Tìm số nguyên x, biết:
x + 5 = -10
( x - 7 ) + ( x + 2 ) = -15
( 5x - 6 ) - ( 2x + 3 ) = -2x + 27
( x + 1 ) . ( x - 3 ) < 0
Giúp mình với !
Bài1: giải các pt sau:
a, 3-4x+24+6x= x+27+3x
b, 5-(6-x)=4(3-2x)
c, x-(x+1)/3 = (2x+1)/5
d,(2x-1)/3 - (5x+2)/7 = x+13
Bài 2:
a, (x-1)(3x+1)=0
b, (x-5)(7-x)=0
c, ( x-1)(x+5)(-3x+8)=0
d, x(x^2 - 1 )=0
Giúp mình 2 bài này với , mình đang cần gấp , CẢM ƠN M.N ạ><
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
tìm x bt
a)x(x-5)-4x+20=0
b)x(x+6)-7x-42=0
c)x^3-5x^2-x+5=0
d)4x^2-25-(2x-5)(3x+7)=0
e)x^3+27+(x+3)(x-9)=0
giúp mk vs ah!!1
a) x(x - 5) - 4x + 20 = 0
\(\Leftrightarrow\) x(x - 5) - (4x + 20)
\(\Leftrightarrow\) x(x - 5) - 4(x - 5) = 0
\(\Leftrightarrow\) (x - 5)(x - 4)
Khi x - 5 = 0 hoặc x - 4 = 0
\(\Leftrightarrow\) x = 5 \(\Leftrightarrow\) x = 4
Vậy S = \(\left\{5;4\right\}\)
b) x(x + 6) - 7x - 42 = 0
\(\Leftrightarrow\) x(x + 6) - (7x - 42) = 0
\(\Leftrightarrow\) x(x + 6) - 7(x + 6) = 0
\(\Leftrightarrow\) (x + 6)(x - 7) = 0
Khi x - 6 = 0 hoặc x - 7 = 0
\(\Leftrightarrow\) x = 6 \(\Leftrightarrow\) x = 7
Vậy S = \(\left\{6;7\right\}\)
c) x3 - 5x2 - x + 5 = 0
\(\Leftrightarrow\) (x3 - 5x2) - (x + 5) = 0
\(\Leftrightarrow\) x2 (x - 5) - (x - 5) = 0
\(\Leftrightarrow\) (x - 5)(x2 - 1) = 0
\(\Leftrightarrow\) (x - 5)(x - 1)(x + 1) = 0
Khi x - 5 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0
\(\Leftrightarrow\) x = 5 \(\Leftrightarrow\) x = 1 \(\Leftrightarrow\) x = -1
Vậy S = \(\left\{5;1;-1\right\}\)
d) 4x2 - 25 - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x)2 - 52 - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x - 5)(2x + 5) - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x - 5) \([\left(2x+5\right)-\left(3x+7\right)]\) = 0
\(\Leftrightarrow\) (2x - 5) ( 2x + 5 - 3x + 7) = 0
\(\Leftrightarrow\) (2x - 5)( -x + 12) = 0
Khi 2x - 5 = 0 hoặc -x + 12 = 0
\(\Leftrightarrow\) 2x = 5 \(\Leftrightarrow\) -x = -12
\(\Leftrightarrow\) x = \(\dfrac{5}{2}\) \(\Leftrightarrow\) x = 12
Vậy S = \(\left\{\dfrac{5}{2};12\right\}\)
e) x3 + 27 + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) x3 - 33 + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) (x - 3)(x2 - 3x + 9) + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) (x - 3) \(\left[\left(x^2-3x+9\right)+\left(x-9\right)\right]\) = 0
\(\Leftrightarrow\) (x - 3) ( x2 - 3x + 9 + x - 9) = 0
\(\Leftrightarrow\) (x - 3)(x2 - 2x) = 0
\(\Leftrightarrow\) (x - 3)x(x - 2)
Khi x - 3 = 0 hoặc x = 0 hoặc x - 2 = 0
\(\Leftrightarrow\) x = 3 \(\Leftrightarrow\) x = 2
Vậy S = \(\left\{3;0;2\right\}\)
Chúc bạn học tốt
a) Ta có: \(x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
b) Ta có: \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
c) Ta có: \(x^3-5x^2-x+5=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\\x=-1\end{matrix}\right.\)
d) Ta có: \(4x^2-25-\left(2x-5\right)\left(3x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5-3x-7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)
Bài 1:Tìm x biết:
a.,2x(x-3)-15+5x=0
b,x^3-7x
c,(2x-3)^2-(x+5)^2=0
d,x^3-x^2-4x^2+8x-4=0
giúp mk với mk cầm gấp lắm TvT
a, \(2x\left(x-3\right)-15+5x=0\\ \Rightarrow2x\left(x-3\right)-\left(15-5x\right)=0\\ \Rightarrow2x\left(x-3\right)-5\left(3-x\right)=0\\ \Rightarrow\left(2x+5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=3\end{matrix}\right.\)
b, \(x^3-7x=0\\ \Rightarrow x\left(x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm7\end{matrix}\right.\)
