nghiệm nguyên

S

st.phamvucuong@gmail.com

1. 156 chia hết cho 3 và 2 nên x = 3k ( k thuôc Z)
y = 2t (t tuộc z )
t + k=26
x= (26-t).3
y= 2t
vậy pt có vô số cặp đôi x,y thoả mãn
 
S

st.phamvucuong@gmail.com

2 $ (x-y)(x^2+y^2+xy)=91 $

th1 $$ \left\{\begin{matrix} x-y=1\\x^2+y^2+xy=91 \end{matrix}\right. $$
th2 $$ \left\{\begin{matrix} x-y=-1\\x^2+y^2+xy=-91 \end{matrix}\right. $$
th3 $$ \left\{\begin{matrix} x-y=91\\x^2+y^2+xy=1 \end{matrix}\right. $$
th4 $$ \left\{\begin{matrix} x-y=91\\x^2+y^2+xy=-1 \end{matrix}\right. $$
 
S

st.phamvucuong@gmail.com

3
(x-5)(x-y-1)=3

th1 $$ \left\{\begin{matrix} x-5=1\\x-y-1=3 \end{matrix}\right. $$
th2 $$ \left\{\begin{matrix} x-5=-1\\x-y-1=-3 \end{matrix}\right. $$
th3 $$ \left\{\begin{matrix} x-5=3\\x-y-1=1 \end{matrix}\right. $$
th4 $$ \left\{\begin{matrix} x-5=-3\\x-y-1=-1 \end{matrix}\right. $$
 
S

st.phamvucuong@gmail.com

4
(2x-y)(x+2y)=7
th1 $$ \left\{\begin{matrix} 2x-y=1\\x+2y=7 \end{matrix}\right. $$
th1 $$ \left\{\begin{matrix} 2x-y=-1\\x+2y=-7 \end{matrix}\right. $$
th1 $$ \left\{\begin{matrix} 2x-y=7\\x+2y=1 \end{matrix}\right. $$
th1 $$ \left\{\begin{matrix} 2x-y=-1\\x+2y=1 \end{matrix}\right. $$
 
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