Câu hỏi của Ninh Thanh Tú Anh

Question

Giải phương trình$\left(x^2+x+1\right)\left(6-2x\right)=0$$\left(8x-4\right)\left(x^2+2x+2\right)=0$$x^3-7x+6=0$$x^5-5x^3+4x=0$

Edogawa Conan · Accepted Answer

(x2 + x + 1)(6 - 2x) = 0<=> 6 - 2x = 0 (do x2 + x + 1 > 0)<=> 2x = 6<=> x = 3Vậy S = {3}(8x - 4)(x2 + 2x + 2) = 0<=> 8x - 4 = 0 (vì x2 + 2x + 2 > 0)<=> 8x = 4<=> x = 1/2 Vậy S = {1/2}x3 - 7x + 6 = 0<=> x3 - x - 6x + 6 = 0<=> x(x2 - 1) - 6(x - 1) = 0<=> x(x - 1)(x + 1) - 6(x - 1) = 0<=> (x2 + x - 6)(x - 1) = 0<=> (x2 + 3x - 2x - 6)(x - 1) = 0<=> (x + 3)(x - 2)(x - 1) = 0<=> x + 3 = 0hoặc x - 2 = 0hoặc x - 1 = 0<=> x = -3hoặc x = 2hoặc x = 1Vậy S = {-3; 1; 2}x5 - 5x3 + 4x = 0<=> x(x4 - 5x2 + 4) = 0<=> x(x4 - x2 - 4x2 + 4) = 0<=> x[x2(x2 - 1) - 4(x2 - 1)] = 0<=> x(x - 2)(x + 2)(x - 1)(x + 1) = 0<=> x = 0 hoặc x - 2 = 0 hoặc x + 2 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0<=> x = 0 hoặc x = 2 hoặc x = -2 hoặc x = 1 hoặc x = -1Vậy S = {-2; -1; 0; 1; 2}Câu trả lời: (x2 + x + 1)(6 - 2x) = 0<=> 6 - 2x = 0 (do x2 + x + 1 > 0)<=> 2x = 6<=> x = 3Vậy S = {3}(8x - 4)(x2 + 2x + 2) = 0<=> 8x - 4 = 0 (vì x2 + 2x + 2 > 0)<=> 8x = 4<=> x = 1/2 Vậy S = {1/2}x3 - 7x + 6 = 0<=> x3 - x - 6x + 6 = 0<=> x(x2 - 1) - 6(x - 1) = 0<=> x(x - 1)(x + 1) - 6(x - 1) = 0<=> (x2 + x - 6)(x - 1) = 0<=> (x2 + 3x - 2x - 6)(x - 1) = 0<=> (x + 3)(x - 2)(x - 1) = 0<=> x + 3 = 0hoặc x - 2 = 0hoặc x - 1 = 0<=> x = -3hoặc x = 2hoặc x = 1Vậy S = {-3; 1; 2}x5 - 5x3 + 4x = 0<=> x(x4 - 5x2 + 4) = 0<=> x(x4 - x2 - 4x2 + 4) = 0<=> x[x2(x2 - 1) - 4(x2 - 1)] = 0<=> x(x - 2)(x + 2)(x - 1)(x + 1) = 0<=> x = 0 hoặc x - 2 = 0 hoặc x + 2 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0<=> x = 0 hoặc x = 2 hoặc x = -2 hoặc x = 1 hoặc x = -1Vậy S = {-2; -1; 0; 1; 2}

