Cho biểu thức A=1203+a.Với giá trị nào của a trong các giá trị dưới đây thì A chia hết cho 5 ?
Giúp mình lẹ nha
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Nguyễn Tất Phúc ko đăng linh tinh
Tl cho đàng hoàng
Còn ko biết TL thì ko nên TL
HT
xem lại đề nka bạn
sự thật là đề k có chữ nào ghi " a " mà câu hỏi lại có "a "
Giá trị của biểu thức 480 - 120 : 4 chia hết cho 2.
xét hiệu: 4.(9a+b+4c)-(3a+4b+5c)
rùi làm như bình thường ngọc nhé,hà phg đây
a) Cho x2 - x + 5=0 =>x={ \(\frac{1}{2}+\frac{\sqrt{19}}{2}i;\frac{1}{2}-\frac{\sqrt{19}}{2}i\) }
Thay giá trị của x là \(\frac{1}{2}+\frac{\sqrt{19}}{2}i\)hoặc \(\frac{1}{2}-\frac{\sqrt{19}}{2}i\) vừa tìm được vào x4 - x3 + 6x2- x sẽ luôn được kết quả là -5
=>-5 +a=0 => a=5
b) Cho x+2=0 => x=-2
Thay giá trị của x vào biểu thức 2x3 - 3x2 + x sẽ được kết quả là -30
=> -30 + a=0 => a=30
a) Cho 3n +1 =0 => n= \(\frac{-1}{3}\)
Thay n= \(\frac{-1}{3}\)vào biểu thức 3n3 + 10n2 -5 sẽ được kết quả -4
Vậy n = -4
b) Cho n-1=0 => n=1
Thay n=1 vào biểu thức 10n2 + n -10 sẽ được kết quả là 1
Vậy n = 1
để A chia hết cho 5 thì chữ số tận cùng của A phải là 0 hoặc 5
nên hoặc \(\orbr{\begin{cases}a=2\\a=7\end{cases}}\)