2023
A=\(\dfrac{2x+4}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}-\dfrac{2}{\sqrt{x}-1}=\dfrac{2x+4+\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-2\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{2x+4+x+\sqrt{x}-2-2x-2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{-x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
27 tháng 10 2021
\(ĐK:x>0;x\ne1\\ A=\dfrac{2+x-1-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}+1}\\ A=\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}+1}=\dfrac{\left(1-\sqrt{x}\right)\left(2-\sqrt{x}\right)}{\sqrt{x}+1}\)
27 tháng 10 2021
\(A=\left(\dfrac{2+x-1-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}+1}\)
\(=\dfrac{-\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
VD
17 tháng 3 2022
\(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Rightarrow A=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Rightarrow A=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Rightarrow A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
H
17 tháng 3 2022
\(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(ĐKXĐ:x\ne4;x\ne9\right)\)
\(=\dfrac{2\sqrt{x}-9}{x-3\sqrt{x}-2\sqrt{x}+6}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x-2}\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
Vậy với ĐKXĐ \(x\ne4;x\ne9\) thì biểu thức \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
27 tháng 10 2021
Bài 1:
a: \(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\dfrac{2x}{x-1}\)
NT
0
GD
1
H
5 tháng 3 2023
\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\)
\(\Leftrightarrow\left[\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}+1}{x\sqrt{x}-1}\right].\dfrac{2}{\sqrt{x}-1}\)
\(\Leftrightarrow\left[\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}\right].\dfrac{2}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{x+1-2\sqrt{x}}{x\sqrt{x}-1}.\dfrac{2}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right)^2}{x\sqrt{x}-1}.\dfrac{2}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right)2}{x+\sqrt{x}+1}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-2}{x+\sqrt{x}+1}\)
5 tháng 5 2021
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: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
5 tháng 5 2021
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
`A = ((x sqrtx-1)/(x-sqrtx) - (xsqrtx+1)/(x+sqrtx))/((x+2)(x-2))`.
`= (((sqrtx-1)(x+sqrtx+1))/(sqrtx(sqrtx-1)) - ((sqrtx+1)(x-sqrtx+1))/(sqrtx(sqrtx+1))/((x+2)(x-2))`
`= (x + sqrtx+1)/(sqrtx) - (x - sqrtx+1)/(sqrtx)/((x+2)(x-2))`
`= (2sqrtx)/sqrtx/((x+2)(x-2))`
`= 2/((x+2)(x-2))`.
hộ mình bài này vs ạ
cho hình vuông ABCD,E thuộc BC qua A kẻ tia Ax vuông góc AE cắt CD tại F.trung tuyết Ay của tam giác AEF cắt CD ở K a,chứng minh rằng AF^2 = FK . FC b,chứng minh rằng khi E di chuyển trên cạnh BC thì chu vi tam giác EKC có giá trị không đổi