Câu hỏi của Yumei

Question

1) (x2-4x+16) (x+4)-x(x+1) (x+2)+3x2=02) (8x+2) (1-3x)+(6x-1) (4x-10)=-503) (x2+2x+4) (2-x)+x(x-3) (x+4)-x2+24=04) ($\dfrac{x}{2}$x2+3) (5-6x)+(12x-2) ($\dfrac{x}{4}$x4+3)=0

Huỳnh Thị Thanh Ngân · Accepted Answer

1)(x2-4x+16)(x+4)-x(x+1)(x+2)+3x2=0$\Rightarrow$(x3+64)-x(x2+2x+x+2)+3x2=0$\Rightarrow$x3+64-x3-2x2-x2-2x+3x2=0$\Rightarrow$-2x+64=0$\Rightarrow$-2x=-64$\Rightarrow$x=$\dfrac{-64}{-2}$$\Rightarrow x=32$Câu trả lời: 1)(x2-4x+16)(x+4)-x(x+1)(x+2)+3x2=0$\Rightarrow$(x3+64)-x(x2+2x+x+2)+3x2=0$\Rightarrow$x3+64-x3-2x2-x2-2x+3x2=0$\Rightarrow$-2x+64=0$\Rightarrow$-2x=-64$\Rightarrow$x=$\dfrac{-64}{-2}$$\Rightarrow x=32$

Huỳnh Thị Thanh Ngân · Answer

2)(8x+2)(1-3x)+(6x-1)(4x-10)=-50$\Rightarrow$8x-24x2+2-6x+24x2-60x-4x+10=50$\Rightarrow$-62x+12=50$\Rightarrow$-62x=50-12$\Rightarrow$-62x=38$\Rightarrow$x=$-\dfrac{38}{62}=-\dfrac{19}{31}$Câu trả lời: 2)(8x+2)(1-3x)+(6x-1)(4x-10)=-50$\Rightarrow$8x-24x2+2-6x+24x2-60x-4x+10=50$\Rightarrow$-62x+12=50$\Rightarrow$-62x=50-12$\Rightarrow$-62x=38$\Rightarrow$x=$-\dfrac{38}{62}=-\dfrac{19}{31}$

Huỳnh Thị Thanh Ngân · Answer

3)(x2+2x+4)(2-x)+x(x-3)(x+4)-x2+24=0$\Rightarrow$8-x3+x(x2+4x-3x-12)-x2+24=0$\Rightarrow$8-x3+x3+4x2-3x2-12x-x2+21=0$\Rightarrow$-12x+29=0$\Rightarrow$-12x=-29$\Rightarrow$x=$\dfrac{-29}{-12}=\dfrac{29}{12}$Câu trả lời: 3)(x2+2x+4)(2-x)+x(x-3)(x+4)-x2+24=0$\Rightarrow$8-x3+x(x2+4x-3x-12)-x2+24=0$\Rightarrow$8-x3+x3+4x2-3x2-12x-x2+21=0$\Rightarrow$-12x+29=0$\Rightarrow$-12x=-29$\Rightarrow$x=$\dfrac{-29}{-12}=\dfrac{29}{12}$

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