TÌM x,y thuộc Z BIẾT
xy + 5x - 2y=13
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Ta có:
S=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)
S=\(1-\frac{1}{n+3}\)
=>S<1
Vậy S<1
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)
1. x thuộc -6, -5, -4, -3, -2
2. x thuộc -2, -1, 0, 1, 2
3. x thuộc -1, 0, 1, 2, 3 ,4, 5, 6
4. x thuộc -5, -4, -3, -2, -1, 0, 2, 3, 4, 5
Ta có
\(\frac{1}{2}< \frac{2}{3},\frac{3}{4}< \frac{4}{5},...,\frac{1599}{1600}< \frac{1600}{1601}\)
Do đó ta có
A=\(\frac{1}{2}\times\frac{3}{4}\times...\times\frac{1599}{1600}< \frac{2}{3}\times\frac{4}{5}\times...\times\frac{1600}{1601}\)
#Châu's ngốc
a)12+x=-14
x=-14-12
x=-26
vậy x=-26
b)60-(5x-5)=45
5x-5=60-45
5x-5=15
5x=15:5
5x=3
x=\(\frac{3}{5}\)
vậy \(x=\frac{3}{5}\)
c)|15+3x|=32.5
|15+3x|=9.5
|15+3x|=45
* 15+3x=45 * 15+3x=-45
3x=45-15 3x=-45-15
3x=30 3x=-60
x=30:3 x=-60:3
x=10 x=-20
vậy x=10 hoặc x=-20
a, 12 + x = -14
=> x = -14 - 12
=> x = -26
b, 60 - ( 5x - 5 ) = 45
=> 5x - 5 = 60 - 45
=> 5x - 5 = 15
=> 5x = 20
=> x = 4
c, | 15 + 3x | = 32. 5
=> | 15 + 3x | = 45
\(\Rightarrow\orbr{\begin{cases}15+3x=45\\15+3x=-45\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=30\\3x=-60\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=10\\x=-20\end{cases}}}\)
Bài giải
Ta có : \(\left(\frac{1}{31}\right)^7=\frac{1}{31^7}\)
\(\left(\frac{1}{15}\right)^9=\frac{1}{15^9}\)
Ta so sánh : \(31^7\text{ và }15^9\)
\(31^7=27512614111\)
\(15^9=38443359375\)
Mà \(31^7< 15^9\text{ nên }\left(\frac{1}{31}\right)^7< \left(\frac{1}{15}\right)^9\)
\(\frac{x-2012}{2}+\frac{x-2008}{3}+\frac{x-2002}{4}+\frac{x-1994}{5}=10\)
\(\Leftrightarrow\frac{x-2012}{2}-1+\frac{x-2008}{3}-2+\frac{x-2002}{4}-3+\frac{x-1994}{5}-4=0\)
\(\Leftrightarrow\frac{x-2014}{2}+\frac{x-2014}{3}+\frac{x-2014}{4}+\frac{x-2014}{5}=0\)
<=> x = 2014(vì 1/2 + 1/3 + 1/4 + 1/5 khác 0)
tìm x E Z biết
a, 0 : x =0
\(\Rightarrow x=\frac{0}{0}\)
\(\Rightarrow x\in\varnothing\)
b, 4 mũ x =64
\(\Rightarrow4^x=4^3\)
\(\Rightarrow x=3\)
c, 2 mũ x =16
\(\Rightarrow2^x=2^4\)
\(\Rightarrow x=4\)
d, 9 mũ x-1=9
\(\Rightarrow x-1=1\)
\(\Rightarrow x=2\)
e,x mũ 4 =16
\(\Rightarrow x^4=2^4\)
\(\Rightarrow x=2\)
g, 2 mũ x : 2 mũ 5 =1
\(\Rightarrow2^{x-5}=1\)
\(\Rightarrow x-5=0\)
\(\Rightarrow x=5\)
giúp mk với mk đang cần
Ta có : xy + 5x - 2y = 13
=> x(y + 5) - 2y = 13
=> x(y + 5) - 2y - 10 = 13 - 10
=> x(y + 5) - 2(y + 5) = 3
=> (x - 2)(y + 5) = 3
Với \(x;y\inℤ\Rightarrow\hept{\begin{cases}x-2\inℤ\\y+5\inℤ\end{cases}}\)
mà 3 = 1.3 = (-1) . (-3)
Lập bảng xét các trường hợp
Vậy các cặp (x ; y) thỏa mãn là : (3 ; -2) ; (5 ; -4) ; (1; - 8) ; (-1;-6)