Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(ab=bc=ca\Rightarrow\frac{ab}{abc}=\frac{bc}{abc}=\frac{ca}{abc}\)
\(\Rightarrow\frac{1}{c}=\frac{1}{a}=\frac{1}{b}\Rightarrow a=b=c\)
\(\Rightarrow M=\left(\frac{a}{b}\right)^{2016}-\left(\frac{c}{a}\right)^{2017}\)
\(=\left(\frac{a}{a}\right)^{2016}-\left(\frac{a}{a}\right)^{2017}\)
\(=1^{2016}-1^{2017}\)
\(=1-1=0\)
Câu hỏi của ✨♔♕ Saiko ♕♔✨ - Toán lớp 6 - Học toán với OnlineMath
Ta có :
\(A=1+5+5^2+...+5^{32}\)
\(A=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{30}+5^{31}+5^{32}\right)\)
\(A=31+5^3\left(1+5+5^2\right)+...+5^{30}\left(1+5+5^2\right)\)
\(A=31+31.5^3+...+31.5^{30}\)
\(A=31\left(1+5^3+...+5^{30}\right)\) chia hết cho 31
Vậy \(A\) chia hết cho 31
\(a)\) Ta có :
\(\frac{a}{b}< \frac{a+c}{b+c}\)
\(\Leftrightarrow\)\(a\left(b+c\right)< b\left(a+c\right)\)
\(\Leftrightarrow\)\(ab+ac< ab+bc\)
\(\Leftrightarrow\)\(ac< bc\)
\(\Leftrightarrow\)\(a< b\)
Mà \(a< b\) \(\Rightarrow\) \(\frac{a}{b}< 1\)
Vậy ...
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
=>a=b=c
\(A=\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{b}{b}\right)\left(1+\dfrac{c}{c}\right)=8\)
\(A=3+3^2+...+3^{50}\)
\(\Rightarrow3A=3^2+3^3+...+3^{50}+3^{51}\)
\(\Rightarrow3A-A=3^{51}-3\)
\(\Rightarrow2A=3^{51}-3\)
\(\Rightarrow A=\frac{3^{51}-3}{2}\)
\(B=2-2^2+2^3-2^4+...+2^{2019}-2^{2020}\)
\(2B=2^2-2^3+2^4-2^5+...+2^{2020}-2^{2021}\)
\(B+2B=2-2^{2021}\)
\(3B=2-2^{2021}\)
\(B=\frac{2-2^{2021}}{3}\)
\(C=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2008.2009}\)
\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2008}-\frac{1}{2009}\)
\(C=1-\frac{1}{2009}\)
\(C=\frac{2008}{2009}\)
\(D=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(D=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(D=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(D=\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(D=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)
1,
a, Để \(\frac{8}{x+2}\) nhận giá trị là số tự nhiên \(\Rightarrow\)\(8⋮x+2\Rightarrow x+2\in\text{Ư}\left(8\right)=\left\{1;2;4;8\right\}\)
\(\Rightarrow x\in\left\{-1;0;2;6\right\}\)
Vì \(x\in N\Rightarrow x\in\text{ }\left\{0;2;6\right\}\)
Vậy \(x\in\left\{0;2;6\right\}\)
b, Để \(\frac{x+3}{x+1}\) nhận giá trị là số tự nhiên\(\Rightarrow\left\{{}\begin{matrix}x+3⋮x+1\\x+1⋮x+1\end{matrix}\right.\Rightarrow x+3-x+1⋮x+1\Rightarrow2⋮x+1\)
\(\Rightarrow x+1\in\text{Ư}\left(2\right)=\left\{1;2\right\}\)\(\Rightarrow x\in\left\{0;1\right\}\)
Vậy \(x\in\left\{0;1\right\}\)
- Bài 2:
b) S = 1 + 2 + 22 +.... + 211
= (1+23) + (2 + 24) +..... + (28+ 211)
= (1+23) + 2(1+23)+....+28(1+23)
= 9 + 2.9 + .... + 28.9
= 9.(1+2+...+28) ⋮ 9
Vậy S ⋮ 9