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6x - 5 = 613
=> 6x = 613 + 5 = 618
=> x = 618 : 6 = 103.
0 : x = 0 => x là số nguyên nhưng x phải khác 0.
(x - 47) - 115 = 0
=> x - 47 = 115
=> x = 115 + 47 = 162
\(a,\Rightarrow12x-91=101\\ \Rightarrow12x=192\\ \Rightarrow x=16\\ b,\Rightarrow x:23+45=133\\ \Rightarrow x:23=88\\ \Rightarrow x=\dfrac{88}{23}\\ c,\Rightarrow\left(6x-39\right):7=3\\ \Rightarrow6x-39=21\\ \Rightarrow6x=60\\ \Rightarrow x=10\\ d,\Rightarrow3x-24=\dfrac{148}{73}\\ \Rightarrow3x=\dfrac{1900}{73}\\ \Rightarrow x=\dfrac{1900}{219}\\ e,\Rightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\\ f,\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\\ d,\left(9-x\right)^3=64=4^3\\ \Rightarrow9-x=4\\ \Rightarrow x=5\\ h,\Rightarrow x=27\\ i,\Rightarrow6x=312\cdot12=624\cdot6\\ \Rightarrow x=624\\ j,\Rightarrow\left(19x+104\right):14=25-42=-17\\ \Rightarrow19x+104=-238\\ \Rightarrow19x=-342\\ \Rightarrow x=-18\)
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
a ) x ∈ ± 23 28 . b ) x ∈ − 23 6 ; 35 6 . c ) x ∈ − 17 60 ; 97 60 . d ) x ∈ 67 140 ; 143 140 .
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6-0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
f) \(25+\left(15-x\right)=30\)
\(\Rightarrow25+15-x=30\)
\(\Rightarrow40-x=30\)
\(\Rightarrow x=40-30\)
\(\Rightarrow x=10\)
g) \(43-\left(24-x\right)=20\)
\(\Rightarrow43-24+x=20\)
\(\Rightarrow19+x=20\)
\(\Rightarrow x=20-19\)
\(\Rightarrow x=1\)
h) \(2\left(x-5\right)-17=25\)
\(\Rightarrow2\left(x-5\right)=17+25\)
\(\Rightarrow x-5=21\)
\(\Rightarrow x=21+5\)
\(\Rightarrow x=26\)
i) \(3\left(x+7\right)-15=27\)
\(\Rightarrow3\left(x+7\right)=27+15\)
\(\Rightarrow x+7=14\)
\(\Rightarrow x=14-7\)
\(\Rightarrow x=7\)
j) \(15+4\left(x-2\right)=95\)
\(\Rightarrow4\left(x-2\right)=95-15\)
\(\Rightarrow4\left(x-2\right)=80\)
\(\Rightarrow x-2=20\)
\(\Rightarrow x=20+2\)
\(\Rightarrow x=22\)
k) \(20-\left(x+14\right)=5\)
\(\Rightarrow x+14=20-5\)
\(\Rightarrow x+14=15\)
\(\Rightarrow x=15-14\)
\(\Rightarrow x=1\)
l) \(14+3\left(5-x\right)=27\)
\(\Rightarrow3\left(5-x\right)=27-14\)
\(\Rightarrow3\left(5-x\right)=13\)
\(\Rightarrow5-x=\dfrac{13}{3}\)
\(\Rightarrow x=5-\dfrac{13}{3}\)
\(\Rightarrow x=\dfrac{2}{3}\)
a) 1254:253= (53)3: (52)3=59:56=53
b) 414:528=414: 52.14=414:(52)14= 414:2514
c) 12n:22n=
a) \(\left(314-x\right)-42=0\)
\(314-x=42\)
\(x=272\)
vay \(x=272\)
b) \(540+\left(345-x\right)=740\)
\(345-x=200\)
\(x=145\)
vay \(x=145\)
c) \(\left(x-72\right):36=418\)
\(x-72=15048\)
\(x=15120\)
vay \(x=15120\)
d) \(2010\left(x-14\right)=0\)
\(x-14=0\)
\(x=14\)
vay \(x=14\)
e) \(\left(x-2\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}\)
vay \(\orbr{\begin{cases}x=2\\x=4\end{cases}}\)
f) \(114-\left(x-47\right)=0\)
\(x-47=114\)
\(x=161\)
vay \(x=161\)
g) \(7272:\left(120x-91\right)=12\)
\(120x-91=606\)
\(120x=697\)
\(x=\frac{697}{120}\)
vay \(x=\frac{697}{120}\)
Despacito đã có dấu'' x= '' thì ko cần kết luận là '' vậy''