Giải phương trình : \(\sqrt{x+1}\) - 4x2 = \(\sqrt{3x}\) -1
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19 tháng 1
ĐKXĐ: x>=-1
\(4x^2-2\sqrt{x+1}=x+2\)
=>\(4x^2-2\sqrt{x+1}-x-2=0\)
=>\(4x^2+3x-4x-3+1-2\sqrt{x+1}=0\)
=>\(\left(4x+3\right)\left(x-1\right)+1-\sqrt{4x+4}=0\)
=>\(\left(4x+3\right)\left(x-1\right)+\dfrac{1-4x-4}{1+\sqrt{4x+4}}=0\)
=>\(\left(4x+3\right)\left(x-1\right)-\dfrac{4x+3}{1+\sqrt{4x+4}}=0\)
=>\(\left(4x+3\right)\left(x-1-\dfrac{1}{1+\sqrt{4x+4}}\right)=0\)
=>4x+3=0
=>x=-3/4(nhận)
KC
1
AH
Akai Haruma
Giáo viên
16 tháng 5 2021
Lời giải:
ĐKXĐ: $x\geq \frac{-1}{3}$
PT $\Leftrightarrow \frac{x}{\sqrt{x+2}}=\sqrt{3x+1}-\sqrt{x+1}$
$\Leftrightarrow \frac{x}{\sqrt{x+2}}=\frac{2x}{\sqrt{3x+1}+\sqrt{x+1}}$
$\Leftrightarrow x\left(\frac{1}{\sqrt{x+2}}-\frac{2}{\sqrt{3x+1}+\sqrt{x+1}}\right)=0$
Xét các TH:
TH1: $x=0$ (thỏa mãn)
TH2: $\frac{1}{\sqrt{x+2}}-\frac{2}{\sqrt{3x+1}+\sqrt{x+1}}$
$\Leftrightarrow \sqrt{3x+1}+\sqrt{x+1}=2\sqrt{x+2}$
$\Rightarrow 4x+2+2\sqrt{(3x+1)(x+1)}=4(x+2)$
$\Leftrightarrow \sqrt{(3x+1)(x+1)}=3$
$\Rightarrow (3x+1)(x+1)=9$
$\Leftrightarrow 3x^2+4x-8=0$
$\Rightarrow x=\frac{-2\pm 2\sqrt{7}}{3}$
Kết hợp với ĐKXĐ suy ra $x=\frac{-2+2\sqrt{7}}{3}$
Vậy............
Y
3
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 5 2021
ĐKXĐ: ...
\(\Leftrightarrow3x-1-x\sqrt{3x-1}+x\sqrt{x+1}-\sqrt{\left(x+1\right)\left(3x-1\right)}=0\)
\(\Leftrightarrow\sqrt{3x-1}\left(\sqrt{3x-1}-x\right)-\sqrt{x+1}\left(\sqrt{3x-1}-x\right)=0\)
\(\Leftrightarrow\left(\sqrt{3x-1}-\sqrt{x+1}\right)\left(\sqrt{3x-1}-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x-1}=\sqrt{x+1}\\\sqrt{3x-1}=x\end{matrix}\right.\)
\(\Leftrightarrow...\)
NT
1
4 tháng 2 2021
Hệ \(\Leftrightarrow x+1+3x-1+3\sqrt[3]{\left(x+1\right)\left(3x-1\right)}\left(\sqrt[3]{x+1}+\sqrt[3]{3x-1}\right)=x-1\)
\(\Leftrightarrow3x+1+3\sqrt[3]{\left(x+1\right)\left(3x-1\right)\left(x-1\right)}=0\)
\(\Leftrightarrow3x+1=-3\sqrt[3]{\left(x+1\right)\left(3x-1\right)\left(x-1\right)}\)
\(\Leftrightarrow27x^3+9x+27x^2+1=-27\left(x^2-1\right)\left(3x-1\right)\)
\(\Leftrightarrow27x^3+9x+27x^2+1+81x^3-81x-27x^2+27=0\)
\(\Leftrightarrow108x^3-72x+28=0\)
\(\Leftrightarrow x^3-\dfrac{2}{3}x+\dfrac{7}{27}=0\)
- AD công thức các đa nô :
\(\Rightarrow x=\sqrt[3]{-\dfrac{-\dfrac{2}{3}}{2}+\sqrt{\dfrac{\left(-\dfrac{2}{3}\right)^2}{4}+\dfrac{\left(\dfrac{7}{27}\right)^3}{27}}}+\sqrt[3]{-\dfrac{-\dfrac{2}{3}}{2}-\sqrt{\dfrac{\left(-\dfrac{2}{3}\right)^2}{4}+\dfrac{\left(\dfrac{7}{27}\right)^3}{27}}}\)
\(\Rightarrow x\approx-0,96685\)
DL
1
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 7 2021
ĐKXĐ: \(x\ge1\)
\(\sqrt{5x-1}=\sqrt{3x-2}+\sqrt{x-1}\)
\(\Leftrightarrow5x-1=3x-2+x-1+2\sqrt{\left(3x-2\right)\left(x-1\right)}\)
