Giải các phương trình sau: a, x-30/10 +x-35/15 +x-47/27=-3 b, x^3-4x^2+x-4=0 25/11/2021 Bởi Madelyn Giải các phương trình sau: a, x-30/10 +x-35/15 +x-47/27=-3 b, x^3-4x^2+x-4=0
`~rai~` $\begin{array}{I}a)\dfrac{x-30}{10}+\dfrac{x-35}{15}+\dfrac{x-47}{27}=-3\\\Leftrightarrow \dfrac{27(x-30)+18(x-35)+10(x-47)}{270}=-3\\\Leftrightarrow 27x-810+18x-630+10x-470=-810\\\Leftrightarrow 55x=1100\\\Leftrightarrow x=20.\\\text{Vậy x=20.}\\b)x^3-4x+x-4=0\\\Leftrightarrow (x^3+x)-(4x+4)=0\\\Leftrightarrow x(x^2+1)-4(x^2+1)=0\\\Leftrightarrow (x-4)(x^2+1)=0\\\Leftrightarrow x-4=0\quad(\text{vì}\quad x^2+1 >0\forall x\in\mathbb{R})\\\Leftrightarrow x=4.\\\text{Vậy x=4.}\end{array}$ Bình luận
Đáp án: b) x=4 Giải thích các bước giải: \(\begin{array}{l}a)\dfrac{{x – 30}}{{10}} + \dfrac{{x – 35}}{{15}} + \dfrac{{x – 47}}{{27}} = – 3\\ \to \dfrac{{27\left( {x – 30} \right) + 18\left( {x – 35} \right) + 10\left( {x – 47} \right)}}{{270}} = 0\\ \to 27x – 810 + 18x – 630 + 10x – 470 = 0\\ \to 55x = 1910\\ \to x = \dfrac{{382}}{{11}}\\b){x^3} – 4{x^2} + x – 4 = 0\\ \to {x^2}\left( {x – 4} \right) + \left( {x – 4} \right) = 0\\ \to \left( {x – 4} \right)\left( {{x^2} + 1} \right) = 0\\ \to x – 4 = 0\left( {do:{x^2} + 1 > 0\forall x} \right)\\ \to x = 4\end{array}\) Bình luận
`~rai~`
$\begin{array}{I}a)\dfrac{x-30}{10}+\dfrac{x-35}{15}+\dfrac{x-47}{27}=-3\\\Leftrightarrow \dfrac{27(x-30)+18(x-35)+10(x-47)}{270}=-3\\\Leftrightarrow 27x-810+18x-630+10x-470=-810\\\Leftrightarrow 55x=1100\\\Leftrightarrow x=20.\\\text{Vậy x=20.}\\b)x^3-4x+x-4=0\\\Leftrightarrow (x^3+x)-(4x+4)=0\\\Leftrightarrow x(x^2+1)-4(x^2+1)=0\\\Leftrightarrow (x-4)(x^2+1)=0\\\Leftrightarrow x-4=0\quad(\text{vì}\quad x^2+1 >0\forall x\in\mathbb{R})\\\Leftrightarrow x=4.\\\text{Vậy x=4.}\end{array}$
Đáp án:
b) x=4
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{x – 30}}{{10}} + \dfrac{{x – 35}}{{15}} + \dfrac{{x – 47}}{{27}} = – 3\\
\to \dfrac{{27\left( {x – 30} \right) + 18\left( {x – 35} \right) + 10\left( {x – 47} \right)}}{{270}} = 0\\
\to 27x – 810 + 18x – 630 + 10x – 470 = 0\\
\to 55x = 1910\\
\to x = \dfrac{{382}}{{11}}\\
b){x^3} – 4{x^2} + x – 4 = 0\\
\to {x^2}\left( {x – 4} \right) + \left( {x – 4} \right) = 0\\
\to \left( {x – 4} \right)\left( {{x^2} + 1} \right) = 0\\
\to x – 4 = 0\left( {do:{x^2} + 1 > 0\forall x} \right)\\
\to x = 4
\end{array}\)