(x-3)(x^2+3x+9)-(3x-17)=x^3-12
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\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(\Rightarrow x\left(x^2+3x+9\right)-3\left(x^2+3x+9\right)-3x+17=x^3-12\)
\(\Rightarrow x^3+3x^2+9x-3x^2-9x-27-3x+17=x^3-12\)
\(\Rightarrow x^3+\left(3x^2-3x^2\right)+\left(9x-9x\right)-3x-10=x^3+12\)
\(\Rightarrow x^3-3x-10=x^3+12\)
\(\Rightarrow x^3-3x-10-12=x^3\)
\(\Rightarrow x^3-3x-22=x^3\)
\(\Rightarrow3x-22=0\)
\(\Rightarrow3x=22\Rightarrow x=\dfrac{22}{3}\)
(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
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(\Leftrightarrow x^3-27-3x+17-x^3+12=0\)
\(\Leftrightarrow2-3x=0\)
\(\Leftrightarrow3x=2\Rightarrow x=\dfrac{2}{3}\)
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(\Leftrightarrow x^3-27-3x+17-x^3+12=0\)
\(\Leftrightarrow-3x+2=0\)
\(\Rightarrow x=\dfrac{2}{3}\)
Vậy .............
(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...
(x−3)(x2+3x+9)−(3x−17)=x3−12
⇔x3−27−3x+17−x3+12=0
⇔2−3x=0
⇔3x=2⇒x=23
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(x^3+3x^2+9x-3x^2-9x-37-3x+17=x^3-12\)
\(x^3+3x^2+9x-3x^2-9x-3x-x^3=37-17+12\)
\(-3x=32\)
\(x=\frac{32}{-3}=-\frac{32}{3}\)
Vậy x = \(-\frac{32}{3}\)
(x-3)(x\(^2\)+3x+9)-(3x-17)=x\(^3\)-12
(x\(^3\)-3\(^3\))-(3x-17)=x\(^3\)-12
3\(^3\)-3\(^3\)-3x+17=x\(^3\)-12
x\(^3\)-27-3x+17=x\(^3\)-12
x\(^3\)-27-3x+17-x\(^3\)=-12
-27-3x+17=-12
-27-3x=-29
3x=2
x=\(\frac{2}{3}\)
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
17 - 2x = -15
2x = 17 - (-15)
2x = 32
x = 32 : 2
x = 16
--------
54 : (x + 2) = -6
x + 2 = 54 : (-6)
x + 2 = -9
x = -9 - 2
x = -11
--------
12.(3 - x) = 72
3 - x = 72 : 12
3 - x = 6
x = 3 - 6
x = -3
-------
-3(x + 5) + 18 = -27
-3(x + 5) = -27 - 18
-3(x + 5) = -45
x + 5 = -45 : (-3)
x + 5 = 15
x = 15 - 5
x = 10
-------
(x + 5)² = 9
x + 5 = 3 hoặc x + 5 = -3
*) x + 5 = 3
x = 3 - 5
x = -2
*) x + 5 = -3
x = -3 - 5
x = -8
Vậy x = -8; x = -2
--------
(3 - x)³ = 27
(3 - x)³ = 3³
3 - x = 3
x = 3 - 3
x = 0
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(\Leftrightarrow x^3-27-3x+17=x^3-12\)
\(\Leftrightarrow-10-3x=-12\)
\(\Leftrightarrow3x=2\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
Vậy...
\((x-3)(x^2+3x+9)-(3x-17)=x^3-12 \)
\(pt\Leftrightarrow x^3-27-3x+17=x^3-12\)
\(\Leftrightarrow x^3-3x-10-x^3+12=0\)
\(\Leftrightarrow2-3x=0\)\(\Leftrightarrow x=\dfrac{2}{3}\)