3) \(-12+2x-9+x=0\\ -21+3x=0\\ 3x=21\\ x=7\)
4)
\(11+\left(15-x\right)=1\)
\(15-x=1-11\)
\(15-x=-10\)
\(x=15-\left(-10\right)\)
\(x=25\)
5)
\(4-\left(27-3\right)=x-\left(13-4\right)\)
\(4-24=x-9\)
\(x-9=-20\)
\(x=-20+9\)
\(x=-11\)
\(3.-12+\left(2x-9\right)+x=0.\)
\(\Leftrightarrow-12+2x-9+x=0.\Leftrightarrow3x=21.\Leftrightarrow x=7.\)
Vậy \(x=7.\)
\(4.11+\left(15-x\right)=1.\Leftrightarrow11+15-x=1.\Leftrightarrow26-x=1.\Leftrightarrow x=25.\)
Vậy \(x=25.\)
\(5.4-\left(27-3\right)=x-\left(13-4\right).\Leftrightarrow4-24=x-9.\Leftrightarrow-20=x-9.\Leftrightarrow x=-11.\)
Vậy \(x=-11.\)
\(6.8-\left(x-10\right)=23-\left(-4+12\right).\Leftrightarrow8-x+10=23-8.\Leftrightarrow18-x=15.\Leftrightarrow x=3.\)
Vậy \(x=3.\)
\(7.105-5\left(10-5x\right)=-20.\Leftrightarrow105-50+25x=-20.\Leftrightarrow25x=-75.\Leftrightarrow x=-3.\)
Vậy \(x=-3.\)
\(8.\left(x-1\right)\left(8-2x\right)\left(3x+123\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\8-2x=0.\\3x+123=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=4.\\x=-41.\end{matrix}\right.\)
Vậy \(x\in\left\{1;4;-41\right\}.\)
\(9.\left(x^2-25\right)\left(x+10\right)=0.\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+10\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0.\\x+5=0.\\x+10=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=-5.\\x=-10.\end{matrix}\right.\)
Vậy \(x\in\left\{5;-5;-10\right\}.\)
\(10.x\left(x^2+5\right)=0.\Leftrightarrow x=0.\)
3.−12+(2x−9)+x=0.3.−12+(2x−9)+x=0.
⇔−12+2x−9+x=0.⇔3x=21.⇔x=7.⇔−12+2x−9+x=0.⇔3x=21.⇔x=7.
Vậy x=7.x=7.
4.11+(15−x)=1.⇔11+15−x=1.⇔26−x=1.⇔x=25.4.11+(15−x)=1.⇔11+15−x=1.⇔26−x=1.⇔x=25.
Vậy x=25.x=25.
5.4−(27−3)=x−(13−4).⇔4−24=x−9.⇔−20=x−9.⇔x=−11.5.4−(27−3)=x−(13−4).⇔4−24=x−9.⇔−20=x−9.⇔x=−11.
Vậy x=−11.x=−11.
6.8−(x−10)=23−(−4+12).⇔8−x+10=23−8.⇔18−x=15.⇔x=3.6.8−(x−10)=23−(−4+12).⇔8−x+10=23−8.⇔18−x=15.⇔x=3.
Vậy x=3.x=3.
7.105−5(10−5x)=−20.⇔105−50+25x=−20.⇔25x=−75.⇔x=−3.7.105−5(10−5x)=−20.⇔105−50+25x=−20.⇔25x=−75.⇔x=−3.
Vậy x=−3.x=−3.
8.(x−1)(8−2x)(3x+123)=0.8.(x−1)(8−2x)(3x+123)=0.
⇔⎡⎢⎣x−1=0.8−2x=0.3x+123=0.⇔⎡⎢⎣x=1.x=4.x=−41.⇔[x−1=0.8−2x=0.3x+123=0.⇔[x=1.x=4.x=−41.
Vậy x∈{1;4;−41}.x∈{1;4;−41}.
9.(x2−25)(x+10)=0.9.(x2−25)(x+10)=0.
⇔(x−5)(x+5)(x+10)=0.⇔(x−5)(x+5)(x+10)=0.
⇔⎡⎢⎣x−5=0.x+5=0.x+10=0.⇔⎡⎢⎣x=5.x=−5.x=−10.⇔[x−5=0.x+5=0.x+10=0.⇔[x=5.x=−5.x=−10.
Vậy x∈{5;−5;−10}.
