Đáp án:
(x+3)(y+1)−1=2(x+3)(y+1)-1=2
⇔(x+3)(y+1)=3⇔(x+3)(y+1)=3
Vì x;y∈Z⇒(x+3);(y+1)∈Zx;y∈Z⇒(x+3);(y+1)∈Z
⇔(x+3);(y+1)∈Ư(3)={±1;±3}⇔(x+3);(y+1)∈Ư(3)={±1;±3}
+)x+3∈Ư(3)={±1;±3}+)x+3∈Ư(3)={±1;±3}
⇒x+3=1⇔x=−2⇒x+3=1⇔x=-2
x+3=−1⇔x=−4x+3=-1⇔x=-4
x+3=3⇔x=0x+3=3⇔x=0
x+3=−3⇔x=−6x+3=-3⇔x=-6
+)y+1∈Ư(3)={±1;±3}+)y+1∈Ư(3)={±1;±3}
⇒ y+1=1⇔y=0⇒ y+1=1⇔y=0
y+1=−1⇔y=−2y+1=-1⇔y=-2
y+1=3⇔y=2y+1=3⇔y=2
y+1=−3⇔y=−4y+1=-3⇔y=-4
Vậy (x;y)∈(−2;0);(−4;2);(0;2);(−6;−4)(x;y)∈(-2;0);(-4;2);(0;2);(-6;-4)
2/
2|x−5|−3=52|x-5|-3=5
⇔2∣x−5∣=8⇔2∣x−5∣=8
⇔∣x−5∣=4⇔∣x−5∣=4
⇔⇔[x−5=4x−5=−4[x−5=4x−5=−4
⇔⇔[x=9x=1[x=9x=1
Vậy x∈{1;9}
Giải thích các bước giải:
