Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
c: Ta có: 11+x:5=13
\(\Leftrightarrow x:5=2\)
hay x=10
d: Ta có: \(13+2\left(x+1\right)=15\)
\(\Leftrightarrow2x+2=2\)
\(\Leftrightarrow2x=0\)
hay x=0
e: Ta có: 2x+21=41
\(\Leftrightarrow2x=20\)
hay x=10
f: Ta có: \(12+3\left(x-2\right)=60\)
\(\Leftrightarrow3\left(x-2\right)=48\)
\(\Leftrightarrow x-2=16\)
hay x=18
g: Ta có: \(24x-11\cdot13=11\cdot11\)
\(\Leftrightarrow24x=11\cdot24\)
hay x=11
h: Ta có: \(17-\left(x-4\right):2=3\)
\(\Leftrightarrow\left(x-4\right):2=14\)
\(\Leftrightarrow x-4=28\)
hay x=32
B3 a) x=4 b) x=-7 c) x=5 d) x=4
B2 a) -3+ -2+ -1+0+1+2+3+4=4
b) -6+ -5+ -4+ -3+ -2+ -1+0+1+2+3+4=-11
c) -18+-17+-16+-15+-14+-13+-12+-11+-10+-9+-8+-7+-6+-5+-4+3+-2+-1+0+1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19=19
a, | x+3 | = -27/11 × 22/-9
=> |x + 3| = 6
=> x + 3 = 6 hoặc x + 3 = -6
=> x = 3 hoặc x = -9
vậy_
b, (x-3) × (2x - 7) = 0
=> x - 3 = 0 hoặc 2x - 7 = 0
=> x = 3 hoặc x = 7/2
vậy_
c, (x-1) + (x-2) + (x-3) + ... + (x-100) = 4950
=> x - 1 + x - 2 + x - 3 + ... + x - 100 = 4950
=> 100x - (1 + 2 + 3 + ... + 100) = 4950
=> 100x - (1 + 100).100 : 2 = 4950
=> 100x - 5050 = 4950
=> 100x = 10000
=> x = 10