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a, 7\(x\).(2\(x\) + 10) =0
\(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\) {-5; 0}
b, -9\(x\) : (2\(x\) - 10) = 0
9\(x\) = 0
\(x\) = 0
c, (4 - \(x\)).(\(x\) + 3) = 0
\(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(x\in\) {-3; 4}
x + 10 = 0 - 10
x + 10 = - 10
x = -10 - 10
x = -20
k mk nha
\(\dfrac{x+8}{12}+\dfrac{x+9}{11}+\dfrac{x+10}{10}+3=0\\ \Leftrightarrow\dfrac{x+8}{12}+1+\dfrac{x+9}{11}+1+\dfrac{x+10}{10}+1=0\\ \Leftrightarrow\dfrac{x+20}{12}+\dfrac{x+20}{11}+\dfrac{x+20}{10}=0\\ \Leftrightarrow\left(x+20\right)\left(\dfrac{1}{12}+\dfrac{1}{11}+\dfrac{1}{10}\right)=0\\ \Leftrightarrow x+20=0\Leftrightarrow x=-20\\ KL:...\)
`<=>((x+8)/12+1)+((x+9)/11+1)+((x+10)/10+1)=0`
`<=>(x+20)/12+(x+20)/11+(x+20)/10=0`
`<=>(x+20)(1/12+1/11+1/10)=0`
Vì `1/12+1/11+1/10 ≠ 0`
`=>x+20=0`
`=>x=0-20`
`=>x=-20`
a: =>10+3x-3=6x+10
=>3x-3=6x
=>-3x=3
=>x=-1
b: =>x+1=0 hoặc x-2=0
=>x=-1 hoặc x=2
a) \(10+3\left(x-1\right)=10+6x\)
\(\Rightarrow10+3x-3=10+6x\)
\(\Rightarrow3x-6x=10-10+3\)
\(\Rightarrow-3x=3\)
\(\Rightarrow x=-\dfrac{3}{3}\)
\(\Rightarrow x=-1\)
b) \(\left(x+1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Ta có :\(\frac{x+8}{12}+\frac{x+9}{11}+\frac{x+10}{10}+3=0\)
=> \(\left(\frac{x+8}{12}+1\right)+\left(\frac{x+9}{11}+1\right)+\left(\frac{x+10}{10}+1\right)=0\)
=> \(\frac{x+20}{12}+\frac{x+20}{11}+\frac{x+20}{10}=0\)
=> \(\left(x+20\right)\left(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\right)=0\)
Vì \(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\ne0\)
=> x + 20 = 0
=> x = -20
Vậy x = -20
\(\frac{x+8}{12}+\frac{x+9}{11}+\frac{x+10}{10}+3=0\)
\(\Leftrightarrow\left(\frac{x+8}{12}+1\right)+\left(\frac{x+9}{11}+1\right)+\left(\frac{x+10}{10}+1\right)=0\)
\(\Leftrightarrow\frac{x+20}{12}+\frac{x+20}{11}+\frac{x+20}{10}=0\)
\(\Leftrightarrow\left(x+20\right)\left(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\right)=0\)
Vì \(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\ne0\)
\(\Rightarrow x+20=0\Rightarrow x=-20\)
\(\left\{{}\begin{matrix}x\in B\left(10\right)\\10< x< 60\end{matrix}\right.\\ \Rightarrow x\in\left\{20;30;40;50\right\}\\ \left\{{}\begin{matrix}x⋮2\\0< x< 20\end{matrix}\right.\\ \Rightarrow x\in\left\{2;4;6;8;10;12;14;16;18\right\}\)
a) -10 - (x - 5) + (3 - x) = -8
=> -10 - x + 5 + 3 -x = -8
=> -2x - 2 = -8
=> -2x = -6
=> x = -6/-2 = 3
b) 10 + 3(x - 1) = 10 + 6x
=> 10 + 3x - 3 = 10 + 6x
=> 3x - 6x = 10 - 7
=> -3x = 3
=> x = 3/-3 = -1
c) (x + 1)(x - 2) = 0
=> \(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
\(\text{( 2x - 10 ) ( 3x + 27 ) = 0 }\)
\(\Rightarrow\orbr{\begin{cases}2x-10=0\\3x+27=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=0+10\\3x=0-27\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=10\\3x=-27\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=10:2\\x=\left(-27\right):3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-9\end{cases}}\)
vậy_____
-17 ( x + 10 ) > 0
=> -17 và x+10 cùng dấu
mà -17 mang dấu (-)
=> x+10 cũng mang dấu (-)
=> x < (-10)
=> x thuộc {-11;-12;-13;.....}
vậy_____
mấy câu này dễ, em chỉ cần nhắm vững lí thuyết là được
\(\left(2x-10\right)\left(3x+27\right)=0\)
\(2.\left(x-5\right).3.\left(x+9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+9=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-9\end{cases}}\)
vậy...
\(-17.\left(x+10\right)>0\)
\(\Rightarrow x+10< 0\) ( vì \(-17< 0\))
\(\Rightarrow x< -10\)
vậy \(x< -10\)
`10.x+(-10)=0`
`10.x=10`
`x=10:10`
`x=1`
10.x+(-10)=0
10.x = 10
x = 10 : 10
x = 1