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18 tháng 2 2022
1: \(P=\dfrac{3a+3\sqrt{a}-3-a+1-a+4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\)
2: Để P nguyên thì \(\sqrt{a}-1+2⋮\sqrt{a}-1\)
\(\Leftrightarrow\sqrt{a}-1\in\left\{1;-1;2\right\}\)
hay \(a\in\left\{4;0;9\right\}\)
19 tháng 8 2023
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
1 tháng 7 2021
a,bn viết đúng đề xíu nhé \(\dfrac{\sqrt{a}+2}{\sqrt{a+3}}\) sửa \(\dfrac{\sqrt{a}+2}{\sqrt{a}+3}\)
đk: \(a\ge0,a\ne4\)
=>\(P=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{1}{\sqrt{a}-2}\)
\(=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)\(=\dfrac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b, \(P=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}=1+\dfrac{-2}{\sqrt{a}-2}\) nguyên\(< =>\sqrt{a}-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(=>a\in\left\{9;1;16;0\right\}\)(TM)
5 tháng 7 2021
a) P = \(\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}+\dfrac{1}{2-\sqrt{a}}\left(ĐKXĐ:a\ge0;a\ne4\right)\)
P = \(\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}-\dfrac{1}{\sqrt{a}-2}\)
P = \(\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
P = \(\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
P = \(\dfrac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
P = \(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b) Ta có: P = \(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) = 1 - \(\dfrac{2}{\sqrt{a}-2}\)
Để \(P\in Z\) <=> 1 - \(\dfrac{2}{\sqrt{a}-2}\) \(\in Z\) <=> \(\sqrt{a}-2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Ta có bảng sau:
| \(\sqrt{a}-2\) | 1 | -1 | 2 | -2 |
| \(\sqrt{a}\) | 3 | 1 | 4 | 0 |
| a | 9 (TM) | 1 (TM) | 16 (TM) | 0 (TM) |
Vậy để \(P\in Z\) thì \(a\in\left\{0;1;9;16\right\}\)
6 tháng 12 2023
\(P=\left(\dfrac{\sqrt{a}-b}{\sqrt{a}+b}-\dfrac{\sqrt{a}+b}{\sqrt{a}-b}\right)\cdot\left(\sqrt{a^3}-\dfrac{ab^2}{\sqrt{a}}\right)\)
\(=\dfrac{\left(\sqrt{a}-b\right)^2-\left(\sqrt{a}+b\right)^2}{\left(\sqrt{a}+b\right)\left(\sqrt{a}-b\right)}\cdot\dfrac{\sqrt{a^4}-ab^2}{\sqrt{a}}\)
\(=\dfrac{\left(\sqrt{a}-b-b-\sqrt{a}\right)\left(\sqrt{a}-b+b+\sqrt{a}\right)}{\left(\sqrt{a}+b\right)\left(\sqrt{a}-b\right)}\cdot\dfrac{a^2-ab^2}{\sqrt{a}}\)
\(=\dfrac{\left(-2b\right)\cdot\left(2\sqrt{a}\right)}{a-b^2}\cdot\dfrac{a\left(a-b^2\right)}{\sqrt{a}}\)
\(=\dfrac{-4b\sqrt{a}}{\sqrt{a}}\cdot a=-4ba\)
8 tháng 6 2021
Đk:\(a>2\)
\(\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}}{3+\sqrt{a-2}}+\dfrac{a+7}{11-a}\right):\left(\dfrac{3\sqrt{a-2}+1}{a-3\sqrt{a-2}-2}-\dfrac{1}{\sqrt{a-2}}\right)\)
Đặt \(b=\sqrt{a-2}\Leftrightarrow a=b^2+2\)
Biểu thức \(\Leftrightarrow\dfrac{b+2}{3}\left(\dfrac{b}{3+b}+\dfrac{b^2+2+7}{11-b^2-2}\right):\left(\dfrac{3b+1}{b^2-3b}-\dfrac{1}{b}\right)\)
\(=\dfrac{b+2}{3}\left[\dfrac{b}{3+b}-\dfrac{b^2+9}{b^2-9}\right]:\left[\dfrac{3b+1}{b\left(b-3\right)}-\dfrac{b-3}{b\left(b-3\right)}\right]\)
