Tìm ĐKXĐ: \(\sqrt{x+2}=\sqrt{6-x}\)
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DN
Duy Nam
28 tháng 8 2021
-2 ≤ x ≤6
#HT
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Những câu hỏi liên quan
27 tháng 8 2021
ĐKXĐ: \(\left\{{}\begin{matrix}x+2\ge0\\6-x\ge0\end{matrix}\right.\)\(\Leftrightarrow6\ge x\ge-2\)
23 tháng 10 2021
ĐKXĐ: \(\left[{}\begin{matrix}x\ge\sqrt{5}\\x\le-\sqrt{5}\end{matrix}\right.\)
b: ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x< -12\end{matrix}\right.\)
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
H
7 tháng 4 2021
\(T=\sqrt{\dfrac{3\sqrt{x}}{\sqrt{x}-6}\cdot\dfrac{x-6\sqrt{x}}{\sqrt{x}-1}}=\sqrt{\dfrac{3\sqrt{x}}{\sqrt{x}-6}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-6\right)}{\sqrt{x}-1}}\\ =\sqrt{\dfrac{3\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}-1}}=\sqrt{\dfrac{3x}{\sqrt{x}-1}}\\ =\sqrt{\dfrac{3\left(x-1\right)+3}{\sqrt{x}-1}}=\sqrt{3\left(\sqrt{x}+1\right)+\dfrac{3}{\sqrt{x}-1}}\\ =\sqrt{3\left(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}\right)+6}\)
Áp dụng bất đẳng thức Cosi ta có:
\(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}\ge2\)
\(\Rightarrow T\ge\sqrt{3\cdot2+6}=2\sqrt{3}\)
Dấu = xảy ra khi x=4
Y
26 tháng 6 2023
\(a,\dfrac{3}{\sqrt{12x-1}}\) xác định \(\Leftrightarrow12x-1>0\Leftrightarrow12x>1\Leftrightarrow x>\dfrac{1}{12}\)
\(b,\sqrt{\left(3x+2\right)\left(x-1\right)}\) xác định \(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}3x+2\ge0\\x-1\ge0\end{matrix}\right.\\\left[{}\begin{matrix}3x+2\le0\\x-1\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-\dfrac{2}{3}\\x\ge1\end{matrix}\right.\\\left[{}\begin{matrix}x\le-\dfrac{2}{3}\\x\le1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\le-\dfrac{2}{3}\\x\ge1\end{matrix}\right.\)
\(c,\sqrt{3x-2}.\sqrt{x-1}\) xác định \(\Leftrightarrow\left[{}\begin{matrix}3x-2\ge0\\x-1\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{2}{3}\\x\ge1\end{matrix}\right.\) \(\Leftrightarrow x\ge1\)
\(d,\sqrt{\dfrac{-2\sqrt{6}+\sqrt{23}}{-x+5}}\) xác định \(\Leftrightarrow-x+5>0\Leftrightarrow x< 5\)
H
6 tháng 8 2023
\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(\text{đ}k\text{x}\text{đ}:x\ge3\right)\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{2\sqrt{x}-9-\left(x-9\right)-\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9-2x+4\sqrt{x}-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5\sqrt{x}-3x+2}{x-5\sqrt{x}+6}\)
__
Để \(M\in Z\) thì \(x-5\sqrt{x}+6\) thuộc ước của \(5\sqrt{x}-3x+2\)
\(\Rightarrow x-5\sqrt{x}+6=-5\sqrt{x}-3x+2\\ \Leftrightarrow x-5\sqrt{x}+6+5\sqrt{x}+3x-2=0\\ \Leftrightarrow4x-4=0\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)
H9
HT.Phong (9A5)
CTVHS
7 tháng 8 2023
a) ĐKXĐ: \(x\ge0;x\ne9;x\ne4\)
\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(M=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(M=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b) Ta có M ϵ Z thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3}{\sqrt{x}-3}+\dfrac{4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)
Phải thuộc Z vậy:
4 ⋮ \(\sqrt{x}-3\)
\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Mà: \(x\ge0,x\ne4,x\ne9\) nên \(\sqrt{x}-3\in\left\{1;2;-2;4\right\}\)
\(\Rightarrow x\in\left\{16;25;1;49\right\}\)
NT
0
AH
Akai Haruma
Giáo viên
17 tháng 12 2021
Bài 1:
ĐKXĐ: $3-2x\geq 0\Leftrightarrow x\leq \frac{3}{2}$
Bài 2:
a. ĐKXĐ: $x\geq \frac{1}{3}$
PT $\Leftrightarrow 3x-1=2^2=4$
$\Leftrightarrow x=\frac{5}{3}$ (tm)
b. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{x-2}+2\sqrt{x-2}=6$
$\Leftrightarrow 3\sqrt{x-2}=6$
$\Leftrightarrow \sqrt{x-2}=2$
$\Leftrightarrow x-2=4$
$\Leftrightarrow x=6$ (tm)
VC
3
27 tháng 8 2021
`sqrt{4x+20}-3sqrt{5+x}+4/3sqrt{9x+15}=6(x>=-5)`
`<=>sqrt{4(x+5)}-3sqrt{x+5}+4/3sqrt{9(x+5)}=6`
`<=>2sqrt{x+5}-3sqrt{x+5}+4sqrt{x+5}=6`
`<=>3sqrt{x+5}=6`
`<=>sqrt{x+5}=2`
`<=>x+5=4`
`<=>x=-1(tm)`
Vậy `x=-1`