Toán Tìm x: $(x+1)$$(3x^{2}$$-x+1)$$+$ $x^{2}$$(4-3x)$$=$ $\frac{5}{2}$ 10/07/2021 By Mary Tìm x: $(x+1)$$(3x^{2}$$-x+1)$$+$ $x^{2}$$(4-3x)$$=$ $\frac{5}{2}$
Đáp án: Giải thích các bước giải: `(x+1)(3x^2-x+1)+x^2(4-3x)=5/2` `⇔3x^3-x^2+x+3x^2-x+1+4x^2-3x^3=5/2` `⇔6x^2=5/2-1` `⇔6x^2=3/2` `⇔x^2=1/4` `⇔x^2-1/4=0` `⇔x^2-(1/2)^2=0` `⇔(x-1/2)(x+1/2)=0` `⇔`\(\left[ \begin{array}{l}x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}x=\frac{1}{2}\\x=-\frac{1}{2}\end{array} \right.\) Vậy `x∈{1/2;-1/2}` ` Trả lời
Đáp án:
Giải thích các bước giải:
`(x+1)(3x^2-x+1)+x^2(4-3x)=5/2`
`⇔3x^3-x^2+x+3x^2-x+1+4x^2-3x^3=5/2`
`⇔6x^2=5/2-1`
`⇔6x^2=3/2`
`⇔x^2=1/4`
`⇔x^2-1/4=0`
`⇔x^2-(1/2)^2=0`
`⇔(x-1/2)(x+1/2)=0`
`⇔`\(\left[ \begin{array}{l}x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\frac{1}{2}\\x=-\frac{1}{2}\end{array} \right.\)
Vậy `x∈{1/2;-1/2}`
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