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BH
12 tháng 8 2019
\(\Leftrightarrow2\left(x-3\right)+\sqrt{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(2\sqrt{x-3}-\sqrt{x+3}\right)=0\)
\(\Rightarrow\sqrt{x-3}\left(2\sqrt{x-3}-\sqrt{8}-\sqrt{x+3}+\sqrt{8}\right)=0\)
\(\Leftrightarrow\sqrt{x-3}\cdot\sqrt{x-5}\left(\frac{1}{2\sqrt{x-3}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{x+3}}\right)=0\)
\(\Rightarrow\left(\frac{1}{2\sqrt{x-3}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{x+3}}\right)>0\left(\forall x\right)\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)
BH
12 tháng 8 2019
\(\Leftrightarrow\sqrt{x-3}\left(2\sqrt{x-3}+\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\sqrt{x-3}\left(2\sqrt{x-3}-\sqrt{8}+\sqrt{x+3}+\sqrt{8}\right)=0\)
\(\Leftrightarrow\sqrt{x-3}\cdot\sqrt{x-5}\cdot\left(\frac{1}{2\sqrt{x-3}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{x+3}}\right)=0\)
\(\left(\frac{1}{2\sqrt{x-3}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{x+3}}\right)>0\left(\forall x\right)\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)
25 tháng 6 2023
a/ x <hoac= -23/4
b/ x=2
a/ có 2xcăn6 > 2x2=4
=> 2 căn 6 > 3+1
<=> 2 căn 6 - 3 >1
b/ có 3 căn 2 > 3
=> 3 căn 2 - 9 > -6
=> 6 > 9- 3 căn 2
DT
Dang Tung
CTVHS
16 tháng 6 2023
\(\left(a\right):2x-7\sqrt{x}+3=0\left(x\ge0\right)\\ < =>\left(2x-6\sqrt{x}\right)-\left(\sqrt{x}-3\right)=0\\ < =>2\sqrt{x}\left(\sqrt{x}-3\right)-\left(\sqrt{x}-3\right)=0\\ < =>\left(2\sqrt{x}-1\right)\left(\sqrt{x}-3\right)=0\\ =>\left[{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}-3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{4}\left(TM\right)\\x=9\left(TM\right)\end{matrix}\right.\)
\(\left(b\right):3\sqrt{x}+5< 6\\ < =>3\sqrt{x}< 1\\ < =>\sqrt{x}< \dfrac{1}{3}\\ < =>0\le x< \dfrac{1}{9}\)
\(\left(c\right):x-3\sqrt{x}-10< 0\\ < =>\left(x-5\sqrt{x}\right)+\left(2\sqrt{x}-10\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}-5\right)+2\left(\sqrt{x}-5\right)< 0\\ < =>\left(\sqrt{x}-5\right)\left(\sqrt{x}+2\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}-5< 0\\\sqrt{x}+2>0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}0\le x< 25\\x\ge0\end{matrix}\right.< =>0\le x< 25\)
\(\left(d\right):x-5\sqrt{x}+6=0\left(x\ge0\right)\\ < =>\left(x-2\sqrt{x}\right)-\left(3\sqrt{x}-6\right)=0\\ < =>\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=0\\ < =>\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}-3=0\\\sqrt{x}-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=9\\x=4\end{matrix}\right.\left(TM\right)\)
\(\left(e\right):x+5\sqrt{x}-14< 0\\ < =>\left(x+7\sqrt{x}\right)-\left(2\sqrt{x}+14\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}+7\right)-2\left(\sqrt{x}+7\right)< 0\\ < =>\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}+7>0\\\sqrt{x}-2< 0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x\ge0\\0\le x< 4\end{matrix}\right.< =>0\le x< 4\)
DT
Đỗ Thanh Hải
CTVVIP
29 tháng 6 2021
a) \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\dfrac{1}{\sqrt{x+3}}\)(\(x\ge0,x\ne9\))
b) \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-3}=\sqrt{x}-2\left(x\ge0,x\ne9\right)\)
AT
29 tháng 6 2021
a) \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\dfrac{1}{\sqrt{x}+3}\)
b) \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)
c) \(6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(3-x\right)^2}=6-2x-\left|3-x\right|\)
mà \(x< 3\Rightarrow3-x>0\Rightarrow6-2x-\left|3-x\right|=6-2x-3+x=3-x\)
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
18 tháng 5 2021
b)đk:\(x\ge\dfrac{1}{2}\)
Có: \(\sqrt{2x^2-1}\le\dfrac{2x^2-1+1}{2}=x^2\)
\(x\sqrt{2x-1}=\sqrt{\left(2x^2-x\right)x}\le\dfrac{2x^2-x+x}{2}=x^2\)
=>\(\sqrt{2x^2-1}+x\sqrt{2x-1}\le2x^2\)
Dấu = xảy ra\(\Leftrightarrow x=1\)
Vậy....
c) đk: \(x\ge0\)
\(\Leftrightarrow\sqrt{x}=\sqrt{x+9}-\dfrac{2\sqrt{2}}{\sqrt{x+1}}\)
\(\Rightarrow x=x+9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
\(\Leftrightarrow0=9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
Đặt \(a=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\left(a>0\right)\)
\(\Leftrightarrow\dfrac{a^2-2}{2}=\dfrac{8}{x+1}\)
pttt \(9+\dfrac{a^2-2}{2}-4a=0\) \(\Leftrightarrow a=4\) (TM)
\(\Rightarrow4=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\) \(\Leftrightarrow16=\dfrac{2\left(x+9\right)}{x+1}\) \(\Leftrightarrow x=\dfrac{1}{7}\) (TM)
Vậy ...
18 tháng 5 2021
a)ĐKXĐ: x≥-1/3; x≤6
<=>\(\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{x-6}+1}+\left(x-5\right)\cdot\left(3x+1\right)=0\Leftrightarrow\left(x-5\right)\cdot\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{x-6}+1}+3x+1\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)(nhận)
(vì x≥-1/3 nên3x+1≥0 )
11 tháng 10 2020
Bài 1:
\(\frac{x-9}{\sqrt{x}+3}+\frac{2\sqrt{x}-6}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}+\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)
\(=\sqrt{x}-3+2=\sqrt{x}-1\)
Bài 2:
a) Không rõ đề
b) \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(\Leftrightarrow\left|x-3\right|=\sqrt{3}+1\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=\sqrt{3}+1\\x-3=-\sqrt{3}-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4+\sqrt{3}\\x=2-\sqrt{3}\end{cases}}\)
PT
1
H9
HT.Phong (9A5)
CTVHS
12 tháng 8 2023
a) \(\sqrt{x}-x-0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\1-\sqrt{x}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
b) \(x-\sqrt{2x-9}=6\)
\(\Leftrightarrow\sqrt{2x-9}=x-6\) (ĐK: \(x\ge\dfrac{9}{2}\))
\(\Leftrightarrow2x-9=\left(x-6\right)^2\)
\(\Leftrightarrow2x-9=x^2-12x+36\)
\(\Leftrightarrow x^2-14x+45=0\)
\(\Leftrightarrow x^2-5x-9x+45=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\)
c) \(3x-\sqrt{6x-\left(3-2\right)}=0\) (ĐK: \(x\ge\dfrac{1}{6}\))
\(\Leftrightarrow3x-\sqrt{6x-1}=0\)
\(\Leftrightarrow\sqrt{6x-1}=3x\)
\(\Leftrightarrow6x-1=9x^2\)
\(\Leftrightarrow9x^2-6x+1=0\)
\(\Leftrightarrow\left(3x-1\right)^2=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\left(tm\right)\)
CD
0
\(\Leftrightarrow\sqrt{x-3}\left(2\sqrt{x-3}+\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\sqrt{x-3}\left(2\sqrt{x-3}-\sqrt{8}+\sqrt{x+3}+\sqrt{8}\right)=0\)
\(\Leftrightarrow\sqrt{x-3}\cdot\sqrt{x-5}\cdot\left(\frac{1}{2\sqrt{x-3}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{x+3}}\right)=0\)
\(\left(\frac{1}{2\sqrt{x-3}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{x+3}}\right)>0\left(\forall x\right)\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)