17-(3x+9)=2-(-3+x)
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\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(\Leftrightarrow x^3-27-3x+17=x^3-12\)
\(\Leftrightarrow-10-3x=-12\)
\(\Leftrightarrow3x=2\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
Vậy...
\((x-3)(x^2+3x+9)-(3x-17)=x^3-12 \)
\(pt\Leftrightarrow x^3-27-3x+17=x^3-12\)
\(\Leftrightarrow x^3-3x-10-x^3+12=0\)
\(\Leftrightarrow2-3x=0\)\(\Leftrightarrow x=\dfrac{2}{3}\)
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(\Rightarrow x\left(x^2+3x+9\right)-3\left(x^2+3x+9\right)-3x+17=x^3-12\)
\(\Rightarrow x^3+3x^2+9x-3x^2-9x-27-3x+17=x^3-12\)
\(\Rightarrow x^3+\left(3x^2-3x^2\right)+\left(9x-9x\right)-3x-10=x^3+12\)
\(\Rightarrow x^3-3x-10=x^3+12\)
\(\Rightarrow x^3-3x-10-12=x^3\)
\(\Rightarrow x^3-3x-22=x^3\)
\(\Rightarrow3x-22=0\)
\(\Rightarrow3x=22\Rightarrow x=\dfrac{22}{3}\)
(x−3)(x2+3x+9)−(3x−17)=x3−12
⇔x3−27−3x+17=x3−12
⇔−10−3x=−12
⇔3x=2
⇔x=23
Vậy........
Nếu đúng thì k mk nha >.<
(x−3)(x2+3x+9)−(3x−17)=x^3−12
⇔x^3−27−3x+17−x^3+12=0
⇔2−3x=0
⇔3x=2⇒x=2/3
(x−3)(x2+3x+9)−(3x−17)=x3−12
⇒x(x2+3x+9)−3(x2+3x+9)−3x+17=x3−12
⇒x3+3x2+9x−3x2−9x−27−3x+17=x3−12
⇒x3+(3x2−3x2)+(9x−9x)−3x−10=x3+12
⇒x3−3x−10=x3+12
⇒x3−3x−10−12=x3
⇒x3−3x−22=x3
⇒3x−22=0
⇒3x=22⇒x=223
(x−3)(x^2+3x+9)−(3x−17)=x^3−12
⇔x^3−27−3x+17=x^3−12
⇔−10−3x=−12
⇔3x=2
⇔x=2/3
Vậy...
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(\Leftrightarrow x^3-27-3x+17-x^3+12=0\)
\(\Leftrightarrow2-3x=0\)
\(\Leftrightarrow3x=2\Rightarrow x=\dfrac{2}{3}\)
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(\Leftrightarrow x^3-27-3x+17-x^3+12=0\)
\(\Leftrightarrow-3x+2=0\)
\(\Rightarrow x=\dfrac{2}{3}\)
Vậy .............
\(\left(2-x\right)^3+\left(3+x\right)\left(9-3x+x^2\right)+6x\left(1-x\right)=17\\ \Leftrightarrow8-3.2^2.x+3.2.x^2-x^3+3^3+x^3+6x-6x^2-17=0\\ \Leftrightarrow x^3-x^3+6x^2-6x^2-12x+6x=17-27-8\\ \Leftrightarrow-6x=-18\\ \Leftrightarrow x=\dfrac{-18}{-6}=3\\ Vậy:x=3\)
(x−3)(x2+3x+9)−(3x−17)=x3−12
⇔x3−27−3x+17−x3+12=0
⇔2−3x=0
⇔3x=2⇒x=23
a) 3x – 15 = 25 – 5x
=> 3x + 5x = 25 + 15
=> 8x = 40
=> x = 5
b) 3x - 17 = 2x – 7
=> 3x - 2x = -7 + 17
=> x = 10
c) 2x – 17 = – (3x – 18)
=> 2x - 17 = -3x + 18
=> 2x + 3x = 18 + 17
=> 5x = 35
=> x = 7
d) 3x – 14 = 2(x – 9) + 1
=> 3x - 14 = 2x - 18 + 1
=> 3x - 2x = -18 + 1 + 14
=> x = -3
f) (x – 5)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
a) Ta có: \(3x-15=25-5x\)
\(\Leftrightarrow3x-15-25+5x=0\)
\(\Leftrightarrow8x-40=0\)
\(\Leftrightarrow8x=40\)
hay x=5
Vậy: x=5
b) Ta có: \(3x-17=2x-7\)
\(\Leftrightarrow3x-17-2x+7=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
Vậy: x=10
c) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=-3x+18\)
\(\Leftrightarrow2x-17+3x-18=0\)
\(\Leftrightarrow5x-35=0\)
\(\Leftrightarrow5x=35\)
hay x=7
Vậy: x=7
d) Ta có: \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14-2x+18-1=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: x=-3
f) Ta có: \(\left(x-5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)