Câu hỏi của Lê Hương Giang

Question

a) (x2 - 5x)2 + 10(x2 - 5x) + 24 = 0b) (2x + 1)2 - 2x - 1 = 2c) x(x - 1)(x2 - x + 1) - 6 = 0d) (x2 + 1)2 + 3x(x2 + 1) + 2x2 = 0

Nguyễn Lê Phước Thịnh · Accepted Answer

a) Ta có: $\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0$$\Leftrightarrow\left(x^2-5x\right)^2+4\left(x^2-5x\right)+6\left(x^2-5x\right)+24=0$$\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x+4\right)+6\left(x^2-5x+4\right)=0$$\Leftrightarrow\left(x^2-5x+6\right)\left(x^2-5x+4\right)=0$$\Leftrightarrow\left(x^2-2x-3x+6\right)\left(x^2-x-4x+4\right)=0$$\Leftrightarrow\left[x\left(x-2\right)-3\left(x-2\right)\right]\left[x\left(x-1\right)-4\left(x-1\right)\right]=0$$\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=0$$\Leftrightarrow\left[{}\begin{matrix}x-1=0\x-2=0\x-3=0\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\x=2\x=3\x=4\end{matrix}\right.$Vậy: S={1;2;3;4}b) Ta có: $\left(2x+1\right)^2-2x-1=2$$\Leftrightarrow\left(2x+1\right)^2-\left(2x+1\right)-2=0$$\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)+\left(2x+1\right)-2=0$$\Leftrightarrow\left(2x+1\right)\left(2x+1-2\right)+\left(2x+1-2\right)=0$$\Leftrightarrow\left(2x+1+1\right)\left(2x-1\right)=0$$\Leftrightarrow\left(2x+2\right)\left(2x-1\right)=0$$\Leftrightarrow\left[{}\begin{matrix}2x+2=0\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\x=\dfrac{1}{2}\end{matrix}\right.$Vậy: $S=\left\{-1;\dfrac{1}{2}\right\}$c) Ta có: $x\left(x-1\right)\left(x^2-x+1\right)-6=0$$\Leftrightarrow x\left(x^3-x^2+x-x^2+x-1\right)-6=0$$\Leftrightarrow x\left(x^3-2x^2+2x-1\right)-6=0$$\Leftrightarrow x^4-2x^3+2x^2-x-6=0$$\Leftrightarrow x^4-2x^3+2x^2-4x+3x-6=0$$\Leftrightarrow x^3\left(x-2\right)+2x\left(x-2\right)+3\left(x-2\right)=0$$\Leftrightarrow\left(x-2\right)\left(x^3+2x+3\right)=0$$\Leftrightarrow\left(x-2\right)\left(x^3-x+3x+3\right)=0$$\Leftrightarrow\left(x-2\right)\left[x\left(x^2-1\right)+3\left(x+1\right)\right]=0$$\Leftrightarrow\left(x-2\right)\left[x\left(x-1\right)\left(x+1\right)+3\left(x+1\right)\right]=0$$\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2-x+3\right)=0$mà $x^2-x+3>0\forall x$nên (x-2)(x+1)=0$\Leftrightarrow\left[{}\begin{matrix}x-2=0\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\x=-1\end{matrix}\right.$Vậy: S={2;-1}d) Ta có: $\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0$$\Leftrightarrow\left(x^2+1\right)^2+2x\left(x^2+1\right)+x\left(x^2+1\right)+2x^2=0$$\Leftrightarrow\left(x^2+1\right)\left(x^2+1+2x\right)+x\left(x^2+1+2x\right)=0$$\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2+x+1\right)=0$mà $x^2+x+1>0\forall x$nên x+1=0hay x=-1Vậy: S={-1}Câu trả lời: a) Ta có: $\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0$$\Leftrightarrow\left(x^2-5x\right)^2+4\left(x^2-5x\right)+6\left(x^2-5x\right)+24=0$$\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x+4\right)+6\left(x^2-5x+4\right)=0$$\Leftrightarrow\left(x^2-5x+6\right)\left(x^2-5x+4\right)=0$$\Leftrightarrow\left(x^2-2x-3x+6\right)\left(x^2-x-4x+4\right)=0$$\Leftrightarrow\left[x\left(x-2\right)-3\left(x-2\right)\right]\left[x\left(x-1\right)-4\left(x-1\right)\right]=0$$\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=0$$\Leftrightarrow\left[{}\begin{matrix}x-1=0\x-2=0\x-3=0\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\x=2\x=3\x=4\end{matrix}\right.$Vậy: S={1;2;3;4}b) Ta có: $\left(2x+1\right)^2-2x-1=2$$\Leftrightarrow\left(2x+1\right)^2-\left(2x+1\right)-2=0$$\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)+\left(2x+1\right)-2=0$$\Leftrightarrow\left(2x+1\right)\left(2x+1-2\right)+\left(2x+1-2\right)=0$$\Leftrightarrow\left(2x+1+1\right)\left(2x-1\right)=0$$\Leftrightarrow\left(2x+2\right)\left(2x-1\right)=0$$\Leftrightarrow\left[{}\begin{matrix}2x+2=0\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\x=\dfrac{1}{2}\end{matrix}\right.$Vậy: $S=\left\{-1;\dfrac{1}{2}\right\}$c) Ta có: $x\left(x-1\right)\left(x^2-x+1\right)-6=0$$\Leftrightarrow x\left(x^3-x^2+x-x^2+x-1\right)-6=0$$\Leftrightarrow x\left(x^3-2x^2+2x-1\right)-6=0$$\Leftrightarrow x^4-2x^3+2x^2-x-6=0$$\Leftrightarrow x^4-2x^3+2x^2-4x+3x-6=0$$\Leftrightarrow x^3\left(x-2\right)+2x\left(x-2\right)+3\left(x-2\right)=0$$\Leftrightarrow\left(x-2\right)\left(x^3+2x+3\right)=0$$\Leftrightarrow\left(x-2\right)\left(x^3-x+3x+3\right)=0$$\Leftrightarrow\left(x-2\right)\left[x\left(x^2-1\right)+3\left(x+1\right)\right]=0$$\Leftrightarrow\left(x-2\right)\left[x\left(x-1\right)\left(x+1\right)+3\left(x+1\right)\right]=0$$\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2-x+3\right)=0$mà $x^2-x+3>0\forall x$nên (x-2)(x+1)=0$\Leftrightarrow\left[{}\begin{matrix}x-2=0\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\x=-1\end{matrix}\right.$Vậy: S={2;-1}d) Ta có: $\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0$$\Leftrightarrow\left(x^2+1\right)^2+2x\left(x^2+1\right)+x\left(x^2+1\right)+2x^2=0$$\Leftrightarrow\left(x^2+1\right)\left(x^2+1+2x\right)+x\left(x^2+1+2x\right)=0$$\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2+x+1\right)=0$mà $x^2+x+1>0\forall x$nên x+1=0hay x=-1Vậy: S={-1}

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