NM
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
13 tháng 10 2021
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
24 tháng 11 2021
\(a,ĐK:x>0;x\ne1\\ b,M=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ M=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x}{\sqrt{x}+1}=\sqrt{x}-1\\ c,M< 0\Leftrightarrow\sqrt{x}< 1\Leftrightarrow0< x< 1\)
28 tháng 9 2021
Câu 2:
a: Ta có: \(P=3x-\sqrt{x^2-10x+25}\)
\(=3x-\left|x-5\right|\)
\(=\left[{}\begin{matrix}3x-x+5=2x+5\left(x\ge5\right)\\3x+x-5=4x-5\left(x< 5\right)\end{matrix}\right.\)
b: Vì x=2<5 nên \(P=4\cdot2-5=8-5=3\)
TA
6 tháng 8 2021
a) ĐK: `x>=0; x \ne 1`
b) \(P=\left(3+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(3-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\\ =\dfrac{3\sqrt{x}+3+x+\sqrt{x}}{\sqrt{x}+1}.\dfrac{3\sqrt{x}-3-x+\sqrt{x}}{\sqrt{x}-1}\\ =\dfrac{x+4\sqrt{x}+1}{\sqrt{x}+1}.\dfrac{-x+4\sqrt{x}-3}{\sqrt{x}-1}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ =\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)=x-9\)
29 tháng 8 2021
a: ĐKXĐ: \(x>0\)
b: Ta có: \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\)
\(=x+\sqrt{x}-2\sqrt{x}-1+1\)
\(=x-\sqrt{x}\)
4 tháng 9 2021
\(a,b,M=\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x}\left(x\ge0;x\ne0;x\ne1\right)\\ M=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{x}{\sqrt{x}+1}\\ M=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\dfrac{x}{\sqrt{x}+1}\\ M=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\cdot\dfrac{x}{\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}-1\right)\)
\(c,M=\sqrt{x}\left(\sqrt{x}-1\right)=x-\sqrt{x}\\ =x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu \("="\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\)
4 tháng 9 2021
\(M=\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x}\)
ĐKXĐ: \(x>0;x\ne1\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1}{x}\)
\(=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right).\dfrac{x}{\sqrt{x}+1}\)
\(=\dfrac{x-1}{x}.\dfrac{x}{\sqrt{x}+1}\)
\(=\sqrt{x}-1\)
18 tháng 10 2021
\(a,ĐK:x>0;x\ne1\\ b,A=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ c,x=4\Leftrightarrow\sqrt{x}=2\Leftrightarrow A=\dfrac{2-1}{2}=\dfrac{1}{2}\)
18 tháng 10 2021
tìm điều kiện xác định có thể rõ ràng chút được không ạ, chỗ này mình không hiểu lắm ý
A
1
4 tháng 7 2021
a, ĐKXĐ : \(\left\{{}\begin{matrix}\dfrac{3x-5}{x-1}\ge0\\x-1\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x-5\ge0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}3x-5\le0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{3}\\x>1\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{5}{3}\\x< 1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{3}\\x< 1\end{matrix}\right.\)
Vậy ...
b, Ta có : \(A=\sqrt{\dfrac{3x-5}{x-1}}=3\)
\(\Leftrightarrow3x-5=9x-9\)
\(\Leftrightarrow x=\dfrac{2}{3}\left(TM\right)\)
Vậy ...
29 tháng 6 2021
`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`
29 tháng 6 2021
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 ...
\(P=\left(\frac{1}{x+1}+\frac{1}{x-1}\right):\frac{2x}{x-1}\)
a) Điều kiện xác định:
\(\hept{\begin{cases}x+1\ne0\\x-1\ne0\\2x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne0-1\\x\ne0+1\\x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne-1\\x\ne1\\x\ne0\end{cases}}\)
Vậy để P có nghĩa thì \(x\ne-1;x\ne1\) và \(x\ne0.\)
b) Rút gọn:
\(P=\left(\frac{1}{x+1}+\frac{1}{x-1}\right):\frac{2x}{x-1}\)
\(P=\left(\frac{1.\left(x-1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{1.\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}\right):\frac{2x}{x-1}\)
\(P=\left(\frac{x-1}{\left(x-1\right).\left(x+1\right)}+\frac{x+1}{\left(x-1\right).\left(x+1\right)}\right):\frac{2x}{x-1}\)
\(P=\left(\frac{x-1+x+1}{\left(x-1\right).\left(x+1\right)}\right):\frac{2x}{x-1}\)
\(P=\frac{2x}{\left(x-1\right).\left(x+1\right)}:\frac{2x}{x-1}\)
\(P=\frac{2x}{\left(x-1\right).\left(x+1\right)}.\frac{x-1}{2x}\)
\(P=\frac{2x.\left(x-1\right)}{2x.\left(x-1\right).\left(x+1\right)}\)
\(P=\frac{1}{x+1}.\)