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6xy-4x+3y=-53
=>2x(3y-2)+3y-2=-55
=>(3y-2)(2x+1)=-55
=>\(\left(2x+1\right)\left(3y-2\right)=1\cdot\left(-55\right)=\left(-1\right)\cdot55=\left(-55\right)\cdot1=55\cdot\left(-1\right)=5\cdot\left(-11\right)=\left(-11\right)\cdot5=\left(-5\right)\cdot11=11\cdot\left(-5\right)\)
=>\(\left(2x+1;3y-2\right)\in\){(1;-55);(-1;55);(-55;1);(55;-1);(5;-11);(-11;5);(-5;11);(11;-5)}
=>\(\left(x;y\right)\in\left\{\left(0;-\dfrac{53}{3}\right);\left(-1;19\right);\left(-28;1\right);\left(27;\dfrac{1}{3}\right);\left(2;-3\right);\left(-6;\dfrac{7}{3}\right);\left(-3;\dfrac{13}{3}\right);\left(5;-1\right)\right\}\)
mà x,y nguyên
nên \(\left(x;y\right)\in\left\{\left(-1;19\right);\left(-28;1\right);\left(2;-3\right);\left(5;-1\right)\right\}\)
c)\(\Leftrightarrow\)(x+1)+2 chia hết x+1
\(\Rightarrow\)2 chia hết x+1
\(\Rightarrow\)x+1 ∈ {1,-1,2,-2}
\(\Rightarrow\)x ∈ {0,-2,1,-3}
c) \(x+3⋮x+1\)
\(\Rightarrow x+1+2⋮x+1\)
\(\Rightarrow2⋮x+1\) ( vì \(x+1⋮x+1\) )
\(\Rightarrow x+1\in\text{Ư}_{\left(2\right)}\)
\(\text{Ư}_{\left(2\right)}=\text{ }\left\{1;-1;2;-2\right\}\)
\(x+1\) | \(1\) | \(-1\) | \(2\) | \(-2\) |
\(x\) | \(0\) | \(-2\) | \(1\) | \(-3\) |
vậy................
\(\frac{5}{n+1}=\frac{n+1}{5}\)
\(\Leftrightarrow\left(n+1\right)^2=5^2\)
\(\Leftrightarrow\sqrt{\left(n+1\right)^2}=\sqrt{5^2}\)
\(\Leftrightarrow n+1=5\)
\(\Leftrightarrow n=5-1\)
\(\Leftrightarrow n=4\)
\(\frac{2x+1}{5}=\frac{x+2}{8}\)
\(\Leftrightarrow8\left(2x+1\right)=5\left(x+2\right)\)
\(\Leftrightarrow16x+8=5x+10\)
\(\Leftrightarrow11x=2\)
\(\Leftrightarrow x=\frac{2}{11}\)
Ha ha ha cu Long ngu qua
Minh cug ngu nhu cu Lomg
Gio van
a) 3y +xy+2x+6=0
3.(y + 2) + x.(y + 2) = 0
(3 + x).(y + 2) = 0
\(\Rightarrow\hept{\begin{cases}3+x=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-3\\y=-2\end{cases}}}\)
Vậy...