\(=\dfrac{b+2}{3}.\dfrac{b\left(b-3\right)-b^2-9}{\left(b-3\right)\left(3+b\right)}:\dfrac{3b+1-\left(b-3\right)}{b\left(b-3\right)}\)
\(=\dfrac{b+2}{3}.\dfrac{-3\left(b+3\right)}{\left(b-3\right)\left(3+b\right)}.\dfrac{b\left(b-3\right)}{2\left(b+2\right)}\)
\(=-\dfrac{b}{2}\)
\(=\dfrac{\sqrt{a-2}}{-2}\)
28 tháng 8 2021
a: Ta có: \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
b: Ta có: \(\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)-\left(2\sqrt{a}-1\right)+1\)
\(=a+\sqrt{a}-2\sqrt{a}+1+1\)
\(=a-\sqrt{a}+2\)
28 tháng 8 2021
a,ĐKXĐ: tự tìm :v
\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(x+2\sqrt{x}+1\right)-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+1\right)^2-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6+2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{9\sqrt{x}-x-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(9\sqrt{x}-9\right)-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{9\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(10-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\dfrac{10-\sqrt{x}}{\sqrt{x}+3}\)
29 tháng 12 2023
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne4\end{matrix}\right.\)
\(P=\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\cdot\left(\sqrt{a}-\dfrac{4}{\sqrt{ }a}\right)\)
\(=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\cdot\dfrac{a-4}{\sqrt{a}}\)
\(=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{\sqrt{a}}=\dfrac{-8\sqrt{a}}{\sqrt{a}}=-8\)
S
29 tháng 12 2023
\(P=\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\) (ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne4\end{matrix}\right.\))
\(=\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}-2-\sqrt{a}-2\right)\left(\sqrt{a}-2+\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}}\right)\)
\(=\dfrac{-4.2\sqrt{a}}{\sqrt{a}}\)
\(=-8\)
#YM
10 tháng 11 2021
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có
18 tháng 5 2021
`P=(sqrta+3)/(sqrta-2)-(sqrta-1)/(sqrta+2)+(4sqrta-4)/(4-a)`
`đk:x>=0,x ne 4`
`P=(a+5sqrta+6-a+3sqrta-2-4sqrta+4)/(a-4)`
`=(4sqrta+8)/(a-4)`
`=4/(sqrta-2)`
`b)a=9`
`=>P=4/(3-2)=4`
18 tháng 5 2021
a) Ta có: \(P=\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}+\dfrac{4\sqrt{a}-4}{4-a}\)
\(=\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{4\sqrt{a}+8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{4\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}=\dfrac{4}{\sqrt{a}-2}\)
b) Thay a=9 vào P, ta được:
\(P=\dfrac{4}{\sqrt{9}-2}=\dfrac{4}{3-2}=\dfrac{4}{1}=4\)
Vậy: khi a=9 thì P=4
Lời giải:
ĐKXĐ: $a\geq 0; a\neq 1$
\(P=\frac{a+3\sqrt{a}+2}{a+\sqrt{a}-2}=\frac{(a+\sqrt{a})+(2\sqrt{a}+2)}{(a-\sqrt{a})+(2\sqrt{a}-2)}\\ =\frac{\sqrt{a}(\sqrt{a}+1)+2(\sqrt{a}+1)}{\sqrt{a}(\sqrt{a}-1)+2(\sqrt{a}-1)}\\ =\frac{(\sqrt{a}+1)(\sqrt{a}+2)}{(\sqrt{a}-1)(\sqrt{a}+2)}=\frac{\sqrt{a}+1}{\sqrt{a}-1}\)