Tìm A, B, C:
A= 1992- 199
B= 2004+ 100
C= 3302+ 330
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b: Đặt \(\dfrac{a}{5}=\dfrac{b}{7}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5k\\b=7k\end{matrix}\right.\)
Ta có: ab=140
nên \(35k^2=140\)
\(\Leftrightarrow k^2=4\)
Trường hợp 1: k=2
\(\Leftrightarrow\left\{{}\begin{matrix}a=5k=10\\b=7k=14\end{matrix}\right.\)
Trường hợp 2: k=-2
\(\Leftrightarrow\left\{{}\begin{matrix}a=5k=-10\\b=7k=-14\end{matrix}\right.\)
\(a,856:214+1284:214\\ =\left(856+1284\right):214\\ =2140:214=10\\ b,1778:254+3302:254\\ =\left(1778+3302\right):254\\ =5080:254\\ =20\)
Bài 2:
a) \(856
:
214+1284
:
214=\left(856+1284\right)
:
214=2140
:
214=10\)
b) \(1778
:
254+3302
:
254=\left(1778+3302\right)
:
254=5080
:
254=20\)
Mình nghĩ \(A=\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{101\cdot400}\)
\(299A=\frac{299}{1\cdot300}+\frac{299}{2\cdot301}+\frac{299}{3\cdot302}+...+\frac{299}{101\cdot400}\)
\(299A=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\)
\(299A=\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)=C\)
\(A=\frac{C}{299}\)
Lại có;
\(B=\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+....+\frac{1}{299\cdot400}\)
\(101B=\frac{101}{1\cdot102}+\frac{101}{2\cdot103}+...+\frac{101}{299\cdot400}\)
\(101B=1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\)
\(101B=\left(1+\frac{1}{2}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+\frac{1}{103}+...+\frac{1}{400}\right)=C\)
\(B=\frac{C}{101}\)
Vậy \(\frac{A}{B}=\frac{C}{299}:\frac{C}{101}=\frac{101}{299}\)
a:b=b:c=c:a
hay \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\)
đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=k\)
\(\Rightarrow\)a = bk ; b = ck ; c = ak
\(\Rightarrow\)abc = abck3
\(\Rightarrow\)k3 = 1
\(\Rightarrow\)k = 1
Từ đó suy ra : a = b = c
Ta co\(\frac{a}{b}\)=\(\frac{b}{c}\)=\(\frac{c}{a}\)
Ap dung tinh chat day cac ti so bang nhau ta co
\(\frac{a}{b}\)=\(\frac{b}{c}\)=\(\frac{c}{a}\)=\(\frac{a+b+c}{b+c+a}\)=1
\(\Rightarrow\)\(\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\)\(\Rightarrow\)a=b=c(dpcm)
a) 2004100 + 200499
= 200499 + 1 + 200499
= 200499.2004 + 200499
= 200499(2004 + 1)
= 200499.2005
Vì 2005 ⋮ 2005 nên 200499.2005 ⋮ 2005
Hay 2004100 + 200499 ⋮ 2005
⇒ đccm
b) 31994 + 31993 - 31992
= 31992 + 2 + 31992 + 1 - 31992
= 31992.32 + 31992.3 - 31992
= 31992(32 + 3 - 1)
= 31992.11
Vì 11 ⋮ 11 nên 31992.11 ⋮ 11
Hay 31994 + 31993 - 31992 ⋮ 11
⇒ đccm
d) 12566 - 5197 + 2598
= \(\left(5^3\right)^{66}-5^{197}+\left(5^2\right)^{98}\)
= 5198 - 5197 + 5196
= 5195 + 3 - 5195 + 2 + 5195 + 1
= 5195.53 - 5195.52 + 5195.5
= 5195(53 - 52 + 5)
= 5195.105
Vì 105 ⋮ 105 nên 5195.105 ⋮ 105
Hay 12566 - 5197 + 2598 ⋮ 105
⇒ đccm
A= 39402
B= 1600000100
C= 109230
A=199
B=1599999900
C=330