Câu hỏi của Bùi Phươngg Thảo

Question

Giải BPT
x6 - 14x4 + 49x2 > 36

Trần Thị Hồng Ngát · Accepted Answer

$x^6-14x^4+49x^2>36$ $\Leftrightarrow x^6-x^5+x^5-x^4-13x^4+13x^3-13x^3+13x^2+36x^2-36x+36x-36>0$ $\Leftrightarrow x^5\left(x-1\right)+x^4\left(x-1\right)-13x^3\left(x-1\right)-13x^2\left(x-1\right)+36x\left(x-1\right)+36\left(x-1\right)>0$ $\Leftrightarrow\left(x-1\right)\left(x^5+x^4-13x^3-13x^2+36x+36\right)>0$ $\Leftrightarrow\left(x-1\right)\left[x^4\left(x+1\right)-13x^2\left(x+1\right)+36\left(x+1\right)\right]>0$ $\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^4-13x^2+36\right)>0$ $\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^4-9x^2-4x^2+36\right)>0$ $\Leftrightarrow\left(x-1\right)\left(x+1\right)\left[x^2\left(x^2-9\right)-4\left(x^2-9\right)\right]>0$ $\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2-9\right)\left(x^2-4\right) >0$ $\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+3\right)\left(x-3\right)>0$ Để $\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+3\right)\left(x-3\right)>0$ $\Rightarrow\left[{}\begin{matrix}x>3\x< -3\end{matrix}\right.$ Vậy để $\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+3\right)\left(x-3\right)>0$ thì x>3 hoặc x<-3Câu trả lời: $x^6-14x^4+49x^2>36$ $\Leftrightarrow x^6-x^5+x^5-x^4-13x^4+13x^3-13x^3+13x^2+36x^2-36x+36x-36>0$ $\Leftrightarrow x^5\left(x-1\right)+x^4\left(x-1\right)-13x^3\left(x-1\right)-13x^2\left(x-1\right)+36x\left(x-1\right)+36\left(x-1\right)>0$ $\Leftrightarrow\left(x-1\right)\left(x^5+x^4-13x^3-13x^2+36x+36\right)>0$ $\Leftrightarrow\left(x-1\right)\left[x^4\left(x+1\right)-13x^2\left(x+1\right)+36\left(x+1\right)\right]>0$ $\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^4-13x^2+36\right)>0$ $\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^4-9x^2-4x^2+36\right)>0$ $\Leftrightarrow\left(x-1\right)\left(x+1\right)\left[x^2\left(x^2-9\right)-4\left(x^2-9\right)\right]>0$ $\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2-9\right)\left(x^2-4\right) >0$ $\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+3\right)\left(x-3\right)>0$ Để $\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+3\right)\left(x-3\right)>0$ $\Rightarrow\left[{}\begin{matrix}x>3\x< -3\end{matrix}\right.$ Vậy để $\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+3\right)\left(x-3\right)>0$ thì x>3 hoặc x<-3

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP