Cho biểu thức P=
\((\dfrac{3}{\sqrt{x}+1})+(\dfrac{3}{\sqrt{x}-1}) . (1-\dfrac{1}{\sqrt{x}})\)
với x>0; x #1 .
a. Rút gọn P.
b. Tìm tất cả các giá trị của x để giá trị của P tương ứng là số nguyên tố có 1 chữ số.
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Những câu hỏi liên quan
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
\(\bigg(\dfrac{1}{\sqrt x-1}-\dfrac{1}{\sqrt x+1}\bigg):\dfrac{x+1}{x-1}\\=\bigg(\dfrac{\sqrt x+1}{(\sqrt x-1)(\sqrt x+1)}-\dfrac{\sqrt x-1}{(\sqrt x-1)(\sqrt x+1)}\bigg.\dfrac{x-1}{x+1}\\=\dfrac{\sqrt x+1-\sqrt x+1}{(\sqrt x-1)(\sqrt x+1)}.\dfrac{(\sqrt x-1)(\sqrt x+1)}{x+1}\\=\dfrac{2}{x+1}\)
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)/
12 tháng 4 2022
a, Với x khác 1
\(A=\dfrac{x^2+x+1-3x^2+2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1-x}{\left(x-1\right)\left(x^2+x+1\right)}=-\dfrac{1}{x^2+x+1}\)
b, Ta có \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\Rightarrow\dfrac{-1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< 0\)
Vậy với x khác 1 thì bth A luôn nhận gtri âm
31 tháng 8 2021
f: \(\dfrac{2}{\sqrt{6}-\sqrt{5}}=2\sqrt{6}+2\sqrt{5}\)
l: \(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\sqrt{10}-\sqrt{7}\)
Q
20 tháng 6 2017
\(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right)\cdot\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\left(\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{x+\sqrt{x}+1-\left(\sqrt{x}+2\right)}{x+\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}+x-\left(x+\sqrt{x}+1\right)}{\left(x+\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(x+\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\left(x+\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{1}{x+\sqrt{x}+1}\cdot\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{x-1}{\left(x+\sqrt{x}+1\right)^2}\)
31 tháng 8 2021
f: \(\dfrac{2}{\sqrt{6}-\sqrt{5}}=2\sqrt{6}+2\sqrt{5}\)
l: \(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\sqrt{10}-\sqrt{7}\)
m: \(\dfrac{1}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\)
