Cho \(F=\left[\dfrac{3\sqrt{a}}{\sqrt{a}+4}+\dfrac{\sqrt{a}}{\sqrt{a}-4}+\dfrac{4\left(a+2\right)}{16-a}\right]:\left(1-\dfrac{2\sqrt{a}+5}{\sqrt{a}+4}\right)\)
a/ Rút gọn
b/ Tìm a thuộc N để F là số nguyên tố
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a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{2}{3}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\\a\ne b\end{matrix}\right.\)
P = \(\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}.\left[\left(\dfrac{a+\sqrt{ab}+b-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right):\dfrac{a-b}{a+\sqrt{ab}+b}\right]\)= \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}.\left[\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)
= \(\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}.\dfrac{\sqrt{a}-\sqrt{b}}{a-b}\)
= \(\dfrac{1}{a-\sqrt{ab}+b}\)
b) có a = 16 và b = 4 (thoả mãn ĐKXĐ)
Thay a = 16, b =4 vào P có:
P = \(\dfrac{1}{16-\sqrt{16.4}+4}\)= \(\dfrac{1}{12}\)
Vậy tại a =16, b = 4 thì P = \(\dfrac{1}{12}\)
`M=sqrt{(3a-1)^2}+2a-3`
`=|3a-1|+2a-3`
`=3a-1+2a-3(do \ a>=1/3)`
`=5a-4`
`N=sqrt{(4-a)^2}-a+5`
`=|4-a|-a+5`
`=a-4-a+5(do \ a>4)`
`=1`
`I=sqrt{(3-2a)^2}+2-7`
`=|3-2a|-5`
`=3-2a-5(do \ a<3/2)`
`=-2-2a`
`K=(a^2-9)/4*sqrt{4/(a-2)^2}`
`=(a^2-9)/4*|2/(a-2)|`
`=(a^2-9)/(2|a-2|)`
Nếu `3>a>2=>|a-2|=a-2`
`=>K=(a^2-9)/(2(a-2))`
Nếu `a<2=>|a-2|=2-a`
`=>K=(a^2-9)/(2(2-a))`
\(M=\left|3a-1\right|+2a-3\)
Mà \(a-\dfrac{1}{3}\ge0\)
\(\Rightarrow M=3a-1+2a-3=5a-4\)
\(N=\left|4-a\right|-a+5\)
Mà \(4-a< 0\)
\(\Rightarrow N=a-4-a+5=1\)
\(I=\left|3-2a\right|-5\)
Mà \(a-\dfrac{3}{2}< 0\)
\(\Rightarrow I=3-2a-5=-2a-2\)
K, Ta có : \(a-3< 0\)
\(\Rightarrow K=\dfrac{2\left(a^2-9\right)}{4\left|a-2\right|}=\dfrac{\left(a-3\right)\left(a+3\right)}{\left|2a-4\right|}\)
\(E=a^{12-4}.b^{3-7}=\dfrac{a^8}{b^4}\)
\(E=a^{4-6}.b^{3.4}=\dfrac{b^{12}}{a^2}\)
\(F=\dfrac{a^{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}}{a^{\left(\sqrt{5}-3\right)+\left(4-\sqrt{5}\right)}}=\dfrac{a^2}{a^1}=a\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
b) Ta có: \(M=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\dfrac{a-1}{2\sqrt{a}}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\sqrt{a}\left[\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2\right]}{2\sqrt{a}}\)
\(=\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{2}\)
\(=\dfrac{-4\sqrt{a}}{2}=-2\sqrt{a}\)
c) Để M=-4 thì \(-2\sqrt{a}=-4\)
\(\Leftrightarrow\sqrt{a}=2\)
hay a=4(thỏa ĐK)
a: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}+4}{x\sqrt{x}-3x+2\sqrt{x}}-\dfrac{3\sqrt{x}+3}{-x+\sqrt{x}+2}\right):\left(\dfrac{x-\sqrt{x}-6}{x-3\sqrt{x}}-\dfrac{x-2\sqrt{x}}{x-4\sqrt{x}+4}\right)+\sqrt{x}\)
\(=\left(\dfrac{\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{3}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)+\sqrt{x}\)
\(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-4-x}{\sqrt{x}\left(\sqrt{x}-2\right)}+\sqrt{x}\)
\(=\dfrac{4\left(\sqrt{x}-1\right)}{-4}+\sqrt{x}\)
\(=-\sqrt{x}-1+\sqrt{x}\)
=-1
a: \(F=\dfrac{3a-12\sqrt{a}+a+4\sqrt{a}-4a-8}{a-16}:\dfrac{\sqrt{a}+4-2\sqrt{a}-5}{\sqrt{a}+4}\)
\(=\dfrac{-8\left(\sqrt{a}+1\right)}{a-16}\cdot\dfrac{\sqrt{a}+4}{-\sqrt{a}-1}\)
\(=\dfrac{8}{\sqrt{a}-4}\)
b: Để F là số nguyên tố thì 8/căn a-4 là số nguyên tố
=>F=2
=>căn a-4=2
=>căn a=6
=>a=36