chứng tỏ \(\frac{1111...11}{n}\)-\(\frac{222...22}{n}\)=\(\frac{\left(333...33\right)^2}{n}\)
Những câu hỏi liên quan
Chứng minh: 1111...11 (2n chữ số 1) - 222..222 (n chữ số 2)=333...332 (n cs 3)
so sánh
\(\left(\frac{1}{222}\right)^{333};\left(\frac{1}{333}\right)^{222}\)
-Vì (1/222)^333=(1/222)^3.111=(3/666)^111
(1/333)^222=(1/333)^2.111=(2/666)^111
-Vì 111=111 và 3/666>2/666
=))(1/222)^333>(1/333)^222
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Ta có:
222333 = 111333.2333 = 111222.111111.(23)111 = 111222.111111.8111 = 111222.888111
333222 = 111222.3222 = 111222.(32)111 = 111222.9111
Vì 111222.888111 > 111222.9111
=> 222333 > 333222
=> \(\frac{1}{222^{333}}< \frac{1}{333^{222}}\)
hay \(\left(\frac{1}{222}\right)^{333}< \left(\frac{1}{333}\right)^{222}\)
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Chứng tỏ \(A=\frac{1}{n\times\left(n+1\right)\times\left(n+2\right)}=\frac{\frac{1}{ }}{2}\times\left(\frac{1}{n\times\left(n+1\right)}-\frac{1}{\left(n+1\right)\times\left(n+2\right)}\right)\)với n\(\in\)N*
với n thuộc N* hãy chứng tỏ rằng :
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}\left[\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\right]\)
\(=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)
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chứng tỏ rằng với mọi n thuộc N* ta có :
\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}=\frac{n}{2\left(3n+2\right)}\)
\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\frac{3n}{2.\left(3n+2\right)}\)
\(=\frac{n}{2\left(3n+2\right)}\)
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Bài 1:Tìm x, biết
\(\frac{1}{2.3}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}=\frac{11}{48}\left(x\in N,x\ge2\right)\)
Bài 2:Chứng tỏ rằng với mọi \(n\in Nsao\),ta có
\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right).\left(3n+2\right)}\)
Bài 1:
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}\)\(=\frac{11}{48}\)
\(\frac{1}{4}.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(x-1\right).x}\right)\)\(=\frac{11}{48}\)
\(\frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x-1}-\frac{1}{x}\right)\)\(=\frac{11}{48}\)
\(\frac{1}{4.}.\left(1-\frac{1}{x}\right)=\frac{11}{48}\)
\(1-\frac{1}{x}=\frac{11}{48}:\frac{1}{4}\)
\(1-\frac{1}{x}=\frac{11}{12}\)
\(\frac{1}{x}=1-\frac{11}{12}\)
\(\frac{1}{x}=\frac{1}{12}\)
Vậy x= 12
Bài 2 :
Xét vế trái ta có :
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{\left(3n-1\right).\left(3n+2\right)}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\frac{1}{2\left(3n+2\right)}=\frac{n}{2\left(3n+2\right)}\)
VẾ TRÁI ĐÚNG BẰNG VẾ PHẢI .ĐẲNG THỨC ĐÃ CHỨNG TỎ LÀ ĐÚNG
cHÚC BẠN HỌC TỐT ( -_- )
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a) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}\frac{125}{376}\left(x€N\cdot\right)\)
b) \(\frac{3}{4}x-14\frac{2}{3}:\left(\frac{11}{15}+\frac{1111}{3535}+\frac{111111}{636363}\right)=12\)
1) Với n ϵ N* hãy chứng tỏ :
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}\) = \(\frac{1}{2}.\) ( \(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\frac{1}{2}\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{1}{2}\left(\frac{\left(n+1\right)\left(n+1\right)}{n\left(n+1\right)\left(n+1\right)\left(n+2\right)}-\frac{n\left(n+1\right)}{n\left(n+1\right)\left(n+1\right)\left(n+2\right)}\right)\)
\(\frac{1}{2}\left(\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+2\right)\left(n+1\right)\left(n+2\right)}-\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+2\right)\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
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\(\frac{1}{2}\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{1}{2}\left(\frac{\left(n+1\right)\left(n+1\right)}{n\left(n+1\right)\left(n+1\right)\left(n+2\right)}-\frac{n\left(n+1\right)}{n\left(n+1\right)\left(n+1\right)\left(n+2\right)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+2\right)\left(n+1\right)\left(n+2\right)}-\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+2\right)\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
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7 tháng 9 2015 lúc 18:05
32+2=11
332+22=1111
3332+222=111111
=>333333333332+22222222222=?????
333333333332 + 22222222222 = 1111111111111111111111(22 chữ số 1)
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