Tìm ĐKXĐ của \(\sqrt{\dfrac{x^2}{2},}\sqrt{\dfrac{x-1}{-2}},\sqrt{x^2-4,}\sqrt{\dfrac{x-1}{2x^2}}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
AD
15 tháng 8 2023
\(a,\dfrac{-5}{x+6}\ge0\\ mà\left(-5< 0\right)\\ \Rightarrow x+6< 0\\ \Rightarrow x< -6\\ b,\dfrac{2}{6-x}\ge0\\ mà\left(2>0\right)\\ \Rightarrow6-x>0\\ \Rightarrow x< 6\\ c,\dfrac{-x+3}{-6}\ge0\\ mà-6< 0\\ \Rightarrow-x+3< 0\\ \Rightarrow x>3\\\)
\(d,\dfrac{7x-1}{-9}\ge0\\mà-9< 0\\ \Rightarrow 7x-1\le0\\ \Rightarrow x\le\dfrac{1}{7}\\ e,\dfrac{x+2}{x^2+2x+1}\ge0\\ mà\left(x^2+2x+1\right)>0\forall x\\ \Rightarrow x+2\ge0\\ \Rightarrow x\ge-2\\ f,\dfrac{x-2}{x^2-2x+4}\ge0\\ mà\left(x^2-2x+4\right)>0\forall x\\ \Rightarrow x-2\ge0\\ \Rightarrow x\ge2\)
Chứng minh : \(x^2-2x+4>0\\ x^2-2x+1+3=\left(x-1\right)^2+3\ge3>0\)
15 tháng 8 2023
a: ĐKXĐ: \(\dfrac{-5}{x+6}>=0\)
=>x+6<0
=>x<-6
b: ĐKXĐ: (-2)/(6-x)>=0
=>6-x<0
=>x>6
c: ĐKXĐ: (-x+3)/(-6)>=0
=>-x+3<=0
=>-x<=-3
=>x>=3
d: ĐKXĐ: (7x-1)/-9>=0
=>7x-1<=0
=>x<=1/7
e: ĐKXĐ: (x+2)/(x^2+2x+1)>=0
=>x+2>=0
=>x>=-1
f: ĐKXĐ: (x-2)/(x^2-2x+4)>=0
=>x-2>=0
=>x>=2
23 tháng 10 2021
a: ĐKXĐ: \(x\ge1\)
b: ĐKXĐ: \(x< 0\)
c: ĐKXĐ: \(\left[{}\begin{matrix}x\ge11\\x\le3\end{matrix}\right.\)
23 tháng 10 2021
1) ĐKXĐ: \(\left\{{}\begin{matrix}2x+11\ge0\\x-1\ge0\end{matrix}\right.\)\(\Leftrightarrow x\ge1\)
2) ĐKXĐ: \(\left\{{}\begin{matrix}-5x\ge0\\x\ne0\end{matrix}\right.\)\(\Leftrightarrow x< 0\)
3) ĐKXĐ: \(7x^2+1\ge0\left(đúng\forall x\right)\Leftrightarrow x\in R\)
4) ĐKXĐ: \(x^2-14x+33\ge0\Leftrightarrow\left(x-11\right)\left(x-3\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-11\ge0\\x-3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-11\le0\\x-3\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge11\\x\le3\end{matrix}\right.\)
5) ĐKXĐ:
+) \(-x^2+6x+16\ge0\)
\(\Leftrightarrow-\left(x^2-6x+9\right)+25\ge0\)
\(\Leftrightarrow\left(x-3\right)^2\le25\Leftrightarrow-5\le x-3\le5\)
\(\Leftrightarrow-2\le x\le8\)
+) \(3x^2\ne0\Leftrightarrow x\ne0\)
\(\Rightarrow\left\{{}\begin{matrix}-2\le x\le8\\x\ne0\end{matrix}\right.\)
29 tháng 8 2021
Bài 2:
Ta có: \(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)
3 tháng 2 2021
Điều kiện: x>2
P= \(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{2}+2}{\sqrt{x}-1}\right)\)
P= \(\left(\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
P= \(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
P= \(\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b) P= \(\dfrac{1}{4}\)
⇔\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}\) =\(\dfrac{1}{4}\)
⇔\(4\sqrt{x}-8=3\sqrt{x}\)
⇔\(\sqrt{x}=8\)
⇔x=64 (TM)
Vậy X=64(TMĐK) thì P=\(\dfrac{1}{4}\)
26 tháng 7 2023
a: ĐKXĐ: x>0; x<>4
\(P=\left(2-\sqrt{x}+2\right)\cdot\dfrac{1}{\sqrt{x}-2}=\dfrac{4-\sqrt{x}}{\sqrt{x}-2}\)
b: P=2/3
=>(4-căn x)/(căn x-2)=2/3
=>2căn x-4=12-3căn x
=>5căn x=16
=>x=256/25
c: Khi x=8-2căn 7 thì \(P=\dfrac{4-\sqrt{7}+1}{\sqrt{7}-1-2}=\dfrac{5-\sqrt{7}}{\sqrt{7}-3}=-4-\sqrt{7}\)
TA
10 tháng 6 2021
a) Biểu thức xác định `<=> x^2-2x-1>0`
`<=>(x^2-2x+1)-2>0`
`<=>(x-1)^2-(\sqrt2)^2>0`
`<=>(x-1+\sqrt2)(x-1-\sqrt2)>0`
`<=>` \(\left[{}\begin{matrix}x< 1-\sqrt{2}\\x>1+\sqrt{2}\end{matrix}\right.\)
`D=(-∞; 1-\sqrt2) \cup (1+\sqrt2 ; +∞)`
b) Biểu thức xác định `<=> x-\sqrt(2x+1)>0`
`<=> x>\sqrt(2x+1)`
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\2x+1\ge0\\x^2>2x+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-\dfrac{1}{2}\\\left[{}\begin{matrix}x< 1-\sqrt{2}\\x>1+\sqrt{2}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow x>1+\sqrt{2}\)
`D=(1+\sqrt2 ; +∞)`
H9
HT.Phong (9A5)
CTVHS
31 tháng 8 2023
\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{x-2\sqrt{x}+1}{x-1}\) (ĐK: \(x>0;x\ne4\))
\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(A=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(A=\dfrac{2\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(A=\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(A=\dfrac{2\sqrt{x}+2}{\sqrt{x}-1}\)
11 tháng 12 2020
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
11 tháng 7 2021
a, \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)ĐK : \(x\ge0;x\ne4\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b, Ta có :
\(P=2\Rightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}=2\Rightarrow3\sqrt{x}=2\sqrt{x}+4\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\)( tmđk )
Vậy P = 2 thì x = 16
* \(\sqrt{\dfrac{x^2}{2}}\) . ĐKXĐ: mọi x
* \(\sqrt{\dfrac{x-1}{-2}}\) . ĐKXĐ: \(\dfrac{x-1}{-2}\ge0\Leftrightarrow x-1\le0\) (vì -2<0) <=> x \(\le\) 1
* \(\sqrt{x^2-4}\) . ĐKXĐ: \(x^2-4\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)
* \(\sqrt{\dfrac{x-1}{2x^2}}\) .ĐKXĐ: \(\dfrac{x-1}{2x^2}\ge0\Leftrightarrow x-1\ge0\)(vì 2x^2 > 0 với mọi x) <=> x \(\ge\) 1