Tìm x,y thuộc Z ( x+3) .(x2+1 ) =0 3x +4y -xy =15 10/11/2021 Bởi Bella Tìm x,y thuộc Z ( x+3) .(x2+1 ) =0 3x +4y -xy =15
Đáp án: a) $(x+3)(x^2+1)=0$ $x+3=0$ hoặc $x^2+1=0$ Vì $x^2\geq 0 \Rightarrow x^2+1\geq1>0$ $\Rightarrow x=-3$ b) $3x+4y-xy=15$ $(3x-12)+(4y-xy)=3$ $3(x-4)-y(x-4)=3$ $(x-4)(3-y)=3$ $\Rightarrow x-4, 3-y\in Ư(3)=\pm1,\pm3$ Ta có bảng sau: \begin{array}{|c|c|c|c|c|} \hline x-4&-3&-1&1&3\\ \hline 3-y&-1&-3&3&1\\ \hline x&1&3&5&7\\ \hline y&4&6&0&2\\ \hline \end{array} Do $(x,y) \in Z$ nên $(x;y)=\{(1;4),(3;6),(5;0),(7;2)\}$ Bình luận
`( x+3).(x^2+1 ) =0` `vì x^2+1≥1` `⇒x+3=0` `⇒x=-3` `3x +4y -xy =15` `⇒3x+4y-xy=12+3` `⇒3x-12-4y-xy=3` `⇒(3x − 12) + (4y − xy) = 3` `⇒3(x − 4) − y(x − 4) = 3` `⇒(x − 4)(3 − y) = 3` `⇒ x − 4, 3 − y ∈ Ư(3) = ±1, ±3` Ta có bảng sau: `x − 4 −3 −1 1 3``3 − y −1 −3 3 1``x 1 3 5 7``y 4 6 0 2``⇒(x; y) = {(1; 4),(3; 6),(5; 0),(7; 2)}` Bình luận
Đáp án:
a)
$(x+3)(x^2+1)=0$
$x+3=0$ hoặc $x^2+1=0$
Vì $x^2\geq 0 \Rightarrow x^2+1\geq1>0$
$\Rightarrow x=-3$
b)
$3x+4y-xy=15$
$(3x-12)+(4y-xy)=3$
$3(x-4)-y(x-4)=3$
$(x-4)(3-y)=3$
$\Rightarrow x-4, 3-y\in Ư(3)=\pm1,\pm3$
Ta có bảng sau:
\begin{array}{|c|c|c|c|c|} \hline x-4&-3&-1&1&3\\ \hline 3-y&-1&-3&3&1\\ \hline x&1&3&5&7\\ \hline y&4&6&0&2\\ \hline \end{array}
Do $(x,y) \in Z$ nên $(x;y)=\{(1;4),(3;6),(5;0),(7;2)\}$
`( x+3).(x^2+1 ) =0`
`vì x^2+1≥1`
`⇒x+3=0`
`⇒x=-3`
`3x +4y -xy =15`
`⇒3x+4y-xy=12+3`
`⇒3x-12-4y-xy=3`
`⇒(3x − 12) + (4y − xy) = 3`
`⇒3(x − 4) − y(x − 4) = 3`
`⇒(x − 4)(3 − y) = 3`
`⇒ x − 4, 3 − y ∈ Ư(3) = ±1, ±3`
Ta có bảng sau:
`x − 4 −3 −1 1 3`
`3 − y −1 −3 3 1`
`x 1 3 5 7`
`y 4 6 0 2`
`⇒(x; y) = {(1; 4),(3; 6),(5; 0),(7; 2)}`