NL
Nguyễn Lê Phước Thịnh
19 tháng 10 2023
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HP
3 tháng 9 2021
\(C=\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right).\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}-2}\)
\(=\left[\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{x\left(\sqrt{x}+2\right)-4\left(\sqrt{x}+2\right)}{\sqrt{x}-2}\)
\(=\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\right].\dfrac{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)
\(=\left[\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}-\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\right].\left(\sqrt{x}+2\right)^2\)
\(=\dfrac{6\sqrt{x}}{\sqrt{x}-2}\)
3 tháng 9 2021
\(C=\left[\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{\sqrt{x}\left(x-4\right)+2\left(x-4\right)}{\sqrt{x}-2}\) (\(x\ge0,x\ne4,x\ne9\))
\(C=\left[\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}\right].\dfrac{\left(\sqrt{x}+2\right)\left(x-4\right)}{\sqrt{x}-2}\)
\(C=\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}{\sqrt{x}-2}\)
\(C=\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}.\left(\sqrt{x}+2\right)^2\)
\(C=\dfrac{2}{\sqrt{x}-2}\)
TT
14 tháng 8 2023
\(a,A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\left(dk:x\ge0,x\ne4\right)\\ =\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+10-x}{\sqrt{x}+2}\right)\\ =\dfrac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x-4+10-x}\)
\(=\dfrac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\sqrt{x}-2}.\dfrac{1}{6}\\ =\dfrac{-6}{\left(\sqrt{x}-2\right).6}\\
=-\dfrac{1}{\sqrt{x}-2}\)
\(b,A>0\Leftrightarrow-\dfrac{1}{\sqrt{x}-2}>0\Leftrightarrow\sqrt{x}-2< 0\\
\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
Kết hợp với \(dk:x\ge0,x\ne4\), ta kết luận \(0\le x< 4\)
19 tháng 8 2023
a: \(A=\left(\dfrac{\left(x-4\right)\left(\sqrt{x}+2\right)-x\sqrt{x}+8}{x-4}\right):\dfrac{x-2\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8-x\sqrt{x}+8}{x-4}\cdot\dfrac{\sqrt{x}+2}{x-2\sqrt{x}+4}\)
\(=\dfrac{2x-4\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{1}{x-2\sqrt{x}+4}=\dfrac{2\sqrt{x}}{x-2\sqrt{x}+4}\)
b: \(A-1=\dfrac{2\sqrt{x}-x+2\sqrt{x}-4}{x-2\sqrt{x}+4}\)
\(=\dfrac{-x+4\sqrt{x}-4}{x-2\sqrt{x}+4}=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-1\right)^2+3}< 0\)
=>A<1
c: \(2\sqrt{x}>=0;x-2\sqrt{x}+4=\left(\sqrt{x}-1\right)^2+3>0\)
=>A>=0 với mọi x thỏa mãn ĐKXĐ
mà A<1
nên 0<=A<1
=>Để A nguyên thì A=0
=>x=0
15 tháng 7 2021
`A=((3sqrtx+6)/(x-4)+sqrtx/(sqrtx-2)):(x-9)/(sqrtx-3)(x>=0,x ne 4,x ne 9)`
`=((3(sqrtx+2))/((sqrtx-2)(sqrtx+2))+sqrtx/(sqrtx-2)):((sqrtx-3)(sqrtx+3))/(sqrtx-3)`
`=(3/(sqrtx-2)+sqrtx/(sqrtx-2)):(sqrtx+3)`
`=(sqrtx+3)/(sqrtx-2)*1/(sqrtx+3)`
`=1/(sqrtx-2)`
AT
15 tháng 7 2021
\(A=\left(\dfrac{3\sqrt{x}+6}{x-4}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{x-9}{\sqrt{x}-3}\)
\(=\left(\dfrac{3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}-3}\)
\(=\left(\dfrac{3}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\left(\sqrt{x}+3\right)=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}.\dfrac{1}{\sqrt{x}+3}=\dfrac{1}{\sqrt{x}-2}\)
11 tháng 8 2021
Ta có: \(\dfrac{\sqrt{x}\left(16-\sqrt{x}\right)}{x-4}+\dfrac{3+2\sqrt{x}}{2-\sqrt{x}}-\dfrac{2-3\sqrt{x}}{\sqrt{x}+2}\)
\(=\dfrac{16\sqrt{x}-x-\left(3+2\sqrt{x}\right)\left(\sqrt{x}+2\right)+\left(3\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{16\sqrt{x}-x-3\sqrt{x}-6-2x-4\sqrt{x}+3x-6\sqrt{x}-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{1}{\sqrt{x}+2}\)
1 tháng 2 2022
a, \(A=\dfrac{4\left(3-\sqrt{7}\right)}{2}+2\sqrt{7}=\dfrac{12}{2}=6\)
b, \(B=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\sqrt{x}}\right):\dfrac{2-\sqrt{x}}{x-1}\)
\(=\left(\dfrac{\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{2-\sqrt{x}}{x-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
1 tháng 2 2022
nhờ bạn làm rõ vì sao \(\dfrac{\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{2-\sqrt{x}}{x-1}\) lại bằng \(\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
mình xin cảm ơn
29 tháng 5 2023
Sửa đề: x-4
\(A=\dfrac{x-2\sqrt{x}+x+4\sqrt{x}+4+2x+8}{x-4}=\dfrac{4x+2\sqrt{x}+12}{x-4}\)
10 tháng 7 2021
a) Ta có: \(\dfrac{6}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{10}{\sqrt{5}}\)
\(=\dfrac{6\left(\sqrt{5}-1\right)}{4}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{4}}-2\sqrt{5}\)
\(=\dfrac{3}{2}\left(\sqrt{5}-1\right)+\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-2\sqrt{5}\)
\(=\dfrac{3}{2}\sqrt{5}-\dfrac{3}{2}-2\sqrt{5}+\dfrac{\sqrt{5}+1}{2}\)
\(=-\dfrac{1}{2}\sqrt{5}-\dfrac{3}{2}+\dfrac{1}{2}\sqrt{5}+\dfrac{1}{2}\)
=-1
10 tháng 7 2021
Bài 1:
a) Thay \(x=\dfrac{1}{4}\)vào B, ta được:
\(B=1:\left(\dfrac{1}{4}\cdot\dfrac{1}{2}+27\right)=1:\left(27+\dfrac{1}{8}\right)=\dfrac{8}{217}\)
b) Ta có: \(A=\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(=\dfrac{x-9+\sqrt{x}+3-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+\sqrt{x}-6-x+2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\)
c) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow3-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 3\)
hay x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne4\end{matrix}\right.\)
