Rút gọn A
A={[(x√x-1)/(x-√x) - (x√x +1)/(x+√x)]}/ (x+2)/(x-2)
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15 tháng 5 2022
\(A=\dfrac{1}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)
4 tháng 7 2023
\(=\dfrac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x-1}\)
\(=\dfrac{x+1-1}{x-1}=\dfrac{x}{x-1}\)
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=\(\dfrac{2x+4}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}-\dfrac{2}{\sqrt{x}-1}=\dfrac{2x+4+\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-2\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{2x+4+x+\sqrt{x}-2-2x-2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{-x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
BP
1
H
3 tháng 4 2023
bổ sung \(ĐKXĐ:\left\{{}\begin{matrix}x+1\ne0\\x-1\ne0\\x-2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne1\\x\ne2\end{matrix}\right.\)
3 tháng 4 2023
A=\(\left(\dfrac{1}{x+1}-\dfrac{1}{x^2-1}\right).\dfrac{x+1}{x-2}\)
=\(\left(\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{1}{x^2-1}\right).\dfrac{x+1}{x-2}\)
=\(\left(\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}-\dfrac{1}{x^2-1}\right).\dfrac{x+1}{x-2}\)
=\(\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}.\dfrac{x+1}{x-2}\)
=\(\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x-2\right)}\)
=\(\dfrac{1}{x-1}\)
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 \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
17 tháng 4 2021
\(\left(\dfrac{\dfrac{x}{x+1}}{\dfrac{x^2}{x^2+x+1}}-\dfrac{2x+1}{x^2+x}\right)\dfrac{x^2-1}{x-1}\)ĐK : \(x\ne\pm1\)
\(=\left(\dfrac{x}{x+1}.\dfrac{x^2+x+1}{x^2}-\dfrac{2x+1}{x\left(x+1\right)}\right)\left(x+1\right)=\left(\dfrac{x^2+x-1}{x^2+x}-\dfrac{2x+1}{x\left(x+1\right)}\right)\left(x+1\right)\)
\(=\left(\dfrac{x^2+x-1-2x-1}{x\left(x+1\right)}\right)\left(x+1\right)=\dfrac{x^2-3x-2}{x}\)
17 tháng 4 2021
à xin lỗi mình nhầm dòng cuối
\(=\dfrac{x^2-x-2}{x}=\dfrac{\left(x+1\right)\left(x-2\right)}{x}\)
Để biểu thức trên nhận giá trị dương khi
\(\dfrac{\left(x+1\right)\left(x-2\right)}{x}>0\)bạn tự xét TH cả tử và mẫu nhé, mình đánh trên này bị lỗi
27 tháng 10 2021
Bài 1:
a: \(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\dfrac{2x}{x-1}\)
`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