giải pt 2x^4-x^3-7x^2+3x+3=0 (1+1/x)^3*(1+x^3)=16 30/08/2021 Bởi Anna giải pt 2x^4-x^3-7x^2+3x+3=0 (1+1/x)^3*(1+x^3)=16
Đáp án: Giải thích các bước giải: a) $2x^4-x^3-7x^2+3x+3=0\\\Leftrightarrow (2x^4-2x^3-6x^2+6x)+(x^3-x^2-3x+3)=0\\\Leftrightarrow 2x(x^3-x^2-3x+3)+(x^3-x^2-3x+3)=0\\\Leftrightarrow (2x+1)(x^3-x^2-3x+3)=0\\\Leftrightarrow (2x+1)((x^3-3x)-(x^2-3))=0\\\Leftrightarrow (2x+1)(x(x^2-3)-(x^2-3))=0\\\Leftrightarrow (2x+1)(x-1)(x^2-3)=0\\\Leftrightarrow \left[\begin{array}{l}x=-\cfrac{1}{2}\\x=1\\x=\pm \sqrt{3}\end{array}\right.$ b) $(1+\cfrac{1}{x})^3(1+x^3)=16\\\Leftrightarrow (x+1)^3(1+x^3)=16x^3\\\Leftrightarrow (x+1)^4(x^2-x+1)=16x^3\\\Leftrightarrow x^6+3x^5+3x^4+2x^3+3 x^2+3 x+1=16x^3\\\Leftrightarrow x^6+3x^5+3x^4-14x^3+3x^3+3=0\\\Leftrightarrow (x-1)^2(x+1)^2(x^2+x+1)=0\\\Rightarrow x=1$ Bình luận
Đáp án:
Giải thích các bước giải:
a) $2x^4-x^3-7x^2+3x+3=0\\\Leftrightarrow (2x^4-2x^3-6x^2+6x)+(x^3-x^2-3x+3)=0\\\Leftrightarrow 2x(x^3-x^2-3x+3)+(x^3-x^2-3x+3)=0\\\Leftrightarrow (2x+1)(x^3-x^2-3x+3)=0\\\Leftrightarrow (2x+1)((x^3-3x)-(x^2-3))=0\\\Leftrightarrow (2x+1)(x(x^2-3)-(x^2-3))=0\\\Leftrightarrow (2x+1)(x-1)(x^2-3)=0\\\Leftrightarrow \left[\begin{array}{l}x=-\cfrac{1}{2}\\x=1\\x=\pm \sqrt{3}\end{array}\right.$
b) $(1+\cfrac{1}{x})^3(1+x^3)=16\\\Leftrightarrow (x+1)^3(1+x^3)=16x^3\\\Leftrightarrow (x+1)^4(x^2-x+1)=16x^3\\\Leftrightarrow x^6+3x^5+3x^4+2x^3+3 x^2+3 x+1=16x^3\\\Leftrightarrow x^6+3x^5+3x^4-14x^3+3x^3+3=0\\\Leftrightarrow (x-1)^2(x+1)^2(x^2+x+1)=0\\\Rightarrow x=1$