Tạ Đức Hoàng Anh · Answer

+ Ta có: $\left(x^2+x+1\right).\left(6-2x\right)=0$ - Ta lại có: $x^2+x+1=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x$- Vì $x^2+x+1>0\forall x$mà $\left(x^2+x+1\right).\left(6-2x\right)=0$  $\Rightarrow6-2x=0\Leftrightarrow-2x=-6\Leftrightarrow x=3\left(TM\right)$Vậy $S=\left\{3\right\}$+ Ta có: $\left(8x-4\right).\left(x^2+2x+2\right)=0$ - Ta lại có: $x^2+2x+2=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\ge1>0\forall x$ - Vì $x^2+2x+2>0\forall x$mà $\left(8x-4\right).\left(x^2+2x+2\right)=0$   $\Rightarrow8x-4=0\Leftrightarrow8x=4\Leftrightarrow x=\frac{1}{2}\left(TM\right)$Vậy $S=\left\{\frac{1}{2}\right\}$+ Ta có: $x^3-7x+6=0$       $\Leftrightarrow\left(x^3-x^2\right)+\left(x^2-x\right)+\left(6x-6\right)=0$       $\Leftrightarrow x^2.\left(x-1\right)+x.\left(x-1\right)-6.\left(x-1\right)=0$       $\Leftrightarrow\left(x-1\right).\left(x^2+x-6\right)=0$       $\Leftrightarrow\left(x-1\right).\left[\left(x^2-2x\right)+\left(3x-6\right)\right]=0$        $\Leftrightarrow\left(x-1\right).\left[x.\left(x-2\right)+3.\left(x-2\right)\right]=0$       $\Leftrightarrow\left(x-1\right).\left(x-2\right).\left(x+3\right)=0$       $\Leftrightarrow x=1\left(TM\right)$hoặc $x=2\left(TM\right)$hoặc $x=-3\left(TM\right)$ Vậy $S=\left\{-3;1;2\right\}$ + Ta có: $x^5-5x^3+4x=0$        $\Leftrightarrow x.\left(x^4-5x^2+4\right)=0$       $\Leftrightarrow x.\left[\left(x^4-x^2\right)-\left(4x^2-4\right)\right]=0$       $\Leftrightarrow x.\left[x^2.\left(x^2-1\right)-4.\left(x^2-1\right)\right]=0$       $\Leftrightarrow x.\left(x^2-1\right).\left(x^2-4\right)=0$       $\Leftrightarrow x=0\left(TM\right)$hoặc  $x^2-1=0\Leftrightarrow x^2=1\Leftrightarrow x=\pm1\left(TM\right)$hoặc $x^2-4=0\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\left(TM\right)$Vậy $S=\left\{-2;-1;0;1;2\right\}$!!@@# ^_^ Chúc bạn hok tốt ^_^#@@!!      Câu trả lời: + Ta có: $\left(x^2+x+1\right).\left(6-2x\right)=0$ - Ta lại có: $x^2+x+1=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x$- Vì $x^2+x+1>0\forall x$mà $\left(x^2+x+1\right).\left(6-2x\right)=0$  $\Rightarrow6-2x=0\Leftrightarrow-2x=-6\Leftrightarrow x=3\left(TM\right)$Vậy $S=\left\{3\right\}$+ Ta có: $\left(8x-4\right).\left(x^2+2x+2\right)=0$ - Ta lại có: $x^2+2x+2=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\ge1>0\forall x$ - Vì $x^2+2x+2>0\forall x$mà $\left(8x-4\right).\left(x^2+2x+2\right)=0$   $\Rightarrow8x-4=0\Leftrightarrow8x=4\Leftrightarrow x=\frac{1}{2}\left(TM\right)$Vậy $S=\left\{\frac{1}{2}\right\}$+ Ta có: $x^3-7x+6=0$       $\Leftrightarrow\left(x^3-x^2\right)+\left(x^2-x\right)+\left(6x-6\right)=0$       $\Leftrightarrow x^2.\left(x-1\right)+x.\left(x-1\right)-6.\left(x-1\right)=0$       $\Leftrightarrow\left(x-1\right).\left(x^2+x-6\right)=0$       $\Leftrightarrow\left(x-1\right).\left[\left(x^2-2x\right)+\left(3x-6\right)\right]=0$        $\Leftrightarrow\left(x-1\right).\left[x.\left(x-2\right)+3.\left(x-2\right)\right]=0$       $\Leftrightarrow\left(x-1\right).\left(x-2\right).\left(x+3\right)=0$       $\Leftrightarrow x=1\left(TM\right)$hoặc $x=2\left(TM\right)$hoặc $x=-3\left(TM\right)$ Vậy $S=\left\{-3;1;2\right\}$ + Ta có: $x^5-5x^3+4x=0$        $\Leftrightarrow x.\left(x^4-5x^2+4\right)=0$       $\Leftrightarrow x.\left[\left(x^4-x^2\right)-\left(4x^2-4\right)\right]=0$       $\Leftrightarrow x.\left[x^2.\left(x^2-1\right)-4.\left(x^2-1\right)\right]=0$       $\Leftrightarrow x.\left(x^2-1\right).\left(x^2-4\right)=0$       $\Leftrightarrow x=0\left(TM\right)$hoặc  $x^2-1=0\Leftrightarrow x^2=1\Leftrightarrow x=\pm1\left(TM\right)$hoặc $x^2-4=0\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\left(TM\right)$Vậy $S=\left\{-2;-1;0;1;2\right\}$!!@@# ^_^ Chúc bạn hok tốt ^_^#@@!!

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