Cho biểu thức A = \(\dfrac{1}{2+\sqrt{x}}\) + \(\dfrac{1}{2-\sqrt{x}}\) - \(\dfrac{2\sqrt{x}}{4-x}\) ; Rút gọn biểu thức A và tìm giá trị của x để A = \(\dfrac{1}{4}\) .
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HP
3 tháng 9 2021
a, \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
b, \(A\in Z\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\in Z\)
\(\Leftrightarrow\sqrt{x}+3\inƯ_3=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow\sqrt{x}=0\)
\(\Leftrightarrow x=0\)
3 tháng 9 2021
\(a,A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\left(x\ge0;x\ne9\right)\\ A=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ A=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
\(b,A\in Z\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\in Z\Leftrightarrow-3⋮\sqrt{x}+3\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-6;-4;-2;0\right\}\)
Mà \(\sqrt{x}\ge0\)
\(\Leftrightarrow x\in\left\{0\right\}\)
Vậy \(x=0\) thì A nguyên
21 tháng 6 2021
`a)(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4,x ne 9)`
`=(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`
`=(2sqrtx-9+(sqrtx-3)(sqrtx+3)+(2sqrtx+1)(sqrtx-2))/(x-5sqrtx+6)`
`=(2sqrtx-9+x-9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(3x-sqrtx-20)/
H
8 tháng 4 2021
\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1}{x-1}\\ =\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x-1}{x+1}=\dfrac{2}{x-1}\cdot\dfrac{x-1}{x+1}\\ =\dfrac{2}{x+1}\)
PD
8 tháng 4 2021
\(\bigg(\dfrac{1}{\sqrt x-1}-\dfrac{1}{\sqrt x+1}\bigg):\dfrac{x+1}{x-1}\\=\bigg(\dfrac{\sqrt x+1}{(\sqrt x-1)(\sqrt x+1)}-\dfrac{\sqrt x-1}{(\sqrt x-1)(\sqrt x+1)}\bigg.\dfrac{x-1}{x+1}\\=\dfrac{\sqrt x+1-\sqrt x+1}{(\sqrt x-1)(\sqrt x+1)}.\dfrac{(\sqrt x-1)(\sqrt x+1)}{x+1}\\=\dfrac{2}{x+1}\)
11 tháng 3 2022
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>=0\\a< >1\end{matrix}\right.\)
\(A=\dfrac{1}{2\left(\sqrt{a}+1\right)}-\dfrac{1}{2\left(\sqrt{a}-1\right)}+\dfrac{a^2+1}{a^2-1}\)
\(=\dfrac{\sqrt{a}-1-\sqrt{a}-1}{2\left(a-1\right)}+\dfrac{a^2+1}{a^2-1}\)
\(=\dfrac{-1}{a-1}+\dfrac{a^2+1}{a^2-1}\)
\(=\dfrac{-a-1+a^2+1}{\left(a-1\right)\left(a+1\right)}=\dfrac{a^2-a}{\left(a-1\right)\left(a+1\right)}=\dfrac{a}{a+1}\)
b: Để A-1/3<0 thì \(\dfrac{a}{a+1}-\dfrac{1}{3}< 0\)
=>3a-a-1<0
=>2a-1<0
hay 0<a<1/2
H9
HT.Phong (9A5)
CTVHS
8 tháng 10 2023
a) \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\) \(\left(x\ge0;x\ne4\right)\)
\(=\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b) \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}}\) (\(x>0\))
\(=\left[\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(x+2\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{x\sqrt{x}+2x+\sqrt{x}}\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 10 2023
c) \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\) (\(x\ge0;x\ne1\))
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
d) \(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\) \(\left(a\ne1;a\ge0\right)\)
\(=\left[\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\dfrac{\sqrt{a}+1+\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2-a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}:\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{a+2\sqrt{a}+1-a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}\)
\(=\dfrac{a-a\sqrt{a}+2\sqrt{a}+1}{2\sqrt{a}}\)
H
23 tháng 8 2023
2)
ĐK: \(x\ge0;x\ne4\)
Biểu thức trở thành:
\(\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)}{a-4}-\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-4}-\dfrac{4\sqrt{a}-4}{a-4}\\ =\dfrac{a+2\sqrt{a}+3\sqrt{a}+6}{a-4}-\dfrac{a-2\sqrt{a}-\sqrt{a}+2}{a-4}-\dfrac{4\sqrt{a}-4}{a-4}\\ =\dfrac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{a-4}\\ =\dfrac{4\sqrt{a}+8}{a-4}\\ =\dfrac{4\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\\ =\dfrac{4}{\sqrt{a}-2}\)
24 tháng 8 2023
1:
\(\left(\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}+1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(=\dfrac{x+2\sqrt{x}-7-\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-1-\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+2\sqrt{x}-8-x-4\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{-4}\)
\(=\dfrac{-2\sqrt{x}-11}{-4}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}-3}=\dfrac{\left(2\sqrt{x}+11\right)\left(\sqrt{x}-1\right)}{4\left(\sqrt{x}-3\right)}\)
12 tháng 9 2023
a: ĐKXĐ: x>=0; x<>4
\(Q=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+2\sqrt{x}\left(\sqrt{x}+2\right)-3x-4}{x-4}\cdot\dfrac{\sqrt{x}-2+2}{2}\)
\(=\dfrac{x-2\sqrt{x}+2x+4\sqrt{x}-3x-4}{x-4}\cdot\dfrac{\sqrt{x}}{2}\)
\(=\dfrac{2\sqrt{x}-4}{x-4}\cdot\dfrac{\sqrt{x}}{2}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
b: \(M=P\cdot Q=\dfrac{\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{1-5\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)
\(M\left(M-1\right)=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-5x-x-3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)\left(-6x-2\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)^2\cdot\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{\sqrt{x}\left(5\sqrt{x}-1\right)\left(6x+2\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}+1\right)^2}\)
TH1: M>=căn M
=>M^2>=M
=>M^2-M>=0
=>5*căn x-1>=0
=>x>=1/25 và x<>4
TH2: M<căn M
=>5căn x-1<0
=>x<1/25
Kết hợp ĐKXĐ, ta được: 0<=x<1/25
đkxđ: x≥0; x≠4
\(A=\dfrac{1}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-x}\)
\(=\dfrac{2-\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{2+\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}-\dfrac{2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(=\dfrac{4-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{2\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{2}{2+\sqrt{x}}\)
+) A = 1/4 <=> \(\dfrac{2}{2+\sqrt{x}}=\dfrac{1}{4}\Leftrightarrow2+\sqrt{x}=8\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\)(tm)
Vậy x = 36
đkxđ \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
\(A=\dfrac{2+\sqrt{x}+2-\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}\)
\(A=\dfrac{4-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}\)
\(A=\dfrac{2}{\sqrt{x}+2}\)
để \(A=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x}+2=8\)
\(\Leftrightarrow x=36\left(tm\right)\)
vậy tại x=36 thì A=1/4