Rút gọn biểu thức \(P=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{2}{x-2\sqrt{x}}\), với \(x>0,\) \(x\ne4\).
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Những câu hỏi liên quan
DT
30 tháng 4 2023
\(P=\left(\dfrac{\sqrt{x}-2}{x-4}+\dfrac{\sqrt{x}+2}{x-4}\right).\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{2}\)
\(=\left(\dfrac{\sqrt{x}-2+\sqrt{x}+2}{x-4}\right).\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{2}\)
\(=\dfrac{2\sqrt{x}.\sqrt{x}.\left(\sqrt{x}-2\right)}{2.\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x}{\sqrt{x}+2}\)
1 tháng 2 2022
a, \(A=\dfrac{4\left(3-\sqrt{7}\right)}{2}+2\sqrt{7}=\dfrac{12}{2}=6\)
b, \(B=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\sqrt{x}}\right):\dfrac{2-\sqrt{x}}{x-1}\)
\(=\left(\dfrac{\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{2-\sqrt{x}}{x-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
1 tháng 2 2022
nhờ bạn làm rõ vì sao \(\dfrac{\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{2-\sqrt{x}}{x-1}\) lại bằng \(\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
mình xin cảm ơn
Y
14 tháng 8 2023
\(a,A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\left(dk:x\ge0,x\ne4\right)\\ =\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+10-x}{\sqrt{x}+2}\right)\\ =\dfrac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x-4+10-x}\)
\(=\dfrac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\sqrt{x}-2}.\dfrac{1}{6}\\ =\dfrac{-6}{\left(\sqrt{x}-2\right).6}\\
=-\dfrac{1}{\sqrt{x}-2}\)
\(b,A>0\Leftrightarrow-\dfrac{1}{\sqrt{x}-2}>0\Leftrightarrow\sqrt{x}-2< 0\\
\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
Kết hợp với \(dk:x\ge0,x\ne4\), ta kết luận \(0\le x< 4\)
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)
\(P=\left[\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\left[\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{-4x-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{-\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{-4\sqrt{x}\cdot\sqrt{x}}{-\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{4x}{\sqrt{x}-3}\)
b) \(P=\dfrac{4x}{\sqrt{x}-3}\)
\(P=4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}+24\)
Theo BĐT côsi ta có:
\(P\ge\sqrt{\dfrac{4\left(\sqrt{x}-3\right)\cdot36}{\sqrt{x}-3}}+24=36\)
Vậy: \(P_{min}=36\Leftrightarrow x=36\)
HP
3 tháng 9 2021
\(C=\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right).\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}-2}\)
\(=\left[\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{x\left(\sqrt{x}+2\right)-4\left(\sqrt{x}+2\right)}{\sqrt{x}-2}\)
\(=\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\right].\dfrac{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)
\(=\left[\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}-\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\right].\left(\sqrt{x}+2\right)^2\)
\(=\dfrac{6\sqrt{x}}{\sqrt{x}-2}\)
3 tháng 9 2021
\(C=\left[\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{\sqrt{x}\left(x-4\right)+2\left(x-4\right)}{\sqrt{x}-2}\) (\(x\ge0,x\ne4,x\ne9\))
\(C=\left[\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}\right].\dfrac{\left(\sqrt{x}+2\right)\left(x-4\right)}{\sqrt{x}-2}\)
\(C=\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}{\sqrt{x}-2}\)
\(C=\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}.\left(\sqrt{x}+2\right)^2\)
\(C=\dfrac{2}{\sqrt{x}-2}\)
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: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
6 tháng 12 2023
P = (\(\dfrac{1}{\sqrt{x}-1}\) - \(\dfrac{1}{\sqrt{x}}\)) : (\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\) - \(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)) với 0 < \(x\) ≠ 1; 4
P = \(\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}\): (\(\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right).\left(\sqrt{x-2}\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}\))
P = \(\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\): \(\dfrac{x-1-\left(x-4\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}\)
P = \(\dfrac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\) : \(\dfrac{3}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}\)
P = \(\dfrac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\) \(\times\) \(\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}{3}\)
P = \(\dfrac{\sqrt{x}-2}{3.\sqrt{x}}\)
P = \(\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{3x}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
6 tháng 12 2023
b, P = \(\dfrac{1}{4}\)
⇒ \(\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{3x}\) = \(\dfrac{1}{4}\)
⇒4\(x\) - 8\(\sqrt{x}\) = 3\(x\)
⇒ 4\(x\) - 8\(\sqrt{x}\) - 3\(x\) = 0
\(x\) - 8\(\sqrt{x}\) = 0
\(\sqrt{x}\).(\(\sqrt{x}\) - 8) = 0
\(\left[{}\begin{matrix}x=0\\\sqrt{x}=8\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=64\end{matrix}\right.\)
\(x=0\) (loại)
\(x\) = 64
AH
Akai Haruma
Giáo viên
6 tháng 8 2021
1.
\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}-2)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)(\sqrt{x}-1)-\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2x}{x-1}\)
AH
Akai Haruma
Giáo viên
6 tháng 8 2021
2.
\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)
\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)
29 tháng 12 2022
\(P=\left(\dfrac{1}{2\sqrt{x}}-\dfrac{x}{2\sqrt{x}}\right)^2.\left(\dfrac{\left(\sqrt{x}-1\right)^2}{x-1}-\dfrac{\left(\sqrt{x}+1\right)^2}{x-1}\right)\)
\(=\left(\dfrac{1-x}{2\sqrt{x}}\right)^2.\left(\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\right)\)
\(=\dfrac{\left(1-x\right)^2}{2\sqrt{x}}.\dfrac{-4\sqrt{x}}{-\left(1-x\right)}\)
\(=\left(1-x\right).2\sqrt{x}\)
\(=2\sqrt{x}-2x\sqrt{x}\)
Với `x > 0,x \ne 4` cs:
`P=(1/[\sqrt{x}+2]+1/[\sqrt{x}-2]):2/[x-2\sqrt{x}]`
`P=[\sqrt{x}-2+\sqrt{x}+2]/[(\sqrt{x}+2)(\sqrt{x}-2)].[\sqrt{x}(\sqrt{x}-2)]/2`
`P=[2\sqrt{x}]/[(\sqrt{x}+2)(\sqrt{x}-2)].[\sqrt{x}(\sqrt{x}-2)]/2`
`P=x/[\sqrt{x}+2]`