Giải pt sau :
\(x^2-3x+1+\dfrac{\sqrt{3}}{3}.\sqrt{x^4+x^2+1}=0\)
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NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2021
a.
\(\Leftrightarrow4x^2-6x+1+\dfrac{1}{\sqrt{3}}\sqrt{\left(4x^2-2x+1\right)\left(4x^2+2x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{4x^2-2x+1}=a>0\\\sqrt{4x^2+2x+1}=b>0\end{matrix}\right.\) ta được:
\(2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)
\(\Leftrightarrow\left(a-\dfrac{b}{\sqrt{3}}\right)\left(2a+\sqrt{3}b\right)=0\)
\(\Leftrightarrow a=\dfrac{b}{\sqrt{3}}\)
\(\Leftrightarrow3a^2=b^2\)
\(\Leftrightarrow3\left(4x^2-2x+1\right)=4x^2+2x+1\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2021
b.
\(x^2-3x+1+\dfrac{1}{\sqrt{3}}\sqrt{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x+1}=b>0\end{matrix}\right.\)
\(\Rightarrow2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)
Lặp lại cách làm câu a
ND
Nguyễn Đức Trí
VIP
15 tháng 7 2023
1) \(\sqrt[]{3x+7}-5< 0\)
\(\Leftrightarrow\sqrt[]{3x+7}< 5\)
\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)
\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)
\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)
28 tháng 9 2021
1) \(\Leftrightarrow\sqrt{\left(x+5\right)^2}=4\)
\(\Leftrightarrow\left|x+5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=4\\x+5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)
2) \(ĐK:x\ge2\)
\(\Leftrightarrow\sqrt{x-2}=2\)
\(\Leftrightarrow x-2=4\Leftrightarrow x=6\left(tm\right)\)
3) \(\Leftrightarrow\left(x^2-x+4\right)-\sqrt{x^2-x+4}+\dfrac{1}{4}=\dfrac{9}{4}\)
\(\Leftrightarrow\left(\sqrt{x^2-x+4}-\dfrac{1}{2}\right)^2=\dfrac{9}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}-\dfrac{1}{2}=\dfrac{3}{2}\\\sqrt{x^2-x+4}-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}=2\\\sqrt{x^2-x+4}=-1\left(VLý\right)\end{matrix}\right.\)
\(\Leftrightarrow x^2-x+4=4\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
4) \(ĐK:x\ge0\)
\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)
DT
1
28 tháng 11 2021
a, ĐKXĐ: ...
\(\sqrt{3x^2-2x+6}+3-2x=0\)
\(\Leftrightarrow\sqrt{3x^2-2x+6}=2x-3\)
\(\Leftrightarrow3x^2-2x+6=4x^2-12x+9\)
\(\Leftrightarrow4x^2-10x+3=0\)
.....
b, ĐKXĐ: ...
\(\sqrt{x+1}+\sqrt{x-1}=4\\ \Leftrightarrow x+1+x-1+2\sqrt{\left(x+1\right)\left(x-1\right)}=16\\ \Leftrightarrow2\sqrt{x^2-1}=16-2x\\ \Leftrightarrow\sqrt{x^2-1}=8-x\\ \Leftrightarrow x^2-1=64-16x+x^2\\ \Leftrightarrow65-16x=0\\ \Leftrightarrow x=\dfrac{65}{16}\)
\(x^2-3x+1=\dfrac{\sqrt{3}}{3}\sqrt{x^4+x^2+1}=0\\ \Leftrightarrow x^2-3x+1=-\dfrac{\sqrt{3}}{3}\sqrt{x^4+x^2+1}\\ \Leftrightarrow\left(x^2-3x+1\right)^2=\dfrac{1}{3}\left(x^4+x^2+1\right)\\ \Leftrightarrow x^4+9x^2+1-6x^3-6x+2x^2=\dfrac{1}{3}x^4+\dfrac{1}{3}x^2+\dfrac{1}{3}\\ \Leftrightarrow3x^4+27x^2+3-18x^3-18x+6x^2=x^4+x^2+1\\ \Leftrightarrow2x^4-18x^3+32x^2-18x+2=0\\ \Leftrightarrow x^4-9x^3+16x^2-9x+1=0\\ \Leftrightarrow\left(x^4-2x^3+x^2\right)-\left(7x^3-14x^2+7x\right)+\left(x^2-2x+1\right)=0\\ \Leftrightarrow x^2\left(x^2-2x+1\right)-7x\left(x^2-2x+1\right)+\left(x^2-2x+1\right)=0\\ \Leftrightarrow\left(x^2-7x+1\right)\left(x^2-2x+1\right)=0\\ \Leftrightarrow\left(x^2-7x+\dfrac{49}{4}-\dfrac{45}{4}\right)\left(x^2-2x+1\right)=0\\ \Leftrightarrow\left[\left(x-\dfrac{7}{2}\right)^2-\dfrac{45}{4}\right]\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{2}-\dfrac{3\sqrt{5}}{2}\right)\left(x-\dfrac{7}{2}+\dfrac{3\sqrt{5}}{2}\right)\left(x-1\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{2}-\dfrac{3\sqrt{5}}{2}=0\\x-\dfrac{7}{2}+\dfrac{3\sqrt{5}}{2}=0\\\left(x-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7+3\sqrt{5}}{2}\\x=\dfrac{7-3\sqrt{5}}{2}\\x=1\end{matrix}\right.\)
Vậy.....................
@Akai Haruma