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a) \(x\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\left(-7-x\right)\left(-x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)
c) \(\left(x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)
\(\Rightarrow x=3\)
e) \(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow-1\le x\le2\)
f) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3
2x-(5-7)=x-14-7\(\Rightarrow2\chi+2=\chi-24\)
\(\Rightarrow2\chi:\chi=-24-2\)
\(\Rightarrow x=-26\)
HTDT
Ah . bn oi, mik sửa chút!
\(2\chi:\chi=-24-2\Rightarrow\chi+1=-26\)
\(\Rightarrow\chi=-26-1\Rightarrow\chi=-27\)
kk
a) \(3-2x=x-4\)
\(2x-x=3-4\)
\(1x=-1\)
\(\Leftrightarrow x=-1\)
a)x- [-2] = [-18]
\(x=\left(-18\right)+\left(-2\right)\)
\(x=-20\)
b) 2x- [+14]=[-14]
\(2x=\left(-14\right)+14\)
\(2x=0\)
\(x=0\)
c) [x+4] +5=20-(-12-7)
\(\left(x+4\right)+5=39\)
\(x+4=39-5\)
\(x+4=34\)
\(x=30\)
d)15-[2-x]=(-2)2
\(15-\left(2-x\right)=4\)
\(2-x=11\)
\(x=-9\)
e)[15-x] +[-25]=[-55]
\(15-x=\left(-55\right)-\left(-25\right)\)
\(15-x=-30\)
\(x=15--30\)
\(x=45\)
g)[17-(-4)] +[-24-(-5)]=[-x+3]
\(-x+3=21+\left(-19\right)\)
\(-x+3=2\)
\(x=1\)
chúc bạn học tốt
a,x-[-2]=[-18]
x =18+2
x =20
Vậy x thuộc{20}
b,2x-[+14]=[-14]
2x-14 =14
2x =14+14
2x =28
x =28:2
x =14
Vậy x thuộc{14}
c,[x+4]+5=20-(-12-7)
[x+4]+5=20-(-19)
[x+4]+5=20+19
[x+4]+5=39
[x+4] =39-5
[x+4] =34
TH1:x+4=34
x =34-4
x =30
TH2:x+4=-34
x =-34-4
x =-38
vậy x thuộc{30;-38}
sorry bạn nha mk ko có tg nên bn làm nốt hộ mk nhá
\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)
\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)