Giải phương trình sau:
\(x^3-3x^2+2\sqrt{\left(x+2\right)^3}-6x=0\)
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I
2
10 tháng 12 2021
\(PT\Leftrightarrow\left(x^3+6x^2+12x+8\right)+2\sqrt{\left(x+2\right)^3}+1-9x^2-18x-9=0\\ \Leftrightarrow\left(x+2\right)^3+2\sqrt{\left(x+2\right)^3}+1-9\left(x+1\right)^2=0\\ \Leftrightarrow\left(\sqrt{\left(x+2\right)^3}+1\right)^2-9\left(x+1\right)^2=0\\ \Leftrightarrow\left[\sqrt{\left(x+2\right)^3}-3x-2\right]\left[\sqrt{\left(x+2\right)^3}+3x+4\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{\left(x+2\right)^3}=3x+2\\\sqrt{\left(x+2\right)^3}=-3x-4\end{matrix}\right.\)
\(TH_1:\sqrt{\left(x+2\right)^3}=3x+2\\ \Leftrightarrow x^3+6x^2+12x+8=9x^2+12x+4\left(x\ge-\dfrac{2}{3}\right)\\ \Leftrightarrow x^3-3x^2+4=0\\ \Leftrightarrow x^3+x^2-4x^2+4=0\\ \Leftrightarrow x^2\left(x+1\right)-4\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
\(TH_2:\sqrt{\left(x+2\right)^3}=-3x-4\\ \Leftrightarrow x^3+6x^2+12x+8=9x^2+24x+16\left(x\le-\dfrac{4}{3}\right)\\ \Leftrightarrow x^3-3x^2-12x-8=0\\ \Leftrightarrow x^3+x^2-4x^2-4x-8x-8=0\\ \Leftrightarrow\left(x+1\right)\left(x^2-4x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=2+2\sqrt{3}\left(ktm\right)\\x=2-2\sqrt{3}\left(tm\right)\end{matrix}\right.\)
Vậy PT có nghiệm \(S=\left\{2;2-2\sqrt{3}\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 12 2021
ĐKXĐ: \(x\ge-2\)
\(x^3-3x\left(x+2\right)+2\sqrt{\left(x+2\right)^3}=0\)
Đặt \(\sqrt{x+2}=a\ge0\) pt trở thành:
\(x^3-3x.a^2+2a^3=0\)
\(\Leftrightarrow\left(x-a\right)^2\left(x+2a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=x\\2a=-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}=x\left(x\ge0\right)\\2\sqrt{x+2}=-x\left(x\le0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-x-2=0\\x^2-4x-8=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=2\\x=2+2\sqrt{3}\left(loại\right)\\x=2-2\sqrt{3}\end{matrix}\right.\)
8 tháng 5 2021
a, ĐKXĐ : \(\left[{}\begin{matrix}x\le-3\\x\ge0\end{matrix}\right.\)
TH1 : \(x\le-3\) ( LĐ )
TH2 : \(x\ge0\)
BPT \(\Leftrightarrow x^2+2x+x^2+3x+2\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge4x^2\)
\(\Leftrightarrow\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge x^2-\dfrac{5}{2}x\)
\(\Leftrightarrow2\sqrt{\left(x+2\right)\left(x+3\right)}\ge2x-5\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x^2+20x+24\ge4x^2-20x+25\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0\le x< \dfrac{5}{2}\\x\ge\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow x\ge0\)
Vậy \(S=R/\left(-3;0\right)\)
AH
Akai Haruma
Giáo viên
30 tháng 7 2021
a.
$x^2-11=0$
$\Leftrightarrow x^2=11$
$\Leftrightarrow x=\pm \sqrt{11}$
b. $x^2-12x+52=0$
$\Leftrightarrow (x^2-12x+36)+16=0$
$\Leftrightarrow (x-6)^2=-16< 0$ (vô lý)
Vậy pt vô nghiệm.
c.
$x^2-3x-28=0$
$\Leftrightarrow x^2+4x-7x-28=0$
$\Leftrightarrow x(x+4)-7(x+4)=0$
$\Leftrightarrow (x+4)(x-7)=0$
$\Leftrightarrow x+4=0$ hoặc $x-7=0$
$\Leftrightarrow x=-4$ hoặc $x=7$
AH
Akai Haruma
Giáo viên
30 tháng 7 2021
d.
$x^2-11x+38=0$
$\Leftrightarrow (x^2-11x+5,5^2)+7,75=0$
$\Leftrightarrow (x-5,5)^2=-7,75< 0$ (vô lý)
Vậy pt vô nghiệm
e.
$6x^2+71x+175=0$
$\Leftrightarrow 6x^2+21x+50x+175=0$
$\Leftrightarrow 3x(2x+7)+25(2x+7)=0$
$\Leftrightarrow (3x+25)(2x+7)=0$
$\Leftrightarrow 3x+25=0$ hoặc $2x+7=0$
$\Leftrightarrow x=-\frac{25}{3}$ hoặc $x=-\frac{7}{2}$
UN
0
UN
0
UN
0
JB
5 tháng 2 2021
ở VP "+4" nằm ở ngoài căn,đau bụng nên viết vội còn chạy ra WC :P
MH
1
\(ĐK:x\ge-2\)
\(\Leftrightarrow x^3+6x^2+12x+8+2\sqrt{\left(x+2\right)^3}+1-9x^2-18x-9=0\)
\(\Leftrightarrow\left(x+2\right)^3+2\sqrt{\left(x+2\right)^3}+1-\left(9x^2+18x+9\right)=0\)
\(\Leftrightarrow\left[\left(x+2\right)^3+1\right]^2-9\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left[\left(x+2\right)^3+1\right]^2-9\left(x+1\right)^2=0\)
ta có: ( 2 trường hợp xảy ra )
TH1: \(\left[\left(x+2\right)^3+1\right]^2=9\left(x+1\right)^2\)
\(\Leftrightarrow\left(x+2\right)^3+1=\left(9x+9\right)\)
\(\Leftrightarrow\left(x+2\right)^3-9x=8\)
\(\Leftrightarrow x^3+6x^2+12x+8-9x-8=0\)
\(\Leftrightarrow x^3+6x^2+3x=0\)
\(\Leftrightarrow x\left(x^2+6x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+6x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\\left[{}\begin{matrix}x=-3+\sqrt{6}\left(n\right)\\-3-\sqrt{6}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
TH2:\(\left[{}\begin{matrix}\left(x+3\right)^3+1=0\\9\left(x+1\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)^3=-1\\\left(9x+9\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=-1\\9x=-9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\left(l\right)\\x=-1\left(n\right)\end{matrix}\right.\)
Vậy \(S=\left\{0;-1;-3+\sqrt{6}\right\}\)
( ko bít đúng ko nha bạn ơi )