Kết quả của phép tính \(\left(\frac{25}{2}\right)^{2009}.\left(\frac{5}{2}\right)^{2010}\)là:
A.\(\left(\frac{25}{4}\right)^{2008}\) B.\(\left(\frac{5}{2}\right)^{2008}\) C.\(\left(\frac{5}{2}\right)^{3014}\) D.\(\left(\frac{25}{4}\right)^{3014}\)
Thu gọn
\(A=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(2009^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(2010^4+\frac{1}{4}\right)}\)
\(B=\frac{\left(a+2008\right)!+\left(a+2009\right)!}{\left(a+2008\right)!-\left(a+2009!\right)}\)
Kết quả của \(\left(\frac{2}{5}\right)^{2008}:\left(\frac{4}{25}\right)^{1004}\)
Phép tính này có kết quả là:0
Vì (2/5)2008 = 0
Vì (4/25)1004 = 0
Và 0 : 0 = 0
Chúc bạn may mắn
Phép tính này có kết quả là 0 vì 2 : 5 ^ 2008 = 0
Và 0 chia cho bao nhiêu cũng là 0
Tk mình nhé
Cảm ơn bạn nhiều
tính
a , A = \(\left[\frac{4}{11}.\left(\frac{1}{25}\right)^0+\frac{7}{22}.2\right]^{2010}-\left(\frac{1}{2^2}:\frac{8^2}{4^4}\right)^{2009}\)
b , B =\(\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{17}}+\left(1,2.0,5\right):\frac{4}{5}\)
\(a,A=\left[\frac{4}{11}.\left(\frac{1}{25}\right)^0+\frac{7}{22}.2\right]^{2010}-\left(\frac{1}{2^2}:\frac{8^2}{4^4}\right)^{2009}\)
\(A=\left(\frac{4}{11}.1+\frac{7}{11}\right)^{2010}-\left(\frac{1}{2^2}.2^2\right)^{2009}\)
\(A=1-1=0\)
\(b,B=\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{17}}+\left(1,2.0,5\right):\frac{4}{5}\)
\(B=\frac{0,8:1}{\frac{3}{5}}+\frac{\left(1\right):\frac{4}{7}}{\left(\frac{59}{9}-\frac{13}{4}\right).36}\)
\(B=0,8.\frac{5}{3}+\frac{\frac{7}{4}}{\frac{119}{36}.36}\)
\(B=\frac{4}{3}+\frac{7}{4}.\frac{1}{119}\)
\(B=\frac{4}{3}+\frac{1}{68}=\frac{275}{204}\)
Tìm x
a/\(\frac{x+7}{2003}+\frac{x+4}{2006}=\frac{x-1}{2011}+\frac{x-5}{2015}\)
b/\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
c/\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
a) \(\Leftrightarrow\frac{x+7}{2003}+1+\frac{x+4}{2006}+1-\frac{x-1}{2011}-1-\frac{x-5}{2015}-1=0\)
\(\Leftrightarrow\frac{x+2010}{2003}+\frac{x+2010}{2006}-\frac{x+2010}{2011}-\frac{x+2010}{2015}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2003}+\frac{1}{2006}-\frac{1}{2011}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x+2010=0\) ( vì 1/2003 + 1/2006 -- 1/2011 -- 1/2015 \(\ne\)0)
\(\Leftrightarrow x=-2010\)
câu b làm tương tự (có gì không hiểu hỏi mk nha) >v<
Kết quả của phép tính \(\left(-2\right)\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)...\left(-1\frac{1}{2009}\right)\left(-1\frac{1}{2010}\right)\)là .......
Kết quả của phép tính:
\(\left(-2\right)\times\left(-1\frac{1}{2}\right)\times\left(-1\frac{1}{3}\right)..........\times\left(-1\frac{1}{2009}\right)\times\left(-1\frac{1}{2010}\right)\)
tính
\(\left(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)\left(1+\frac{1}{2}+...+\frac{1}{2008}\right)\)
\(-\left(1+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)\)
câu hỏi hay......nhưng tui xin nhường cho các bn khác
Hãy tích đúng cho tui nha
THANKS
Kết quả của phép tính \(\left(-2\right)\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)...\left(-1\frac{1}{2009}\right)\left(-1\frac{1}{2010}\right)\) là ?
Kết quả của phép tính \(\left(-2\right)\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)...\left(-1\frac{1}{2009}\right)\left(-1\frac{1}{2010}\right)\) là ?
\(\left(-2\right)\left(-1\frac{1}{2}\right)...........\left(-1\frac{1}{2010}\right)=\frac{\left[\left(-2\right)\left(-3\right).....\left(-2010\right)\right].\left(-2011\right)}{\left(2.3.4.............2010\right)}=\frac{\left(-1\right)\left(-2011\right)}{1}=2011\)
\(\left(-2\right).\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right)....\left(-1\frac{1}{2009}\right).\left(-1\frac{1}{2010}\right)=\left(-2\right).\left(-\frac{3}{2}\right).\left(-\frac{4}{3}\right)....\left(-\frac{2010}{2009}\right).\left(-\frac{2011}{2010}\right)=\frac{\left(-2\right).\left(-3\right).\left(-4\right)....\left(-2010\right).\left(-2011\right)}{2.3.4....2009.2010}\)=2011