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a, => x-1 = 0
=> x = 1
b, => x=0 hoặc x+2=0
=> x=0 hoặc x=-2
c, => -12x+60+21-7x = 5
=> 81 - 19x = 5
=> 19x = 81 - 5 = 76
=> x = 76 : 19 = 4
Tk mk nha
a ,( x- 1 ) ^ 2 = 0
x - 1 = 0
x = 0 + 1
x = 1
b , x ( x + 2 ) = 0
=> x hoặc x + 2 = 0
TH1 : x = 0
TH2 : x + 2 = 0
x = 0 - 2
x = -2
Vậy ..
1/\(x.\left(x+7\right)=0\)
\(\Rightarrow\left[\begin{matrix}x=0\\x+7=0\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=0\\x=0-7\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
2/\(\left(x+12\right).\left(x-3\right)=0\)
\(\Rightarrow\left[\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=0-12\\x=0+3\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=-12\\x=3\end{matrix}\right.\)
3/\(\left(-x+5\right).\left(3-x\right)\)
\(\Rightarrow\left[\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}-x=0-5\\x=3-0\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}-x=-5\\x=3\end{matrix}\right.\)
4/\(x.\left(2+x\right).\left(7-x\right)\)
\(\Rightarrow\left[\begin{matrix}x=0\\2+x=0\\7-x=0\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=0\\x=0-2\\x=7-0\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)
5/\(\left(x-1\right).\left(x+2\right).\left(-x-3\right)=0\)
\(\Rightarrow\left[\begin{matrix}x-1=0\\x+2=0\\-x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=0+1\\x=0-2\\-x=0+3\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=1\\x=-2\\-x=3\end{matrix}\right.\)
a) (x+2).(x^2+1).(x+2)=(x+2)2(x2+1)
vì (x+2)2\(\ge\)0;x\(\ge\)0
nên (x+2)(x2+1)>0
vậy ko có giá trị nào của x thỏa mãn (x+2)(x2+1)(x+2)<0
b) -12.(x-5)+7.(3-x)=5
=-12.x+(-12)(-5)+7.3-7.x=5
=-12x+60+21-7.x=5
=(-12x-7x)+60+21=5
=x(-12-7)+81=5
=x(-19) =5-81
=x(-19) =-76
=x =-76:(-19)
=x =4
c)2.x.(x+3)-x.(2.x-1)=6
=2.x.x+2.x.3-2.x.x+x=6
=6.x+x =6
=x(6+1) =6
=x.7 =6
=x =6/7
bài 1: x.(x+7) = 0
Th1:x=0 Th2:x+7=0
=>x=-7
bài 2 (x+12).(x-3)= 0
Th1:x+12=0 Th2:x-3=0
=>x=-12 =>x=3
bài 3 (-x+5).(3-x)=0
Th1 (-x)+5=0 Th2:3-x=0
=>-x=-5 =>x=3
bài 4 x.(2+x).(7-x)=0
Th1:x=0 Th3:7-x=0
Th2:2+x=0 =>x=7
=>x=-2
bài 5 (x-1).(x+2).(-x-3)=0
Th1:x-1=0 Th2:x+2=0
=>x=1 =>x=-2
Th3:-x-3=0
=>-x=-3
1, - 12 ( x - 5 ) + 7 ( 3 - x ) = 5
<=> - 12x + 60 + 21 - 7x = 5
<=> (- 12 x - 7x )+ ( 60 + 21 ) = 5
<=> - 19x + 81 = 5
<=> - 19x = 5 - 81 = 76
<=> x = - 4
Vậy x = - 4
2, Ta có \(\hept{\begin{cases}x\in Z\\\left|x\right|< 3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\in Z\\\left|x\right|< 3\end{cases}}\)
<=> \(\left|x\right|\in\left\{0;1;2\right\}\)
\(\Leftrightarrow x\in\left\{0;1;-1;2;-2\right\}\)
Vậy \(x\in\left\{0;1;-1;2;-2\right\}\)
3, ( x - 3 ) ( x - 5 ) < 0
\(\Leftrightarrow\hept{\begin{cases}x-3< 0\\x-5>0\end{cases}}\) hoặc \(\hept{\begin{cases}x-3>0\\x-5< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x< 3\\x>5\end{cases}}\) ( vô lí ) hoặc \(\hept{\begin{cases}x>3\\x< 5\end{cases}}\)
<=> 3 < x < 5
Mà \(x\in Z\)
<=> x = 4
Vậy x = 4
@@ Học tốt
CHiyuki Fujito