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21 tháng 6 2017
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(\Leftrightarrow x^3-27-3x+17=x^3-12\)
\(\Leftrightarrow-10-3x=-12\)
\(\Leftrightarrow3x=2\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
Vậy...
21 tháng 6 2017
\((x-3)(x^2+3x+9)-(3x-17)=x^3-12 \)
\(pt\Leftrightarrow x^3-27-3x+17=x^3-12\)
\(\Leftrightarrow x^3-3x-10-x^3+12=0\)
\(\Leftrightarrow2-3x=0\)\(\Leftrightarrow x=\dfrac{2}{3}\)
ES
4
TN
7 tháng 9 2017
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(\Leftrightarrow x^3-27-3x+17-x^3+12=0\)
\(\Leftrightarrow2-3x=0\)
\(\Leftrightarrow3x=2\Rightarrow x=\dfrac{2}{3}\)
7 tháng 9 2017
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(\Leftrightarrow x^3-27-3x+17-x^3+12=0\)
\(\Leftrightarrow-3x+2=0\)
\(\Rightarrow x=\dfrac{2}{3}\)
Vậy .............
TB
2
M
29 tháng 7 2019
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(x^3+3x^2+9x-3x^2-9x-37-3x+17=x^3-12\)
\(x^3+3x^2+9x-3x^2-9x-3x-x^3=37-17+12\)
\(-3x=32\)
\(x=\frac{32}{-3}=-\frac{32}{3}\)
Vậy x = \(-\frac{32}{3}\)
29 tháng 7 2019
(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=\(\frac{2}{3}\)
28 tháng 4 2018
a) 5 - 4x = 3x - 9
\(\Leftrightarrow5-4x-3x+9=0\)
\(\Leftrightarrow14-7x=0\)
\(\Leftrightarrow7x=14\Leftrightarrow x=2\)
Vậy \(S=\left\{2\right\}\)
b) \(\left(x-4\right)\left(3x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(S=\left\{-3;4\right\}\)
c) \(\dfrac{x}{x+4}+\dfrac{12}{x-4}=\dfrac{4x+48}{x\cdot x-16}\)(1)
ĐKXĐ: \(x\ne\pm4\)
\(\left(1\right)\Leftrightarrow\dfrac{x\left(x-4\right)+12\left(x+4\right)-4x-48}{\left(x+4\right)\left(x-4\right)}=0\)
\(\Leftrightarrow x^2-4x+12x+48-4x-48=0\)
\(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-4\left(KTM\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
d) \(4-2x=7-x\)
\(\Leftrightarrow4-2x-7+x=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow-x=3\Leftrightarrow x=-3\)
Vậy \(S=\left\{-3\right\}\)
e) \(\left(x+4\right) \left(8-4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\8-4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{-4;2\right\}\)
f) \(\dfrac{x}{x+5}+\dfrac{11}{x-5}=\dfrac{x+55}{x\cdot x-25}\left(2\right)\)
ĐKXĐ: \(x\ne\pm5\)
\(\left(2\right)\Leftrightarrow\dfrac{x\left(x-5\right)+11\left(x+5\right)-x-55}{\left(x+5\right)\left(x-5\right)}=0\)
\(\Leftrightarrow x^2-5x+11x+55-x-55=0\)
\(\Leftrightarrow x^2+5x=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-5\left(KTM\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
g) \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)-3x-1-10-12x}{6}=0\)
\(\Leftrightarrow9x+6-3x-1-10-12x=0\)
\(\Leftrightarrow-6x-5=0\)
\(\Leftrightarrow-6x=5\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
Vậy \(S=\left\{-\dfrac{5}{6}\right\}\)
h) \(2x-\left(3-5x\right)=4\left(x+3\right)\)
\(\Leftrightarrow2x-3+5x-4x-12=0\)
\(\Leftrightarrow3x-15=0\)
\(\Leftrightarrow x=5\)
Vậy \(S=\left\{5\right\}\)
i) \(3x-6+x=9-x\)
\(\Leftrightarrow3x-6+x-9+x=0\)
\(\Leftrightarrow5x-15=0\)
\(\Leftrightarrow x=3\)
Vậy \(S=\left\{3\right\}\)
k)\(2t-3+5t=4t+12\)
\(\Leftrightarrow2t-3+5t-4t-12=0\)
\(\Leftrightarrow3t-15=0\)
\(\Leftrightarrow t=5\)
Vậy \(S=\left\{5\right\}\)
JS
15 tháng 2 2020
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
15 tháng 3 2020
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
15 tháng 3 2020
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
PT
18 tháng 9 2018
a) \(3\left(x^2-2x+1\right)+x\left(2-3x\right)=7\)
\(\Rightarrow3x^2-6x+3+2x-3x^2=7\)
\(\Rightarrow-4x+3=7\)
\(\Rightarrow-4x+3-7=0\)
\(\Rightarrow-4x-4=0\)
\(\Rightarrow-4\left(x+1\right)=0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
b) \(5\left(x-2\right)+2\left(x+3\right)=10\)
\(\Rightarrow5x-10+2x+6=10\)
\(\Rightarrow7x-4=10\)
\(\Rightarrow7x=10+4=14\)
\(\Rightarrow x=\dfrac{14}{7}=2\)
c) \(\left(x+1\right)\left(-3\right)+5\left(x-4\right)=-3\)
\(\Rightarrow-3x-3+5x-20=-3\)
\(\Rightarrow2x-23=-3\)
\(\Rightarrow2x=-3+23=20\)
\(\Rightarrow x=\dfrac{20}{2}=10\)
d) \(2\left(x-1\right)-x\left(3-x\right)=x^2\)
\(\Rightarrow2x-2-3x+x^2=x^2\)
\(\Rightarrow-x-2+x^2-x^2=0\)
\(\Rightarrow-x-2=0\)
\(\Rightarrow-x=2\)
\(\Rightarrow x=-2\)
đ) \(3x\left(x+5\right)-2\left(x+5\right)=3x^2\)
\(\Rightarrow3x^2+15x-2x-10=3x^2\)
\(\Rightarrow3x^2-3x^2+13x-10=0\)
\(\Rightarrow13x-10=0\)
\(\Rightarrow13x=10\)
\(\Rightarrow x=\dfrac{10}{13}\)
e) \(4x\left(x+2\right)+x\left(4-x\right)=3x^2+12\)
\(\Rightarrow4x^2+8x+4x-x^2=3x^2+12\)
\(\Rightarrow3x^2+12x=3x^2+12\)
\(\Rightarrow3x^2+12x-3x^2-12=0\)
\(\Rightarrow12\left(x-1\right)=0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
f) \(\dfrac{1}{3}x\left(3x+6\right)-x\left(x-5\right)=9\)
\(\Rightarrow x^2+2x-x^2+5x=9\)
\(\Rightarrow7x=9\)
\(\Rightarrow x=\dfrac{9}{7}\)
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(\Rightarrow x\left(x^2+3x+9\right)-3\left(x^2+3x+9\right)-3x+17=x^3-12\)
\(\Rightarrow x^3+3x^2+9x-3x^2-9x-27-3x+17=x^3-12\)
\(\Rightarrow x^3+\left(3x^2-3x^2\right)+\left(9x-9x\right)-3x-10=x^3+12\)
\(\Rightarrow x^3-3x-10=x^3+12\)
\(\Rightarrow x^3-3x-10-12=x^3\)
\(\Rightarrow x^3-3x-22=x^3\)
\(\Rightarrow3x-22=0\)
\(\Rightarrow3x=22\Rightarrow x=\dfrac{22}{3}\)