rút gọn
P = \(\left(\dfrac{\sqrt{x}}{x-4}-\dfrac{2}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
K
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Những câu hỏi liên quan
12 tháng 9 2021
a: Ta có: \(P=\left(\dfrac{x-2\sqrt{x}+4}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{x+4}{x-4}\right)\)
\(=\dfrac{x-2\sqrt{x}+4}{\sqrt{x}-2}:\dfrac{x+4\sqrt{x}+4+x-2\sqrt{x}-x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-2\sqrt{x}+4}{1}\cdot\dfrac{\sqrt{x}+2}{x+2\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+4}{\sqrt{x}}\)
b: \(P-2=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}}>0\forall x\) thỏa mãn ĐKXĐ
nên P>2
17 tháng 4 2021
Ta có: \(P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\left(\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\dfrac{8\sqrt{x}-8x+8x}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(=\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
H
17 tháng 4 2021
ta có : \(P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
=\(\left(\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)}{4-x}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-x\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
=\(\dfrac{8\sqrt{x}-4x+8x}{4-x}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\dfrac{8\sqrt{x}+4x}{4-x}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x-2}\right)}\) =\(\dfrac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\dfrac{4\sqrt{x}}{2-\sqrt{x}}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\) =\(\dfrac{4x\left(\sqrt{x}-2\right)}{\left(2-\sqrt{x}\right)\left(3-\sqrt{x}\right)}\)
=\(-\dfrac{4x\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(3-\sqrt{x}\right)}\) =\(-\dfrac{4x}{3-\sqrt{x}}\) =\(\dfrac{4x}{\sqrt{x}-3}\)
này mới đúng !!
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.
\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)
\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)
24 tháng 5 2021
\(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\) (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...
20 tháng 5 2022
\(B=\dfrac{x-\sqrt{x}+4-x+4-x-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-x-2\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-\left(\sqrt{x}+4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\dfrac{-\sqrt{x}-4}{\sqrt{x}+1}\)
1 tháng 11 2023
a: \(Q=\left(\dfrac{4\sqrt{x}}{x+2\sqrt{x}}+\dfrac{8\sqrt{x}}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\left(\dfrac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\left(\dfrac{4}{\sqrt{x}+2}-\dfrac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\dfrac{4\left(\sqrt{x}-2\right)-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\)
\(=\dfrac{-4\sqrt{x}-8}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{-\left(\sqrt{x}-3\right)}=\dfrac{4\sqrt{x}}{\sqrt{x}-3}\)
b: Q<4
=>Q-4<0
=>\(\dfrac{4\sqrt{x}}{\sqrt{x}-3}-4< 0\)
=>\(\dfrac{4\sqrt{x}-4\sqrt{x}+12}{\sqrt{x}-3}< 0\)
=>\(\dfrac{12}{\sqrt{x}-3}< 0\)
=>\(\sqrt{x}-3< 0\)
=>0<=x<9
Kết hợp ĐKXĐ, ta được: 0<x<9 và x<>4
H
1 tháng 11 2023
\(a,Q=\left(\dfrac{4\sqrt{x}}{x+2\sqrt{x}}+\dfrac{8\sqrt{x}}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\\ =\left(\dfrac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2}{\sqrt{x}}\right)\\ =\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{4x-8\sqrt{x}-8x}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\dfrac{-4x-8\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\\ =\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\\ =\dfrac{-4\sqrt{x}}{3-\sqrt{x}}\)
`b,` Để `Q<4` ta có :
\(\dfrac{-4\sqrt{x}}{3-\sqrt{x}}< 4\\ \Leftrightarrow\dfrac{-4\sqrt{x}}{3-\sqrt{x}}-4< 0\\ \Leftrightarrow\dfrac{-4\sqrt{x}-4\left(3-\sqrt{x}\right)}{3-\sqrt{x}}< 0\\ \Leftrightarrow-4\sqrt{x}-12+4\sqrt{x}< 0\\ \Leftrightarrow-12< 0\left(luon.dung\right)\)
NM
2 tháng 3 2021
ĐKXĐ x\(\ge0,x\ne1,x\ne4\)
P=
P=\(\left(\dfrac{\left(x+2\sqrt{x}+4\right)}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}+\dfrac{x+2\sqrt{x}+4}{x-1}\right):\dfrac{3\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)+\sqrt{x}+1+2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
P=\(\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{x+2\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{3\left(x-3\right)}\)
P=\(\dfrac{x-1+\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{3\left(x-3\right)}\)
P=\(\dfrac{x\sqrt{x}+x-9}{3\left(x-3\right)}\)
25 tháng 7 2021
\(A=\left(\sqrt{x}+\dfrac{4\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4}{2\sqrt{x}-x}\right)\)ĐK : x > 0 ; x \(\ne\)4
\(=\left(\dfrac{x+2\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)=\dfrac{x\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(x-4\right)}\)
\(=\dfrac{x}{\sqrt{x}-2}\)
Ta có: \(P=\left(\dfrac{\sqrt{x}}{x-4}-\dfrac{2}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\left(\dfrac{\sqrt{x}-2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\dfrac{-\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\dfrac{-\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+4}\)
\(=\dfrac{-1}{\sqrt{x}-2}\)