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+) (5x-1). (2x+3)-3. (3x-1)=0
10x^2+15x-2x-3 - 9x+3=0
10x^2 +8x=0
2x(5x+4)=0
=> x=0 hoặc x= -4/5
+) x^3 (2x-3)-x^2 (4x^2-6x+2)=0
2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0
-2x^4 + 3x^3-2x^2=0
x^2(-2x^2+x-2)=0
-2x^2(x-1)^2=0
=> x=0 hoặc x=1
+) x (x-1)-x^2+2x=5
x^2 -x -x^2+2x=5
x=5
+) 8 (x-2)-2 (3x-4)=25
8x - 16-6x+8=25
2x=33
x=33/2
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
a)5x ( 1 - 2x ) - 3x ( x + 18 ) = 0
5x - 10x^2 - 3x^2 - 54x =0
-13x^2 - 49x =0
(-13x - 49)x =0
\(\Rightarrow\orbr{\begin{cases}-13x-49=0\\x=0\end{cases}\Rightarrow\orbr{\begin{cases}-13x=49\\x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{49}{13}\\x=0\end{cases}}}\)
Vậy x= -49/13 hoặc x=0
b)5x - 10x^2 - 3x^2 - 54 = 0
(câu b giống câu a)
5x(1 - 2x) - 3x(x + 18) = 0
=> 5x - 10x2 - 3x2 - 54x = 0
=> -13x2 - 49x = 0
=> x.(-13x - 49) = 0
=> x = 0
hoặc -13x - 49 = 0 => -13x = 49 => x = -49/13
Vậy x = 0, x = -49/13
\(5x\left(1-2x\right)-3x\left(x+18\right)=0\)
\(\Rightarrow x\left(5\left(1-2x\right)-3\left(x+18\right)\right)=0\)
\(\Rightarrow x\left(5-10x-3x+54\right)=0\)
\(\Rightarrow x\left(59-13x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\59-13x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\13x=59\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=\frac{59}{13}\end{cases}}}\)
5x(1-2x)-3x(x+18)=0
=> \(5x-10x^2-3x^2-54x=0\)
=> \(-49x-13x^2=0\)
=> \(x\left(-49-13x\right)=0\)
Suy ra
- \(x=0\)
- \(-49-13x=0\)=> \(-13x=49\)=> \(x=\frac{-49}{13}\)
Vậy \(x=\frac{-49}{13}; x=0\)
\(5x\left(1-2x\right)-3x\left(x+18\right)=0\)
\(\Leftrightarrow5x-10x^2-3x^2-54x\)
\(\Leftrightarrow-13x^2-49x=0\)
\(\Leftrightarrow x\left(-13x-49\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-13x=49\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{49}{13}\end{matrix}\right.\)
Vậy ..
\(5x\left(1-2x\right)-3x\left(x+18\right)=0\)
=>\(5x-10x^2-3x^2-54x=0\)
=> \(-49x-13x^2=0\)
=> \(x\left(-49-13x\right)=0\)
=> \(\left[{}\begin{matrix}x=0\\-49-13x=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0\\-49=13x\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0\\x=\dfrac{-49}{13}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-49}{13};0\right\}\)
\(5x\left(1-2x\right)-3x\left(x+18\right)=0\)
\(\Leftrightarrow5x-10x^2-3x^2-54x=0\)
\(\Leftrightarrow-13x^2-49x=0\)
\(\Leftrightarrow-x\left(13x-49\right)=0\)
\(\Rightarrow\hept{\begin{cases}-x=0\Leftrightarrow x=0\\13x-49=0\Leftrightarrow x=\frac{49}{13}\end{cases}}\)
Vậy \(x=0\)hoặc \(x=\frac{49}{13}\)
5 (1-2x) - 3x(x+18) = 0
5x - 10x2 - 3x 2- 54x = 0
- 13x2 - 49x = 0
(-13x - 49)x = 0
TH1
:x = 0
TH2:
-13x - 49 = 0
-13x = 49
x = -49/13
vậy x = 0 và x = -49/13