giải phương trình bậc hai: 1)x^4-13x^2+36=0 2)9x^4+6x^2+1=0 3)2x^4+5x^2+2=0 4)2x^4-7x^2-4=0 5)x^4-5x^2+4=0 04/11/2021 Bởi Iris giải phương trình bậc hai: 1)x^4-13x^2+36=0 2)9x^4+6x^2+1=0 3)2x^4+5x^2+2=0 4)2x^4-7x^2-4=0 5)x^4-5x^2+4=0
$1)x^4-13x^2+36=0\\ \Leftrightarrow x^4-4x^2-9x^2+36=0\\ \Leftrightarrow x^2(x^2-4)-9(x^2-4)=0\\ \Leftrightarrow (x^2-9)(x^2-4)=0\\ \Leftrightarrow (x-3)(x+3)(x-2)(x+2)=0\\ \Leftrightarrow \left[\begin{array}{l} x=3\\x=-3 \\ x=2\\x=-2\end{array} \right.\\ 2)9x^4+6x^2+1=0\\ \Leftrightarrow (3x^2+1)^2=0\\ \Leftrightarrow x\in\varnothing( \text{Do}\, (3x^2+1)^2 >0 \, \forall \, x)\\ 3)2x^4+5x^2+2=0\\ \Leftrightarrow2x^4+4x^2+x^2+2=0\\ \Leftrightarrow2x^2(x^2+2)+x^2+2=0\\ \Leftrightarrow(2x^2+1)(x^2+2)=0\\ \Leftrightarrow x\in\varnothing(\text{Do}\, 2x^2+1;x^2+2 >0 \, \forall \, x)\\ 4)2x^4-7x^2-4=0\\ \Leftrightarrow 2x^4-8x^2+x^2-4=0\\ \Leftrightarrow 2x^2(x^2-4)+x^2-4=0\\ \Leftrightarrow (2x^2+1)(x^2-4)=0\\ \Leftrightarrow (2x^2+1)(x-2)(x+2)=0\\ \Leftrightarrow \left[\begin{array}{l} x=2\\x=-2\end{array} \right.(\text{Do} \,2x^2+1 >0 \, \forall \, x)\\ 5)x^4-5x^2+4=0\\ x^4-x^2-4x^2+4=0\\ \Leftrightarrow x^2(x^2-1)-4(x^2-1)=0\\ \Leftrightarrow (x^2-4)(x^2-1)=0\\ \Leftrightarrow (x-2)(x+2)(x-1)(x+1)=0\\ \Leftrightarrow \left[\begin{array}{l} x=1\\x=-1 \\ x=2\\x=-2\end{array} \right.$ Bình luận
Cách giải: `1,x^2-13x^2+36=0` `->x^2-9x^2-4x^2+36=0` `->x^2(x^2-9)-4(x^2-9)=0` `->(x^2-9)(x^2-4)=0` `->` \(\left[ \begin{array}{l}x^2=4\\x^2=9\end{array} \right.\) `->` \(\left[ \begin{array}{l}x=2\\x=-2\\x=3\\x=-3\end{array} \right.\) `2,9x^4+6x^2+1=0` `->(3x^2+1)^2=0` `->` pt vô nghiệm `3,2x^3+5x^2+2=0` `->2x^4+4x^2+x^2+2=0` `->2x^2(x^2+2)+x^2+2=0` `->(x^2+2)(2x^2+1)=0` `->` pt vô nghiệm `4,2x^4-7x^2-4=0` `->2x^2-8x^2+x^2-4=0` `->2x^2(x^2-4)+x^2-4=0` `->(x^2-4)(2x^2+1)=0` `->x^2-4=0` `->x^2=4` `->` \(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\) `5,x^4-5x^2+4=0` `->x^4-x^2-4x^2+4=0` `->x^2(x^2-1)-4(x^2-1)=0` `->(x^2-1)(x^2-4)=0` `->` \(\left[ \begin{array}{l}x^2=4\\x^2=1\end{array} \right.\) `->` \(\left[ \begin{array}{l}x=2\\x=-2\\x=1\\x=-1\end{array} \right.\) Bình luận
$1)x^4-13x^2+36=0\\ \Leftrightarrow x^4-4x^2-9x^2+36=0\\ \Leftrightarrow x^2(x^2-4)-9(x^2-4)=0\\ \Leftrightarrow (x^2-9)(x^2-4)=0\\ \Leftrightarrow (x-3)(x+3)(x-2)(x+2)=0\\ \Leftrightarrow \left[\begin{array}{l} x=3\\x=-3 \\ x=2\\x=-2\end{array} \right.\\ 2)9x^4+6x^2+1=0\\ \Leftrightarrow (3x^2+1)^2=0\\ \Leftrightarrow x\in\varnothing( \text{Do}\, (3x^2+1)^2 >0 \, \forall \, x)\\ 3)2x^4+5x^2+2=0\\ \Leftrightarrow2x^4+4x^2+x^2+2=0\\ \Leftrightarrow2x^2(x^2+2)+x^2+2=0\\ \Leftrightarrow(2x^2+1)(x^2+2)=0\\ \Leftrightarrow x\in\varnothing(\text{Do}\, 2x^2+1;x^2+2 >0 \, \forall \, x)\\ 4)2x^4-7x^2-4=0\\ \Leftrightarrow 2x^4-8x^2+x^2-4=0\\ \Leftrightarrow 2x^2(x^2-4)+x^2-4=0\\ \Leftrightarrow (2x^2+1)(x^2-4)=0\\ \Leftrightarrow (2x^2+1)(x-2)(x+2)=0\\ \Leftrightarrow \left[\begin{array}{l} x=2\\x=-2\end{array} \right.(\text{Do} \,2x^2+1 >0 \, \forall \, x)\\ 5)x^4-5x^2+4=0\\ x^4-x^2-4x^2+4=0\\ \Leftrightarrow x^2(x^2-1)-4(x^2-1)=0\\ \Leftrightarrow (x^2-4)(x^2-1)=0\\ \Leftrightarrow (x-2)(x+2)(x-1)(x+1)=0\\ \Leftrightarrow \left[\begin{array}{l} x=1\\x=-1 \\ x=2\\x=-2\end{array} \right.$
Cách giải:
`1,x^2-13x^2+36=0`
`->x^2-9x^2-4x^2+36=0`
`->x^2(x^2-9)-4(x^2-9)=0`
`->(x^2-9)(x^2-4)=0`
`->` \(\left[ \begin{array}{l}x^2=4\\x^2=9\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=2\\x=-2\\x=3\\x=-3\end{array} \right.\)
`2,9x^4+6x^2+1=0`
`->(3x^2+1)^2=0`
`->` pt vô nghiệm
`3,2x^3+5x^2+2=0`
`->2x^4+4x^2+x^2+2=0`
`->2x^2(x^2+2)+x^2+2=0`
`->(x^2+2)(2x^2+1)=0`
`->` pt vô nghiệm
`4,2x^4-7x^2-4=0`
`->2x^2-8x^2+x^2-4=0`
`->2x^2(x^2-4)+x^2-4=0`
`->(x^2-4)(2x^2+1)=0`
`->x^2-4=0`
`->x^2=4`
`->` \(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\)
`5,x^4-5x^2+4=0`
`->x^4-x^2-4x^2+4=0`
`->x^2(x^2-1)-4(x^2-1)=0`
`->(x^2-1)(x^2-4)=0`
`->` \(\left[ \begin{array}{l}x^2=4\\x^2=1\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=2\\x=-2\\x=1\\x=-1\end{array} \right.\)