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NT
0
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 )
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: \(x\ge\dfrac{1}{6}\))
\(\Leftrightarrow3x-\sqrt{6x-1}=0\)
\(\Leftrightarrow\sqrt{6x-1}=3x\)
\(\Leftrightarrow6x-1=9x^2\)
\(\Leftrightarrow9x^2-6x+1=0\)
\(\Leftrightarrow\left(3x-1\right)^2=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\left(tm\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 1
ĐKXĐ: \(-\dfrac{1}{3}\le x\le6\)
\(\left(\sqrt{3x+1}-4\right)+\left(1-\sqrt{6-x}\right)+\left(3x^2-14x-5\right)=0\)
\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{x-5}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1\right)=0\)
\(\Leftrightarrow x-5=0\) (do \(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1>0;\forall x\))
\(\Rightarrow x=5\)
17 tháng 1
ĐKXĐ: \(\left\{{}\begin{matrix}3x+1>=0\\6-x>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{3}\\x< =6\end{matrix}\right.\)
\(\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0\)
=>\(\sqrt{3x+1}-4+1-\sqrt{6-x}+3x^2-14x-5=0\)
=>\(\dfrac{3x+1-16}{\sqrt{3x+1}+4}+\dfrac{1-6+x}{1+\sqrt{6-x}}+3x^2-15x+x-5=0\)
=>\(\dfrac{3\cdot\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{6-x}+1}+\left(x-5\right)\left(3x+1\right)=0\)
=>\(\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{6-x}+1}+3x+1\right)=0\)
=>x-5=0
=>x=5(nhận)
vẫn có trường hợp >0 mà bn
min \(min\frac{x}{\sqrt{x}-2}voix>4\)