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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 ( -_- )
bài 1 A là số nguyên <=> 3 chia hết cho (x-1) <=> (x-1) thuộc Ư(3) = { 1;-1;3;-3}
<=> x thuộc {2;0;4;-2}
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}\left[\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\right]\)
\(=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)
\(\frac{1}{2}\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{1}{2}\left(\frac{\left(n+1\right)\left(n+1\right)}{n\left(n+1\right)\left(n+1\right)\left(n+2\right)}-\frac{n\left(n+1\right)}{n\left(n+1\right)\left(n+1\right)\left(n+2\right)}\right)\)
\(\frac{1}{2}\left(\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+2\right)\left(n+1\right)\left(n+2\right)}-\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+2\right)\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(\frac{1}{2}\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{1}{2}\left(\frac{\left(n+1\right)\left(n+1\right)}{n\left(n+1\right)\left(n+1\right)\left(n+2\right)}-\frac{n\left(n+1\right)}{n\left(n+1\right)\left(n+1\right)\left(n+2\right)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+2\right)\left(n+1\right)\left(n+2\right)}-\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+2\right)\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
Bài 1 :
\(P=\frac{9^4.27^3.24}{81^2.3^7.5^2}=\frac{9^4.3^9.3.2^3}{9^4.3^7.3^2.2^3}=3\)
Bài 2 :
\(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\left(đpcm\right)\)
Bài 2 : Ta có : \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\left(đpcm\right)\)