Ta có: (2x+5)(x-3)=(x-4)(3-x)
⇔(2x+5)(x-3)-(x-4)(3-x)=0
⇔(2x+5)(x-3)+(x-4)(x-3)=0
⇔(x-3)(2x+5+x-4)=0
⇔(x-3)(3x+1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\3x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-1}{3};3\right\}\)
a) (2x +1)(3 – x)(4 - 2x) = 0 b)2x(x – 3) + 5(x – 3) = 0
c) (x2 – 4) – (x – 2)(3 – 2x) = 0 d) x2 – 5x + 6 = 0
e) (2x + 5)2 = (x + 2)2 f) 2x3 + 6x2 = x2 + 3x
Bài1:Giải phương trình:
a,(5-x)(3-2x)(3x+4)=0
b,(2x-1)(3x+2)(5-x)=0
c,(2x-1)(x-3)(x+7)=0
Giúp mình với :)
d,(3-2x)(6x+4)(5-8x)=0
Giải phương trình :
a)(2x-5)^3-(3x-4)^x+(x+1)^3=0
b)(x-1)^3+(2x-3)^3+(3x-5)^3 - 3(x-1)(2x-3)(3x-5) = 0
c)(x^2+3x-4)^3 + (3x^2+7x+4)^3 = (4x^2+10x)^3
A, (x-4)(x-5)=12
B,(2x-2)(2x-4)=3
C,(x-5)(x+6)=42
D, (2x-1)(x+3)=-6
A, (x-4)(x-5)=12
B,(2x-2)(2x-4)=3
C,(x-5)(x+6)=42
D, (2x-1)(x+3)=-6
1/(2x-1)(3x+2)(5-x)=0
2/(2x+5)(x-4)=(x-5)(4-x)
3/16x^2-8x+1=4(x+3)(4x-1)
4/27x^2(x+3)-12(×^2+3x)=0
5/2(9x^2+6x+1)=(3x+1)(x-2)
6/(2x-1)^2=49
Bài 1: giải các pt sau:
1,(x-1)^2-(x+1)^2=2(x+3)
2,(2x-1)^2-(2x+1)^2=4(x-3)
3,(2x+3)^2-(2x+3).(2x-4)=-(x-2)^2
4,8x^3-(x+1)^3=3x-3
5,(3x-2).(9x^2+6x+4)-(3x+1).(9x^2-3x+1)=(2x+1).(2x-1)-4x(x-3)
A) x+5/4-2x-3/3=6x-1/8+2x-1/12
B) x+4/5-x+4=x/3-x-2/2
C) x^2-3x+2=0
Giúp em với mọi người:33
1, \(\dfrac{\left(2x-3\right)\cdot\left(2x+3\right)}{8}=\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\)
2, \(x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}=\dfrac{7x-\dfrac{x-3}{2}}{5}\)
3, \(1-\dfrac{x-\dfrac{1+x}{3}}{3}=\dfrac{x}{2}-\dfrac{2x-\dfrac{10-7}{3}}{2}\)
4, \(\dfrac{x+1}{99}+\dfrac{x+3}{97}+\dfrac{x+5}{95}=\dfrac{x+7}{93}+\dfrac{9+x}{91}+\dfrac{x+11}{89}\)
Bài 1 : Giải các phương trình sau:1) 5-(x-6)=4(3-2x) 2) (x-3)(x+4)-2(3x-2)=(x-4)2 3)3-x(1-3x)=5(1-2x) 4)9x-1=(3x+1)(4x+1) 5)2x(x-1)=x2-1 6)x2-5x2+6x=0 7)x2+4x-5=0 8)x3+9x2-4x-36=0
