Tìm số nguyên x, sao cho :
a) (x - 1)2 = 0
b) x (x - 1) = 0
c) (x + 1) (x - 2) = 0
Giúp mình nhé ! 
K
Khách
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13 tháng 7 2021
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
13 tháng 7 2021
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: \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
19 tháng 4 2023
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
=>\(x\in\left\{0;1;-1\right\}\)
13 tháng 1 2022
Bài 2:
a: =>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
PL
1
17 tháng 11 2023
a: x(x+5)=0
=>\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
b: 2x(x+3)=0
=>x(x+3)=0
=>\(\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
c: \(\left(6-x\right)\left(x+10\right)=0\)
=>\(\left[{}\begin{matrix}6-x=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6-0=6\\x=0-10=-10\end{matrix}\right.\)
d: \(\left(5x+20\right)\left(x^2+1\right)=0\)
=>\(5x+20=0\left(x^2+1>=1>0\forall x\right)\)
=>5x=-20
=>x=-4
DL
2
19 tháng 10 2021
\(a,\Rightarrow\left(x-2000\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\\ b,\Rightarrow x\left(x^2-13\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\\ c,\Rightarrow3x\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ d,\Rightarrow\left(x-5\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\\ e,\Rightarrow\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
HP
15 tháng 9 2021
a. (2x + 1)2 - 4x2 + 2x2 - 2 = 0
<=> (2x + 1 - 2x)(2x + 1 + 2x) + 2(x2 - 1) = 0
<=> (4x + 1) + 2x2 - 2 = 0
<=> 4x + 1 + 2x2 - 2 = 0
<=> 2x2 + 4x - 2 + 1 = 0
<=> 2x2 + 4x - 1 = 0
<=> 2x2 + 4x = 1
<=> 2x(x + 2) = 1
Vì 1 chỉ có tích là 1 . 1 nên:
<=> \(\left[{}\begin{matrix}2x=1\\x+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
15 tháng 9 2021
\(a,\Leftrightarrow4x^2+4x+1-4x^2+2x^2-2=0\\ \Leftrightarrow2x^2+4x-1=0\\ \Leftrightarrow2\left(x^2+2x+1\right)-3=0\\ \Leftrightarrow2\left(x+1\right)^2-3=0\\ \Leftrightarrow\left(x+1\right)^2=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{3}{2}}\\x+1=-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{6}}{2}\\x=\dfrac{-2+\sqrt{6}}{2}\end{matrix}\right.\)
\(b,\left(x-2\right)\left(x+2\right)-\left(x+3\right)^2-2x-5=0\\ \Leftrightarrow x^2-4-x^2-6x-9-2x-5=0\\ \Leftrightarrow-8x=18\\ \Leftrightarrow x=-\dfrac{9}{4}\)
a) ( x-1)2 = 0
-> x-1 = 0
-> x = 0+1=1
Vậy x=1
b) x ( x - 1 ) = 0
-> x-1=0
-> x=0+1=1
Vậy x=1
c) (x + 1 ) ( x-2) = 0
-> hoặc x+1=0 -> x=0-1=-1
hoặc x-2=0 -> x= 0+2=2
Vậy x thuộc { -1 ; 2 }
a x=1
b x=1
c x=-1
x=2