(x - 4)(14 - 2x) =0
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`#3107.\text {DN01012007}`
\(\left(x-5\right)\cdot\left(3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\3-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0+5\\x=3-0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
Vậy, \(x\in\left\{3;5\right\}\)
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\(\left(2x-8\right)\cdot\left(5-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-8=0\\5-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=8\\x=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\div2\\x=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
Vậy, \(x\in\left\{4;5\right\}\)
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\(7x\left(2x-14\right)=0\\ \Rightarrow\left[{}\begin{matrix}7x=0\\2x-14=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\2x=14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=14\div2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)
Vậy, \(x\in\left\{0;7\right\}\)
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\(\left(2x-4\right)\cdot\left(6-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-4=0\\6-2x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=4\\2x=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\div2\\x=6\div2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy, \(x\in\left\{2;3\right\}.\)
Suy ra (2x-4)-(3x-3×5)=1 Suy ra(2x-4)-3x+15=1 Suy ra 2x-4-3x+15=1 Suy ra (2x-3x)+(15-4)=1 -1x+11=1 1-11=-1x -1x=-10 X=10
làm dùm bn 1 bài thôi
=( 2x -3 +x+5)(2x-3-x-5)=0
3x + 2=0
x = -2/3
x-8 =0
x = 8
x(2x-7)-4x+14=0
=> x(2x-7)-2(2x-7)=0
=> (x-2)(2x-7)=0
=> x=2 hoặc x=7/2
1) (2x-1)(x+3)(2-x)=0
=>2x-1 =0 hoặc x+3=0 hoặc 2-x=0
=>x=1/2 hoặc x=-3 hoặc x=2
2)x^3 + x^2 + x + 1 = 0
=>.x^2(x+1)+(x+1)=0
=>(x^2+1)(x+1)=0
=>x^2+1=0 hoặc x+1=0
=> x =-1
3) 2x(x-3)+5(x-3) =0
=>(2x+5)(x-3)=0
=>2x+5=0 hoặc x-3=0
=>x=-5/2 hoặc x=3
4)x(2x-7)-(4x-14)=0
=> (x-2)(2x-7)=0
=> x-2 =0 hoặc 2x-7=0
=>x=2 hoặc x=7/2
5)2x^3+3x^2+2x+3=0
=>x^2(2x+3)+2x+3=0
=>(x^2+1)(2x+3)=0
=>x^2+1=0 hoặc 2x+3=0
=> x =-3/2
1) \(\Rightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
2) \(\Rightarrow5\left(x-2\right).3\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
3) \(\Rightarrow2\left(x-4\right)\left(x-7\right)=0\Rightarrow\left[{}\begin{matrix}x=4\\x=7\end{matrix}\right.\)
a) 3x + 12 = 2x - 4
=> 3x - 2x = -4 - 12
=> 1x = -16
=> x = -16
vậy___
b) 14 - 3x = -x + 4
=> 14-4 = -x+3x
=> 10 = 2x
=> x = 10 : 2
=> x = 5
vậy_____
\(c) ( 2x - 8 ) ( x + 6 ) = 0\)
\(\Rightarrow\orbr{\begin{cases}2x-8=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-6\end{cases}}\)
vậy_____
(2x - 7) + 17 = 6
=> 2x - 7 = 6 - 17
=> 2x - 7 = -11
=> 2x = -11 + 7
=> 2x = -4
=> x = -4 : 2
=> x = -2
+) 12 -2(3 - 3x)= -2
=> 2(3 - 3x) = 12 + 2
=> 2(3 - 3x) = 14
=> 3 - 3x = 14 : 2
=> 3 - 3x = 7
=> 3x = 3 - 7
=> 3x = -4
=> x = -4/3
\(\left(x+1\right)\left(x-3\right)=0\)
=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vậy...
a)4x+4-3x+1=14
x+5=14
x=11
b)trường hợp 1 x2-9=0
x2=9
->x=3;-3
-trường hợp 2: x+2=0
x=-2
c)-th1:x2+9=0
x2=-9
->x rỗng
d)xy+2x-y-2=0
(xy-y)+(2x-2)=0
y(x-1)+2(x-1)=0
(y+2)(x-1)=0
th1: y+2=0
y=-2
th2:x-1=0
x=1
(th1: trường hợp 1)
(\(x\) - 4).(14 - 2\(x\)) = 0
\(\left[{}\begin{matrix}x-4=0\\14-2x=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\2x=14\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=14:2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=7\end{matrix}\right.\)
Vậy \(\in\) {4; 7}
(12-2x)(x+5)=0