\(x^3-3x^2-3x-4=0\)
\(\Leftrightarrow x^3-4x^2+x^2-4x+x-4=0\)
\(\Leftrightarrow x^2\left(x-4\right)+x\left(x-4\right)+\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow x-4=0\) (vì \(x^2+x+1>0\))
\(\Leftrightarrow x=4\)
Vậy \(S=\left\{4\right\}\)
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
Giải phương trình tích
1) (4-3x)(10x-5)=0
2) (7-2x)(4+8x)=0
3) (9-7x)(11-3x)=0
4)(7-14x)(x-2)=0
5)(2x+1)(x-3)=0
6)(8-3x)(-3x+5)=0
Mọi người giải hộ mình với ạ !
bài phương trình tích:
(x-1)(3x-6)=0
(2x+5)(1-3x)=0
(x+1)(2x-3)(3x-5)=0
6(x-2)(x-4)(1-7x)=0
(x+1)2(x+2)=0
(3x-2)2(x+1)(x-2)=0
(5-x)2(3x-1)=0
(14-2x)2(3-x)(2x-4)=0
(5x-6)2(x+2)(x+10)=0
(3x-3)3(x+4)=0
Giúp mình với nhé cảm ơn nhiều ^^
giải hộ mk vs
1/2x^4+3x^3-x^2+3x+2=0
2/x^4-5x^3+7x^2-5x-16=0
3/(x+2)^4+(x+4)^4=16
I) giải các pt tích:
1) 3x - 12= 5x(x - 4)
2) 3x - 15= 2x(x - 5)
3) 3x(2x - 3) + 2(2x - 3)= 0
4) (4x - 6) (3 - 3x)= 0
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
a/ (x+2)(2-2x)-(2x+1)(3-x)=0
b/(x-4)(x2-2)=0
c/\(\dfrac{3x+1}{4}-x-1=\dfrac{4-3x}{3}\)
A) (3x+2)(x^2+1)=(9x^2-4)(x+1)
B) x(2x-9)=3x(x-5)
C) (x-3/4)^2+(x-3/4)(x-1/2)=0
1,Giải PT sau
a, (5x+3)(x2+4)(x-4)=0
b, (4x-1)(x-3)-(x-2)(5x+2)=0
c, (x+3)(x-5)+(x+3)(3x-4)=0
d, (x+6)(3x-1)+x2-36=0
e, 0,75x(x+5)=(x+5)(3-1,25x)
Giải phương trình:
a) \(\dfrac{x^2-x-6}{x-3}=0\)
b) \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)
c) \(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\)
d) \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)
e) \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)Thể loại truyện
