Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: ĐKXĐ: x>0; x<>4
b: \(P=\dfrac{\sqrt{x}+5\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{6\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-4-x}\)
\(=\dfrac{-6\sqrt{x}+4}{4}\)
c: Khi \(x=\dfrac{3-\sqrt{5}}{2}=\left(\dfrac{\sqrt{5}-1}{2}\right)^2\) thì \(P=\dfrac{-6\cdot\dfrac{\sqrt{5}-1}{2}+4}{4}=\dfrac{-3\left(\sqrt{5}-1\right)+4}{4}\)
\(=\dfrac{-3\sqrt{5}+7}{4}\)
\(a,ĐK:x>0;x\ne1\\ b,A=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ c,x=4\Leftrightarrow\sqrt{x}=2\Leftrightarrow A=\dfrac{2-1}{2}=\dfrac{1}{2}\)
tìm điều kiện xác định có thể rõ ràng chút được không ạ, chỗ này mình không hiểu lắm ý
a: ĐKXĐ: x>0; x<>1
b: \(A=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2}{x-1}\)
c: A nguyên
=>x-1 thuộc {1;-1;2;-2}
=>x thuộc {2;3}
`a)ĐK:` \(\begin{cases}x \ge 0\\x-\sqrt{x} \ne 0\\x-1 \ne 0\\\end{cases}\)
`<=>` \(\begin{cases}x \ge 0\\x \ne 0\\x \ne 1\\\end{cases}\)
`<=>` \(\begin{cases}x>0\\x \ne 1\\\end{cases}\)
`b)A=(sqrtx/(sqrtx-1)-1/(x-sqrtx)):(1/(1+sqrtx)+2/(x-1))`
`=((x-1)/(x-sqrtx)):((sqrtx-1+2)/(x-1))`
`=(x-1)/(x-sqrtx):(sqrtx+1)/(x-1)`
`=(sqrtx+1)/sqrtx:1/(sqrtx-1)`
`=(x-1)/sqrtx`
`c)A>0`
Mà `sqrtx>0AAx>0`
`<=>x-1>0<=>x>1`
a, ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
b, Ta có : \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{1}{\sqrt{x}-1}=\dfrac{x-1}{\sqrt{x}}\)
c, Ta có : \(A>0\)
\(\Leftrightarrow x-1>0\)
\(\Leftrightarrow x>1\)
Vậy ...
a) \(ĐKXĐ:x\ne4;x\ne9\)
b) \(A=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{-\sqrt{x}+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
c) Ta có: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-3\right)+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\) (ĐK: x thuộc Z)
| \(\sqrt{x}-3\) | 1 | -1 | 2 | -2 | 4 | -4 |
| \(\sqrt{x}\) | 4 | 2 | 5 | 1 | 7 | -1 |
| x | 2 | \(\sqrt{2}\) | \(\sqrt{5}\) | \(\sqrt{1}\) | \(\sqrt{7}\) | \(\varnothing\) |
Vậy để A thuộc Z khi x = {2;\(\sqrt{2};\sqrt{5};\sqrt{1};\sqrt{7}\) }
\(1,\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\in Z\Rightarrow\frac{4}{\sqrt{x}-3}\in Z\)
\(\Rightarrow\sqrt{x}-3\in\left(1;4;-1;-4\right)\)
\(\Rightarrow\sqrt{x}\in\left(4;7;2;-1\right)\)
\(\Rightarrow\sqrt{x}=4\Leftrightarrow x=2\)
\(4,A=x+\sqrt{x}+1\)
\(A=\left(\sqrt{x}\right)^2+2.\frac{1}{2}.\sqrt{x}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\)
\(A=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}\)
\(\Rightarrow A\ge\frac{3}{4}.\left(\sqrt{x}+\frac{1}{2}\right)^2\ge0\)
Dấu "=" xảy ra khi :
\(\sqrt{x}+\frac{1}{2}=0\Leftrightarrow\sqrt{x}=-\frac{1}{2}\)
Vậy Min A = 3/4 khi căn x = -1/2
\(a,b,M=\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x}\left(x\ge0;x\ne0;x\ne1\right)\\ M=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{x}{\sqrt{x}+1}\\ M=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\dfrac{x}{\sqrt{x}+1}\\ M=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\cdot\dfrac{x}{\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}-1\right)\)
\(c,M=\sqrt{x}\left(\sqrt{x}-1\right)=x-\sqrt{x}\\ =x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu \("="\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\)
\(M=\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x}\)
ĐKXĐ: \(x>0;x\ne1\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1}{x}\)
\(=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right).\dfrac{x}{\sqrt{x}+1}\)
\(=\dfrac{x-1}{x}.\dfrac{x}{\sqrt{x}+1}\)
\(=\sqrt{x}-1\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b: Ta có: \(A=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}\)
\(=2\sqrt{x}-1\)