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Mai Trung Hải Phong · Accepted Answer

e) $\left(x^2-8\right)\left(x-2023\right)-x+2023=0$$\Rightarrow\left(x^2-8\right)\left(x-2023\right)-\left(x-2023\right)=0$$\Rightarrow\left(x^2-8-1\right)\left(x-2023\right)=0$$\Rightarrow\left(x^2-9\right)\left(x-2023\right)=0$$\Rightarrow\left[{}\begin{matrix}x^2-9=0\x-2023=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x^2=9\x=2023\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x=\pm3\x=2023\end{matrix}\right.$Vậy PT có cặp nghiệm $S=\left\{\pm3;2023\right\}$g) $x^4-2x^3+10x^2-20x=0$$\Rightarrow\left(x^4-2x^3\right)+\left(10x^2-20x\right)=0$$\Rightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0$$\Rightarrow\left(x^3+10x\right)\left(x-2\right)=0$$\Rightarrow x\left(x^2+10\right)\left(x-2\right)=0$$\Rightarrow\left[{}\begin{matrix}x=0\x^2+10=0\x-2=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x=0\x^2=-10\x=2\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x=0\x=2\end{matrix}\right.$Vậy PT có cặp nghiệm $S=\left\{0;2\right\}$Câu trả lời: e) $\left(x^2-8\right)\left(x-2023\right)-x+2023=0$$\Rightarrow\left(x^2-8\right)\left(x-2023\right)-\left(x-2023\right)=0$$\Rightarrow\left(x^2-8-1\right)\left(x-2023\right)=0$$\Rightarrow\left(x^2-9\right)\left(x-2023\right)=0$$\Rightarrow\left[{}\begin{matrix}x^2-9=0\x-2023=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x^2=9\x=2023\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x=\pm3\x=2023\end{matrix}\right.$Vậy PT có cặp nghiệm $S=\left\{\pm3;2023\right\}$g) $x^4-2x^3+10x^2-20x=0$$\Rightarrow\left(x^4-2x^3\right)+\left(10x^2-20x\right)=0$$\Rightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0$$\Rightarrow\left(x^3+10x\right)\left(x-2\right)=0$$\Rightarrow x\left(x^2+10\right)\left(x-2\right)=0$$\Rightarrow\left[{}\begin{matrix}x=0\x^2+10=0\x-2=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x=0\x^2=-10\x=2\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x=0\x=2\end{matrix}\right.$Vậy PT có cặp nghiệm $S=\left\{0;2\right\}$

Mai Trung Hải Phong · Answer

f) $x^3+27+\left(x-9\right)\left(x+3\right)=0$$\Rightarrow x^3+3^3+\left(x-9\right)\left(x+3\right)=0$$\Rightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x-9\right)\left(x+3\right)=0$$\Rightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0$$\Rightarrow\left(x+3\right)\left(x^2-2x\right)=0$$\Rightarrow x\left(x+3\right)\left(x-2\right)=0$$\Rightarrow\left[{}\begin{matrix}x=0\x+3=0\x-2=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x=0\x=-3\x=2\end{matrix}\right.$Vậy PT có cặp nghiệm $S=\left\{0;-3;2\right\}$h) $x\left(x^2+1\right)-2x\left(x-2\right)-2\left(x^2+1\right)=0$$\Rightarrow\left(x^2+1\right)\left(x-2\right)-2x\left(x-2\right)=0$$\Rightarrow\left(x-2\right)\left(x^2-2x+1\right)=0$$\Rightarrow\left(x-2\right)\left(x-1\right)^2=0$$\Rightarrow\left[{}\begin{matrix}x-2=0\\left(x-1\right)^2=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x=2\x-1=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x=2\x=1\end{matrix}\right.$Vậy PT có cặp nghiệm $S=\left\{2;1\right\}$Câu trả lời: f) $x^3+27+\left(x-9\right)\left(x+3\right)=0$$\Rightarrow x^3+3^3+\left(x-9\right)\left(x+3\right)=0$$\Rightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x-9\right)\left(x+3\right)=0$$\Rightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0$$\Rightarrow\left(x+3\right)\left(x^2-2x\right)=0$$\Rightarrow x\left(x+3\right)\left(x-2\right)=0$$\Rightarrow\left[{}\begin{matrix}x=0\x+3=0\x-2=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x=0\x=-3\x=2\end{matrix}\right.$Vậy PT có cặp nghiệm $S=\left\{0;-3;2\right\}$h) $x\left(x^2+1\right)-2x\left(x-2\right)-2\left(x^2+1\right)=0$$\Rightarrow\left(x^2+1\right)\left(x-2\right)-2x\left(x-2\right)=0$$\Rightarrow\left(x-2\right)\left(x^2-2x+1\right)=0$$\Rightarrow\left(x-2\right)\left(x-1\right)^2=0$$\Rightarrow\left[{}\begin{matrix}x-2=0\\left(x-1\right)^2=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x=2\x-1=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x=2\x=1\end{matrix}\right.$Vậy PT có cặp nghiệm $S=\left\{2;1\right\}$

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