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x.(y - 3 ) = - 12
x.y - x.3 = -12
x.3 - x.y = 12
Suy ra : y < 3 . Tức y là các số : 0,1,2
x sẽ là :
x.( 0 - 3 ) = -12
x.(-3) = -12
x = 4
x.( 1 - 3 ) = -12
x.(-2) = -12
x = 6
x.(2-3) = -12
x.(-1) = -12
x = 12
Vậy y = 0 ; 1 ; 2
x = 12 ; 6 ; 4
\(x\left(y-3\right)=-12\)
\(=xy-3x=-12\)
\(=xy-3x-\left(-12\right)=0\)
\(=xy-3x+12=0\)
=> x = 0
=> y - 3 = 0
=> y = -3
=(-1+2)-(3+4)-(5+6)-........-(2017+2018)
=1-7-11-........-4035
=-1009
đk : x khác -3
\(\left(x+3\right)^2=36\Leftrightarrow\left[{}\begin{matrix}x+3=6\\x+3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-9\end{matrix}\right.\left(tm\right)\)
\(-\dfrac{4}{5}+2x=\dfrac{1}{3}-\dfrac{2}{3}x\\ \Leftrightarrow-\dfrac{4}{5}+2x-\dfrac{1}{3}+\dfrac{2}{3}x=0\\ \Leftrightarrow-\dfrac{17}{15}+\dfrac{8}{3}x=0\\ \Leftrightarrow\dfrac{8}{3}x=\dfrac{17}{15}\\ \Leftrightarrow x=\dfrac{17}{40}\)
=>8/3x=1/3+4/5=5/15+12/15=17/15
=>x=17/15:8/3=17/15x3/8=51/120=17/40
3²ˣ⁺¹ - 20 = 7
3²ˣ⁺¹ = 7 + 20
3²ˣ⁺¹ = 27
3²ˣ⁺¹ = 3³
2x + 1 = 3
2x = 3 - 1
2x = 2
x = 2 : 2
x = 1
Ta có : (2x - 1)3 = 125
=> (2x - 1)3 = 53
=> 2x - 1 = 5
=> 2x = 5 + 1
=> 2x = 6
=> x = 6 : 2
=> x = 3
( 2x - 1 )3 = 125
ta có :
(2x - 1)3 = 53
=> 2x - 1 = 5
2x = 5 + 1
2x = 6
x = 6 : 2
x = 3
VẬY x = 3
\(2x+xy+y=5\)
\(x\left(2+y\right)+\left(y+2\right)-2=5\)
\(\left(2+y\right).\left(x+1\right)=5+2\)
\(\left(2+y\right).\left(x+1\right)=7\)
\(=>\left[{}\begin{matrix}2+y=-7\\x+1=-1\end{matrix}\right.=>\left[{}\begin{matrix}y=-9\\x=-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}2+y=-1\\x+1=-7\end{matrix}\right.=>\left[{}\begin{matrix}y=-3\\x=-8\end{matrix}\right.\)
\(\left[{}\begin{matrix}2+y=1\\x+1=7\end{matrix}\right.=>\left[{}\begin{matrix}y=-1\\x=6\end{matrix}\right.\)
\(\left[{}\begin{matrix}2+y=7\\x+1=1\end{matrix}\right.=>\left[{}\begin{matrix}y=5\\x=0\end{matrix}\right.\)
Vậy ta có 4 cặp x,y: \(x=-2;y=-9\)
\(x=-8;y=-3\)
\(x=6;y=-1\)
\(x=0;y=5\)
\(A=\left(1+3+3^2\right)+...+\left(3^{99}+3^{100}+3^{101}\right)\\ A=\left(1+3+3^2\right)+...+3^{99}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(1+...+3^{99}\right)=13\left(1+...+3^{99}\right)⋮13\)