2x2 - 3x = 0
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\(\Leftrightarrow x\left(2x+3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)
a) \(\left(x^2-3x\right)\left(x^2+7x+10\right)=216\Rightarrow x\left(x-3\right)\left(x+2\right)\left(x+5\right)=216\)
\(\Rightarrow x\left(x+2\right)\left(x-3\right)\left(x+5\right)=216\Rightarrow\left(x^2+2x\right)\left(x^2+2x-15\right)=216\)
Đặt \(t=x^2+2x\Rightarrow\) pt trở thành \(t\left(t-15\right)=216\Rightarrow t^2-15t-216=0\)
\(\Rightarrow\left(t+9\right)\left(t-24\right)=0\Rightarrow\left[{}\begin{matrix}t=-9\\t=24\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2+2x=-9\\x^2+2x=24\end{matrix}\right.\)
\(TH_1:x^2+2x=-9\Rightarrow x^2+2x+9=0\Rightarrow\left(x+1\right)^2+8=0\) (vô lý)
\(TH_2:x^2+2x=24\Rightarrow x^2+2x-24=0\Rightarrow\left(x-4\right)\left(x+6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-6\end{matrix}\right.\)
b) \(\left(2x^2-7x+3\right)\left(2x^2+x-3\right)+9=0\)
\(\Rightarrow\left(x-3\right)\left(2x-1\right)\left(x-1\right)\left(2x+3\right)+9=0\)
\(\Rightarrow\left(x-3\right)\left(2x+3\right)\left(x-1\right)\left(2x-1\right)+9=0\)
\(\Rightarrow\left(2x^2-3x-9\right)\left(2x^2-3x+1\right)+9=0\)
Đặt \(t=2x^2-3x-9\Rightarrow\) pt trở thành \(t\left(t+10\right)+9=0\)
\(\Rightarrow t^2+10t+9=0\Rightarrow\left(t+1\right)\left(t+9\right)=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-9\end{matrix}\right.\)
\(TH_1:t=-1\Rightarrow2x^2-3x-9=-1\Rightarrow2x^2-3x-8=0\)
\(\Delta=\left(-3\right)^2-4\left(-8\right).2=73\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{3-\sqrt{73}}{4}\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+\sqrt{73}}{4}\end{matrix}\right.\)
\(TH_2:t=-9\Rightarrow2x^2-3x-9=-9\Rightarrow2x^2-3x=0\Rightarrow x\left(2x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow2x^2+2x+x+1=0\)
\(\Leftrightarrow2x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy...
2x2 + 3x + 1 =0
<=> (2x2 + 2x) + (x + 1) = 0
<=> 2x(x + 1) + (x+1) = 0
<=> (2x+1)(x+1) = 0
<=> \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=-1\end{matrix}\right.\)
KL: x \(\in\left\{\dfrac{-1}{2};-1\right\}\)
Ta có: \(x^3-2x^2+3x-6>0\)
\(\Leftrightarrow x^3+3x-2x^2-6>0\)
\(\Leftrightarrow x\left(x^2+3\right)-2\left(x^2+3\right)>0\)
\(\Leftrightarrow x-2>0\)
hay x>2
\(\Leftrightarrow\left(x^2+1\right)^2+x\left(x^2+1\right)+2x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2+1+x\right)+2x\left(x^2+1+x\right)=0\)
\(\Leftrightarrow\left(x^2+1+2x\right)\left(x^2+1+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1=0\\x^2+x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô-nghiệm\right)\end{matrix}\right.\)
\(\Leftrightarrow x=-1\)
a) Ta có: \(\left(x^2-2x\right)^2-6x^2+12x+9=0\)
\(\Leftrightarrow\left(x^2-2x\right)^2-6\left(x^2-2x\right)+9=0\)
\(\Leftrightarrow\left(x^2-2x-3\right)^2=0\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow x^2-3x+x-3=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy: S={3;-1}
b) Ta có: \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)
\(\Leftrightarrow\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12=0\)
\(\Leftrightarrow\left(x^2+x\right)^2+5\left(x^2+x\right)-2\left(x^2+x\right)-10=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+5\right)-2\left(x^2+x+5\right)=0\)
\(\Leftrightarrow\left(x^2+x+5\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow x^2+x-2=0\)(Vì \(x^2+x+5>0\forall x\))
\(\Leftrightarrow x^2+2x-x-2=0\)
\(\Leftrightarrow x\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
Vậy: S={-2;1}
2 ý a và b anh CTV nãy đã làm rồi nha, còn câu c này thì làm dài dòng+không chắc :VVV
c)\(\left(2x^2-3x+1\right)\left(2x^2+5x+1\right)-9x^2=0\)
\(\Leftrightarrow\left(2x^2-3x+1\right)\left(2x^2-3x+1+8x\right)-9x^2=0\)
\(\Leftrightarrow\left(2x^2-3x+1\right)^2+8x\left(2x^2-3x+1\right)+16x^2-25x^2=0\)
\(\Leftrightarrow\left(2x^2-3x+1+4x\right)^2-25x^2=0\)
\(\Leftrightarrow\left(2x^2+x+1\right)^2-25x^2=0\)
\(\Leftrightarrow\left(2x^2+x+1-5x\right)\left(2x^2+x+1+5x\right)=0\)
\(\Leftrightarrow\left(2x^2-4x+1\right)\left(2x^2+6x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x^2-4x+1\right)=0\\\left(2x^2+6x+1\right)=0\end{matrix}\right.\)
Rồi đến đây tự giải nhé, không phân tích được thì bấm máy tính là ra nha:vv
help mik vs
\(2x^2-3x=0\Rightarrow x\left(2x-3\right)=0\)
\(\Rightarrow|^{x=0}_{2x-3=0\Rightarrow x=\dfrac{3}{2}}\)