c, \(\left(2x-3\right)^2-\left(x+5\right)^2=0\\ \Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ \Rightarrow\left(x-8\right)\left(3x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Xem lại đề câu d
Bài 1: Tìm x [GIÚPPPPPPPPPPPPPPPPPPP]
1) 5 – (10 – x) = 7
2) - 32 - (x – 5) = 0
3) -12 + (2x – 9) + x= 0
4) 11 + (15 - x) = 1
5) 4 - (27 - 3) = x - (13 - 4)
6) 8 - (x - 10) = 23 - (- 4 +12)
7) 105 – 5(10 – 5x) = -20
8) (x -1)(8-2x)(3x+123) = 0
9) (x2 - 25)(x+ 10) = 0
10) x(x2+5) = 0
\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)
\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)
1)10-x=5-7
x=10-(-2)
x=12
2)x+5=0+32
x=32-5
x=27
3) -12+2x-9+x=0
-12+(2x+x-9)=0
3x-9=0+12
3x=12+9
x=21:3
x=7
4) 15-x=1-11
x=15-(-10)
x=25
5) 4-(27+3)=x-9
4-30=x-9
-26+9=x
x=-17
Cho A= x^3 -2x+n B=x-2 Tìm n để A chia hết cho B
(x-3)^2-2x+6=0
x^2-5x+6=0
Giúp mình nhanh vs ạ
\(A=x^3-2x+n\)
\(B=n-2\)
\(A\text{⋮}B\) ⇒ \(\left(x^3-2x+n\right)\text{⋮}\left(n-2\right)\)
⇒ \(\left[\left(x^3-2x^2\right)+\left(2x^2-4x\right)+\left(2x-4\right)+\left(n+4\right)\right]\text{⋮}\left(n-2\right)\)
⇒ \(\left[x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)+\left(n+4\right)\right]\text{⋮}\left(n-2\right)\)
⇒ \(\left[\left(x-2\right)\left(x^2+2x+2\right)+\left(n+4\right)\right]\text{⋮}\left(x-2\right)\)
Vì \(\left(x-2\right)\left(x^2+2x+2\right)\text{⋮}\left(n-2\right)\)
Để \(A\text{⋮}B\)
⇒ \(n+4=0\)
⇒ \(n=-4\)
tìm số nguyên x biết
a, 2x+1/3=x-5/2 b, 4(x-2) ^2/3=12
25/30=2x+3/6 -7/x+1=6/x+27
a: =>2x-x=-5/2-1/3
=>x=-17/6
b: =>4(x-2)2=36
=>(x-2)2=9
=>x-2=3 hoặc x-2=-3
hay x=5 hoặc x=-1
c: =>2x+1/2=5/6
=>2x=1/3
hay x=1/6
a: =>2x-x=-5/2-1/3
=>x=-17/6
b: =>4(x-2)2=36
=>(x-2)2=9
=>x-2=3 hoặc x-2=-3
hay x=5 hoặc x=-1
c: =>2x+1/2=5/6
=>2x=1/3
hay x=1/6
2.(x-5)-3.(x-4)=-6+15:9-3)
(x+7).(x-9)=0
/2x-5/-7=22
(/2x/-5)-7=22
tìm x,y thuộc z biết:x+y=x.y
Tim X
3) -12 + (2x – 9) + x= 0
4) 11 + (15 - x) = 1
5) 4 - (27 - 3) = x - (13 - 4)
6) 8 - (x - 10) = 23 - (- 4 +12)
7) 105 – 5(10 – 5x) = -20
8) (x -1)(8-2x)(3x+123) = 0
9) (x2 - 25)(x+ 10) = 0
10) x(x2+5) =
3) \(-12+2x-9+x=0\\ -21+3x=0\\ 3x=21\\ x=7\)
4)
\(11+\left(15-x\right)=1\)
\(15-x=1-11\)
\(15-x=-10\)
\(x=15-\left(-10\right)\)
\(x=25\)
5)
\(4-\left(27-3\right)=x-\left(13-4\right)\)
\(4-24=x-9\)
\(x-9=-20\)
\(x=-20+9\)
\(x=-11\)
\(3.-12+\left(2x-9\right)+x=0.\)
\(\Leftrightarrow-12+2x-9+x=0.\Leftrightarrow3x=21.\Leftrightarrow x=7.\)
Vậy \(x=7.\)
\(4.11+\left(15-x\right)=1.\Leftrightarrow11+15-x=1.\Leftrightarrow26-x=1.\Leftrightarrow x=25.\)
Vậy \(x=25.\)
\(5.4-\left(27-3\right)=x-\left(13-4\right).\Leftrightarrow4-24=x-9.\Leftrightarrow-20=x-9.\Leftrightarrow x=-11.\)
Vậy \(x=-11.\)
\(6.8-\left(x-10\right)=23-\left(-4+12\right).\Leftrightarrow8-x+10=23-8.\Leftrightarrow18-x=15.\Leftrightarrow x=3.\)
Vậy \(x=3.\)
\(7.105-5\left(10-5x\right)=-20.\Leftrightarrow105-50+25x=-20.\Leftrightarrow25x=-75.\Leftrightarrow x=-3.\)
Vậy \(x=-3.\)
\(8.\left(x-1\right)\left(8-2x\right)\left(3x+123\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\8-2x=0.\\3x+123=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=4.\\x=-41.\end{matrix}\right.\)
Vậy \(x\in\left\{1;4;-41\right\}.\)
\(9.\left(x^2-25\right)\left(x+10\right)=0.\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+10\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0.\\x+5=0.\\x+10=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=-5.\\x=-10.\end{matrix}\right.\)
Vậy \(x\in\left\{5;-5;-10\right\}.\)
\(10.x\left(x^2+5\right)=0.\Leftrightarrow x=0.\)
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
a,\(x\in\left\{5;1,5;\dfrac{-4}{3}\right\}\)