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5 tháng 4 2022
a: Khi x=4 thì \(A=\left(\dfrac{2+2}{2+1}-\dfrac{2\cdot2-2}{2-1}\right)\cdot\left(4-1\right)=\dfrac{1}{3}\cdot3=1\)
b: \(A=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-2\right)\cdot\left(x-1\right)\)
\(=\dfrac{\sqrt{x}+2-2\sqrt{x}-2}{\sqrt{x}+1}\cdot\left(x-1\right)=-\sqrt{x}\left(\sqrt{x}-1\right)\)
17 tháng 5 2021
`A=(1/(x-sqrtx)+1/(sqrtx-1)):(sqrtx+1)/(sqrtx-1)^2`
`=((sqrtx+1)/(x-sqrtx)).(sqrtx-1)^2/(sqrtx+1)`
`=(sqrtx-1)^2/(x-sqrtx)`
`=(sqrtx-1)/sqrtx`
HP
0
AT
6 tháng 7 2021
a) \(A=\dfrac{3x}{x\sqrt{x}+1}-\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{1}{1+\sqrt{x}}\)
\(=\dfrac{3x}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{1}{1+\sqrt{x}}\)
\(=\dfrac{3x-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(\dfrac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
AT
12 tháng 7 2021
a) \(P=\dfrac{3x+3\sqrt{x}-9}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)
\(=\dfrac{3x+3\sqrt{x}-9}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{3x+3\sqrt{x}-9+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)
b) \(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+2}{\sqrt{x}+2}=3+\dfrac{2}{\sqrt{x}+2}\)
Để \(P\in Z\Rightarrow2⋮\sqrt{x}+2\Rightarrow\sqrt{x}+2=2\left(\sqrt{x}+2\ge2\right)\)
\(\Rightarrow x=0\)
c) Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\Rightarrow3+\dfrac{2}{\sqrt{x}+2}\le4\)
\(\Rightarrow P_{max}=4\) khi \(x=0\)
21 tháng 5 2019
Ta có P=\(\frac{20-x-5\sqrt{x}+4\sqrt{x}}{\sqrt{x}+5}\)
P=\(\frac{\sqrt{x}\left(4-\sqrt{x}\right)+5\left(4-\sqrt{x}\right)}{\sqrt{x}+5}\)
P=\(\frac{\left(\sqrt{x}+5\right).\left(4-\sqrt{x}\right)}{\sqrt{x}+5}\)
P=\(4-\sqrt{x}\)
b) Ta có P=\(4-\sqrt{x}\)\(\le\)4 với mọi x\(\ge0\)
=> P đạt GTLN là 4 khi \(\sqrt{x}=0\)
=> x=0
30 tháng 8 2021
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\)
b: Thay x=16 vào A, ta được:
\(A=\dfrac{3}{4+3}=\dfrac{3}{7}\)
ĐKXĐ: x>=0
\(A=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x-3}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}-\dfrac{x-3}{\sqrt{x}+1}\)
\(=\dfrac{x-\sqrt{x}-x+3}{\sqrt{x}+1}=\dfrac{-\sqrt{x}+3}{\sqrt{x}+1}\)
\(=\dfrac{-\sqrt{x}-1+4}{\sqrt{x}+1}=-1+\dfrac{4}{\sqrt{x}+1}\)
\(\dfrac{4}{\sqrt{x}+1}< =\dfrac{4}{1}=4\)
=>\(\dfrac{4}{\sqrt{x}+1}-1< =4-1=3\)
Dấu = xảy ra khi x=0