\(\Leftrightarrow x+2=2\sqrt{\left(3x-2\right)\left(x-1\right)}\)
\(\Leftrightarrow x^2+4x+4=4\left(3x-2\right)\left(x-1\right)\)
\(\Leftrightarrow11x^2-24x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{11}\left(loại\right)\\x=2\end{matrix}\right.\)
GM
2
9 tháng 11 2021
\(PT\Leftrightarrow\sqrt{3x+1}=\sqrt{x+4}+1\\ \Leftrightarrow3x+1=x+5+2\sqrt{x+4}\\ \Leftrightarrow2x-4=2\sqrt{x+4}\\ \Leftrightarrow x-2=\sqrt{x+4}\\ \Leftrightarrow x^2-4x+4=x+4\\ \Leftrightarrow x^2-5x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Thử lại ta thấy x=0 ko thỏa mãn
Vậy PT có nghiệm x=5
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 11 2021
ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
\(\sqrt{3x+1}=1+\sqrt{x+4}\)
\(\Leftrightarrow3x+1=1+x+4+2\sqrt{x+2}\)
\(\Leftrightarrow x+2-\sqrt{x+2}-4=0\)
Đặt \(\sqrt{x+2}=t\ge0\)
\(\Rightarrow t^2-t-4=0\Rightarrow\left[{}\begin{matrix}t=\dfrac{1+\sqrt{17}}{2}\\t=\dfrac{1-\sqrt{17}}{2}< 0\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+2}=\dfrac{1+\sqrt{17}}{2}\)
\(\Rightarrow x=\dfrac{5+\sqrt{17}}{2}\)
PT
1
13 tháng 8 2021
\(\sqrt{x+11}-\sqrt{10-3x}=\sqrt{1-x}\left(1\ge x\ge-11\right)\)
\(\Leftrightarrow\left(x+11\right)+\left(10-3x\right)-2\sqrt{\left(x+11\right)\left(10-3x\right)}=1-x\\ \Leftrightarrow-2x+21-2\sqrt{-3x^2-23x+110}=1-x\\ \Leftrightarrow-2\sqrt{-3x^2-23x+110}=x-20\\ \Leftrightarrow4\left(-3x^2-23x+110\right)=x^2-40x+400\\ \Leftrightarrow-12x^2-92x+440=x^2-40x+400\\ \Leftrightarrow13x^2+52x-40=0\)
\(\Delta=52^2-4\cdot\left(-40\right)\cdot13=4784>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4\sqrt{299}-52}{26}\\x=\dfrac{4\sqrt{299}-52}{26}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2\sqrt{299}-26}{13}\\x=\dfrac{2\sqrt{299}-26}{13}\end{matrix}\right.\)
Tick nha
\(\sqrt{x+1}-4x^2=\sqrt{3x}-1\left(x\ge0\right)\left(1\right)\)
\(\Leftrightarrow-4x^2+1+\sqrt{x+1}-\dfrac{\sqrt{6}}{2}=\sqrt{3x}-\dfrac{\sqrt{6}}{2}\)
\(\Leftrightarrow-\left(2x-1\right)\left(2x+1\right)+\dfrac{x+1-\dfrac{3}{2}}{\sqrt{x+1}+\dfrac{\sqrt{6}}{2}}=\dfrac{3x-\dfrac{3}{2}}{\sqrt{3x}+\dfrac{\sqrt{6}}{2}}\)
\(\Leftrightarrow-4\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)+\dfrac{x-\dfrac{1}{2}}{\sqrt{x+1}+\dfrac{\sqrt{6}}{2}}-\dfrac{3\left(x-\dfrac{1}{2}\right)}{\sqrt{3x}+\dfrac{\sqrt{6}}{2}}=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)\left[-4\left(x+\dfrac{1}{2}\right)+\dfrac{1}{\sqrt{x+1}+\dfrac{\sqrt{6}}{2}}-\dfrac{3}{\sqrt{3x}+\dfrac{\sqrt{6}}{2}}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\-4\left(x+\dfrac{1}{2}\right)+\dfrac{1}{\sqrt{x+1}+\dfrac{\sqrt{6}}{2}}-\dfrac{3}{\sqrt{3x}+\dfrac{\sqrt{6}}{2}}=0\left(2\right)\end{matrix}\right.\)
\(\left(x\ge0\right)\Rightarrow\left(2\right)< 0\Rightarrow\left(2\right)vô\) \(nghiệm\)
\(\Rightarrow S=\left\{\dfrac{1}{2}\right\}\)
\(\)