|x+2|-|x+7|=0
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\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
1) -12+3.(-x+7)=-18
3.(-x+7)=-18+12
3.(x+7)=-6
x+7=-6:3
x+7=-2
x=-2-7
x=-9
a) x=0 hoặc x+7=0
suy ra x=0 hoặc x=-7
b) x+12=0 hoặc x-3=0
x=-12 hoặc x=3
c) x=0 hoặc x+2=0 hoặc 7-x=0
x=0 hoặc x=-2 hoặc x=7
d) x-1=0 hoặc x+2=0 hoặc -x-3=0
suy ra x=1 hoặc x=-2 hoặc x=-3
Bài làm
x( x + 7 ) = 0
<=> x = 0 hoẵ x + 7 = 0
=> x = 0 hoặc x = -7
Vậy x = 0 hoặc x = -7
( x + 12 )( x - 3 ) = 0
<=> x + 12 = 0 hoặc x - 3 = 0
=> x = -12 hoặc x = 3
Vậy x = -12 hoặc x = 3
( -x + 5 )( 3 - x ) = 0
<=> -x + 5 = 0 hoặc 3 - x = 0
=> x = 5 hoặc x = 3
Vậy x = 5 hoặc x = 3
x( 2 + x )( 7 - x ) = 0
<=> x = 0 hoặc 2 + x = 0 hoặc 7 - x = 0
=> x = 0 hoặc x = -2 hoặc x = 7
Vậy x = 0 hoặc x = -2 hoặc x j 7
( x - 1 )( x + 2 )( -x - 3 ) = 0
<=> ( x - 1 ) = 0 hoặc x + 2 = 0 hoặc ( -x - 3 ) = 0
<=> x = 1 hoăc x = -2 hoặc x = ( -3)
Vậy x = 1 hoặc x = 2 hoặc x = -3
7 x 3 = 21
7 x 8 = 56
7 x 2 = 14
7 x 1 = 7
7 x 4 = 35
7 x 6 = 42
7 x 10 = 70
0 x 7 = 0
7 x 7 = 49
7 x 4 = 28
7 x 9 = 63
7 x 0 = 0
7 x 1 = 7
7 x 8 = 56
7 x 6 = 42
7 x 5 = 35
7x 2 = 14
7 x 9 = 63
7 x 4 = 28
0 x 7 = 0
7x 3 = 21
7 x 7 = 49
7 x 0 = 0
7 x 10 = 70
7 x 2 = 14 7 x 5 = 35 7 x 6 = 42 0 x 7 = 0
7 x 4 = 28 7 x 3 = 21 7 x 9 = 63 7 x 0 = 0
7 x 8 = 56 7 x 1 = 7 7 x 10 =70 1 x 7 = 7
Làm theo công thức: tích bằng 0 thì một trong x thừa số bằng 0 rồi xét các trường hợp
\(1,x.\left(x+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}}\)
\(2,\left(x+12\right).\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+12=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-12\\x=3\end{cases}}}\)
\(3,\left(-x+5\right).\left(3-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x+5=0\\3-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}}\)
4/ \(x.\left(2+x\right).\left(7-x\right)=0\)
\(\hept{\begin{cases}x=0\\2+x=0\\7-x=0\end{cases}}\) => \(\hept{\begin{cases}x=0\\x=-2\\x=7\end{cases}}\)
Vậy \(x=\left\{0,-2,7\right\}\)
5/ \(\left(x-1\right).\left(x+2\right).\left(-x-3\right)=0\)
\(\hept{\begin{cases}x-1=0\\x+2=0\\-x-3=0\end{cases}}\)=> \(\hept{\begin{cases}x=1\\x=-2\\x=-3\end{cases}}\)
a) \(\Rightarrow\left(x-2\right)\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b) \(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
c) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
d) \(\Rightarrow\left(x-7\right)\left(3x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)
\(a,\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\\ c,\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\\ d,\Leftrightarrow\left(x-7\right)\left(3x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)
1/(2.x-5)+17=6
=> 2x - 5 = -11
=> 2x = -6
=> x = 3
vậy_
2/10-2.(4-3x)=-4
=> 2(4 - 3x) = 14
=> 4 - 3x = 7
=> 3x = -3
=> x = -1
3/-12+3.(-x+7)=-18
=> 3(-x+7) = -6
=> -x+7 = -2
=> -x = -9
=> x = 9
4/24:(3.x-2)=-3
=> 3x - 2 = -8
=> 3x = -6
=> x = -2
5/-45:5.(-3-2.x)=3
=> 5(-3 - 2x) = -15
=> -3 - 2x = -3
=> - 2x = 0
=> x = 0
6/x.(x+7)=0
=> x = 0 hoặc x + 7 = 0
=> x = 0 hoặc x = -7
7/(x+12).(x-3)=0
=> x + 12 = 0 hoặc x - 3 = 0
=> x = -12 hoặc x = 3
8/(-x+5).(3-x)=0
=> -x + 5 = 0 hoặc 3 - x = 0
=> x = 5 hoặc x = 3
9/x.(2+x).(7-x)=0
=> x = 0 hoặc 2 + x = 0 hoặc 7 - x = 0
=> x = 0 hoặc x = -2 hoặc x = 7
10/(x-1).(x+2).(-x-3)=0
=> x - 1 = 0 hoặc x + 2 = 0 hoặc -x-3 = 0
=> x = 1 hoặc x = -2 hoặc x = -3
x + 2 = 0
{
x + 7 = 0
x = 2
{
x = 7
Vậy x = { 2 ; 7 }
k cho mình nhaaaa
<=> \(|x+2|=|x+7|\)
<=> \(\orbr{\begin{cases}x+2=x+7\\x+2=-x-7\end{cases}}\)
<=> \(\orbr{\begin{cases}0=5\\2x=-9\end{cases}}\)
<=> \(\orbr{\begin{cases}VL\\x=-\frac{9}{2}\end{cases}}\)
VẬY \(x=-\frac{9}{2}\)