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d: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(3\left(2x-3\right)\left(3x+2\right)-2\left(x+4\right)\left(4x-3\right)+9x\left(4-x\right)=0\)
\(\Leftrightarrow x^2-5x+6=0\)
\(\Leftrightarrow\left(x^2-3x\right)+\left(-2x+6\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
a: Ta có: \(-\left(-3x^2\right)^3+4x-9-27x^6\)
\(=27x^6-27x^6+4x-9\)
=4x-9
=-1
\(5x^2+4x+2x^3+x^4-12=0\)
\(\Leftrightarrow x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)
\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^3+2x^2+x^2+2x+6x+12\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[x^2+2\times\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2+6\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\right]\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\\left(x^2+\dfrac{1}{2}\right)^2+\dfrac{23}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vì \(\left(x^2+\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x^2+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}\forall x\)
\(\Rightarrow\left(x^2+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\) vô nghiệm
Vậy phương trình có tập nghiệm là\(S=\left\{1;-2\right\}\)
\(x^4+2x^3-4x=4\)
\(\Leftrightarrow\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)=0\)
\(\Leftrightarrow x^2-2=0\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
\(\Rightarrow x^4+2x^3-4x-4=0\\ \Rightarrow x^4-2x^2+2x^3-4x+2x^2-4=0\\ \Rightarrow\left(x^2-2\right)\left(x^2+2x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=2\\\left(x+1\right)^2+1=0\left(vô.lí\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
x2+4x+4=0
(x+2)2=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)2+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0 Th2: x+1=0
x=-3 x=-1
vậy ...
(2x-3)^2-4x(x-3)= 0
=> 4x^2-12x+9 - 4x^2 + 12x=0
=> 9=0 ( vô cmn lí )
=> vô nghiệm
Sai or đúng chưa rõ tự kiểm tra oke
\(m^2x+6m=4x-12\)
\(\Leftrightarrow m^2x-4x=-6m-12\)
\(\Leftrightarrow\left(m^2-4\right)x=-6\left(m+2\right)\)
\(\Leftrightarrow\left(m-2\right)\left(m+2\right)x=-6\left(m+2\right)\)
- Nếu \(m=-2\Leftrightarrow m+2=0\) pt trở thành:
\(0.x=-6.0\Leftrightarrow0=0\) (luôn đúng)
Phương trình có vô số nghiệm
- Nếu \(m=2\Rightarrow m-2=0\) pt trở thành:
\(0.x=-6.4\Leftrightarrow0=-24\) (vô lý)
Phương trình vô nghiệm
- Nếu \(m\ne\pm2\Rightarrow\left(m-2\right)\left(m+2\right)\ne0\) ta được:
\(\left(m-2\right)\left(m+2\right)x=-6\left(m+2\right)\)
\(\Leftrightarrow x=\dfrac{-6\left(m+2\right)}{\left(m-2\right)\left(m+2\right)}\)
\(\Leftrightarrow x=-\dfrac{6}{m-2}\)
Vậy:
Nếu \(m=2\) pt vô nghiệm
Nếu \(m=-2\) pt có vô số nghiệm
Nếu \(m\ne\pm2\) pt có nghiệm duy nhất: \(x=-\dfrac{6}{m-2}\)
\(\Leftrightarrow x^4+2x^3+2x^2-2x^2-4x-4=0\)
\(\Leftrightarrow x^2\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)=0\)
\(\Leftrightarrow\left(x^2-2\right)\left(x^2+2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\\left(x+1\right)^2+1=0\left(\text{vô nghiệm}\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\sqrt{2}\)