Tìm x biết x^4 +4=5x ² HELPPPPPPP!!!!!!!!!!!1 13/07/2021 Bởi Maria Tìm x biết x^4 +4=5x ² HELPPPPPPP!!!!!!!!!!!1
`x^4 + 4 = 5x^2` `⇔ x^4 – 5x^2 + 4 = 0` Đặt `x^2 = t(t \ge 0)` `⇔ t^2 – 5t + 4 = 0` `⇔ (t-1)(t-4) = 0` `⇔`\(\left[ \begin{array}{l}t-1=0\\t-4=0\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}t=1\\t=4\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}x^2=4\\x^2=1\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}x=\pm2\\x=\pm1\end{array} \right.\) Vậy `S = {\pm2,\pm1}` Bình luận
Đáp án: `x∈\{1;-1;2;-1\}` Giải thích các bước giải: `x^4+4=5x^2` `⇔x^4+4-5x^2=0` `⇔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` `⇔(x-1)(x+1)(x-2)(x+2)=0` \(⇔\left[ \begin{array}{l}x-1=0\\x+1=0\\x-2=0\\x+2=0\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=1\\x=-1\\x=2\\x=-2\end{array} \right.\) Vậy `x∈\{1;-1;2;-1\}` Bình luận
`x^4 + 4 = 5x^2`
`⇔ x^4 – 5x^2 + 4 = 0`
Đặt `x^2 = t(t \ge 0)`
`⇔ t^2 – 5t + 4 = 0`
`⇔ (t-1)(t-4) = 0`
`⇔`\(\left[ \begin{array}{l}t-1=0\\t-4=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}t=1\\t=4\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x^2=4\\x^2=1\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\pm2\\x=\pm1\end{array} \right.\)
Vậy `S = {\pm2,\pm1}`
Đáp án:
`x∈\{1;-1;2;-1\}`
Giải thích các bước giải:
`x^4+4=5x^2`
`⇔x^4+4-5x^2=0`
`⇔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`
`⇔(x-1)(x+1)(x-2)(x+2)=0`
\(⇔\left[ \begin{array}{l}x-1=0\\x+1=0\\x-2=0\\x+2=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=1\\x=-1\\x=2\\x=-2\end{array} \right.\)
Vậy `x∈\{1;-1;2;-1\}`