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(2x2 + 1)(x-3)=0
\(\Rightarrow\orbr{\begin{cases}2x^2+1=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x^2=-1\Rightarrow x^2=-\frac{1}{2}\left(vl\right)\\x=3\end{cases}}\)
Vậy x=3
48-(15-x)5=48
(15-x)5=48-48
(15-x)5=0
=> 15-x =0
x =15-0
x =15
Vậy x=15
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2x - 3 = 17
2x = 17 + 3
2x = 20
x = 20 : 2
x = 10
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4² - 3x = 27.3 - 77
16 - 3x = 81 - 77
16 - 3x = 4
3x = 16 - 4
3x = 12
x = 12 : 3
x = 4
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48 + 2x = 3.2³
48 + 2x = 3.8
48 + 2x = 24
2x = 24 - 48
2x = -24
x = -24 : 2
x = -12
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(13x - 12²) : 5 = 5
13x - 144 = 5.5
13x - 144 = 25
13x = 25 + 144
13x = 169
x = 169 : 13
x = 13
\(2x-3=17\)
\(2x=17+3\)
\(2x=20\)
\(x=20:2\)
\(x=10\)
____
\(4^2-3x=27.3-77\)
\(16-3x=81-77\)
\(16-3x=4\)
\(3x=16-4\)
\(3x=12\)
\(x=12:3\)
\(x=4\)
_____
\(48+2x=3.2^3\)
\(48+2x=3.8=24\)
\(2x=24-48\)
\(2x=\left(-24\right)\)
\(x=\left(-24\right):2\)
\(x=\left(-12\right)\)
____
\(\left(13x-12^2\right):5=5\)
\(\left(13x-144\right):5=5\)
\(13x-144=5.5\)
\(13x-144=25\)
\(13x=25+144\)
\(13x=169\)
\(x=169:13\)
\(x=13\)
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a, \(2x+3⋮2x-1\)
\(2x-1+4⋮2x-1\)
\(4⋮2x+1\)hay \(2x+1\inƯ\left(4\right)=\left\{1;2;4\right\}\)
2x + 1 | 1 | 2 | 4 |
2x | 0 | 1 | 3 |
x | 0 | 1/2 | 3/2 |
c, \(\left(x+5\right)\left(y-3\right)=15\Leftrightarrow x+5;y-3\inƯ\left(15\right)=\left\{1;3;5;15\right\}\)
x + 5 | 1 | 3 | 5 | 15 |
y - 3 | 15 | 5 | 3 | 1 |
x | -4 | -2 | 0 | 10 |
y | 18 | 8 | 6 | 4 |
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a) => (12x-4)+(6x+15)=16
=>12x-48+6x+15=16
=>(12x+6x)-(15-48)=16
=>18x-(-33)=16
=>18x=16+(-33)
=>18x=-17
=>x=-17/18
KL
b) =>(-20x+(-40))-(12-30x)=48
=>-20x+(-40)-12+30x=48
=>(-20x+30x)+(-40-12)=48
=>10x+(-52)=48
=>10x=48-(-52)
=>10x=4
=>x=4/10
KL
a) 12( x - 4 ) + 3( 2x + 5 ) = 16
<=> 12x - 48 + 6x + 15 = 16
<=> 18x - 33 = 16
<=> 18x = 49
<=> x = 49/18
b) -10( 2x + 4 ) - 2( 6 - 15x ) = 48
<=> -20x - 40 - 12 + 30x = 48
<=> 10x - 52 = 48
<=> 10x = 100
<=> x = 10
\(\left(2x-4\right)^{38}=\left(2x-4\right)^{48}\)
\(\Rightarrow\left(2x-4\right)^{38}-\left(2x-4\right)^{48}=0\)
\(\Rightarrow\left(2x-4\right)^{38}\left[1-\left(2x-4\right)^{10}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-4\right)^{38}=0\\1-\left(2x-4\right)^{10}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4=0\\\left(2x-4\right)^{10}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=4\\2x-4=1\\2x-4=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\2x=5\\2x=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{2;\dfrac{5}{2};\dfrac{3}{2}\right\}.\)
#\(Toru\)
thank you