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\(a.2.x-138=2^3.3^2\)
\(2.x-138=8.9\)
\(2.x-138=72\)
\(2.x\) \(=72+138\)
\(2.x\) \(=210\)
\(x\) \(=210:2\)
\(x\) \(=105\)
\(b.231-\left(x-6\right)=1339:13\)
\(231-\left(x-6\right)=103\)
\(x-6=231-103\)
\(x-6=128\)
\(x\) \(=128+6\)
\(x\) \(=134\)
\(c.\left[\left(6.x-72\right):2-84\right].28=5628\)
\(\left[\left(6.x-72\right):2-84\right]\) \(=5628:28\)
\(\left(6.x-72\right):2-84\) \(=201\)
\(\left(6.x-72\right):2\) \(=201-84\)
\(\left(6.x-72\right):2\) \(=117\)
\(6.x-72\) \(=117.2\)
\(6.x-72\) \(=234\)
\(6.x\) \(=234+72\)
\(6.x\) \(=306\)
\(x\) \(=306:6\)
\(x\) \(=51\)
mik nè
`#3107.101107`
a,
\(\text{A = }\left\{x\in R\text{ | }\left(2x-x^2\right)\left(3x-2\right)=0\right\}\)
`<=> (2x - x^2)(3x - 2) = 0`
`<=>`\(\left[{}\begin{matrix}2x-x^2=0\\3x-2=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x\left(2-x\right)=0\\3x=2\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\2-x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=2\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy, `A = {0; 2; 2/3}`
b,
\(\text{B = }\left\{x\in R\text{ | }2x^3-3x^2-5x=0\right\}\)
`<=> 2x^3 - 3x^2 - 5x = 0`
`<=> x(2x^2 - 3x - 5) = 0`
`<=>`\(\left[{}\begin{matrix}x=0\\2x^2-3x-5=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\2x^2-2x+5x-5=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\\left(2x^2-2x\right)+\left(5x-5\right)=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\2x\left(x-1\right)+5\left(x-1\right)=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\\left(2x+5\right)\left(x-1\right)=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\2x+5=0\\x-1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\\x=1\end{matrix}\right.\)
Vậy, `B = {-5/2; 0; 1}.`
c,
\(\text{C = }\left\{x\in Z\text{ | }2x^2-75x-77=0\right\}\)
`<=> 2x^2 - 75x - 77 = 0`
`<=> 2x^2 - 2x + 77x - 77 = 0`
`<=> (2x^2 - 2x) + (77x - 77) = 0`
`<=> 2x(x - 1) + 77(x - 1) = 0`
`<=> (2x + 77)(x - 1) = 0`
`<=>`\(\left[{}\begin{matrix}2x+77=0\\x-1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=-77\\x=1\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=-\dfrac{77}{2}\\x=1\end{matrix}\right.\)
Vậy, `C = {-77/2; 1}`
d,
\(\text{D = }\left\{x\in R\text{ | }\left(x^2-x-2\right)\left(x^2-9\right)=0\right\}\)
`<=> (x^2 - x - 2)(x^2 - 9) = 0`
`<=>`\(\left[{}\begin{matrix}x^2-x-2=0\\x^2-9=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2+x-2x-2=0\\x^2=9\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}\left(x^2+x\right)-\left(2x+2\right)=0\\x^2=\left(\pm3\right)^2\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x\left(x+1\right)-2\left(x+1\right)=0\\x=\pm3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}\left(x-2\right)\left(x+1\right)=0\\x=\pm3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x-2=0\\x+1=0\\x=\pm3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=2\\x=-1\\x=\pm3\end{matrix}\right.\)
Vậy, `D = {-1; -3; 2; 3}.`
Tu ten suy ra x thuoc UC(84;180)
84=2^2x3x7
180=2^2x3^2x5
=>UCLN(84;180)=2^2x3=12
=>UC(84;180)=U(12)={1;2;3;4;6;12}
Vi x>6 nen suy ra x=12
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a,231 - ( x - 6 ) = 1339 : 13
=> 231 - ( x - 6 ) = 103
=> x = 6 + 231 - 103
=> x = 1 34
b,
( 3x - 21 ) : 4 + 108 = 144
=> ( 3x - 21 ) : 4 = 144 - 108
=> ( 3x - 21 ) :4 = 36
=> 3x - 21 = 36.4
=> 3x - 21 = 144
=> 3x = 144 + 21
=> 3x = 165
=> x = 165 : 3
=> x = 55
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