Tính giá trị của biểu thức :
P = \(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)......\left(1-\frac{1}{1+2+3+...+2008}\right)\)
Tính giá trị của biểu thức:
\(N=\frac{\left(2^4+\frac{1}{4}\right).\left(4^4+\frac{1}{4}\right).\left(6^4+\frac{1}{4}\right)...\left(2008^4+\frac{1}{4}\right)}{\left(1^4+\frac{1}{4}\right).\left(3^4+\frac{1}{4}\right).\left(5^4+\frac{1}{4}\right)...\left(2007^4+\frac{1}{4}\right)}\)
Với mọi n thuộc N* ta có :
\(n^4+\frac{1}{4}=\left(n^4+2.\frac{1}{2}.n^2+\frac{1}{4}\right)-n^2=\left(n^2+\frac{1}{2}\right)^2-n^2\)
\(=\left(n^2+n+\frac{1}{2}\right)\left(n^2-n+\frac{1}{2}\right)\)
\(\Rightarrow N=\frac{\left(2^2+2+\frac{1}{2}\right)\left(2^2-2+\frac{1}{2}\right)...\left(2008^2+2008+\frac{1}{2}\right)\left(2008^2-2008+\frac{1}{2}\right)}{\left(1^2+1+\frac{1}{2}\right)\left(1^2-1+\frac{1}{2}\right)...\left(2007^2+2007+\frac{1}{2}\right)\left(2007^2-2007+\frac{1}{2}\right)}\)
\(=\frac{\left(2.3+\frac{1}{2}\right)\left(1.2+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right)...\left(2008.2009+\frac{1}{2}\right)}{\frac{1}{2}\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)...\left(2007.2008+\frac{1}{2}\right)}\)
\(=\frac{2008.2009+\frac{1}{2}}{\frac{1}{2}}=8068145\)
1) Tính A=\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right)....\left(\frac{1}{2008}-1\right).\left(\frac{1}{2009}-1\right)\)
2)Xác định giá trị của x để A>0 biết A=\(\frac{x-7}{2}\)
3) Xác định giá trị của x để biểu thức sau nhận giá trị âm A=\(\frac{x+3}{x-5}\)
1)\(A=\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right)....\left(\frac{1}{2008}-1\right).\left(\frac{1}{2009}-1\right)=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)...\left(-\frac{2008}{2009}\right)=\frac{1.2.3...2008}{2.3.4....2009}=\frac{1}{2009}\)
2)\(A=\frac{x-7}{2}\)
Do 2>0 =>A>0 <=>x-7>0<=>x>7
Vậy x>7 thì A>0
3)\(A=\frac{x+3}{x-5}\)
Do x+3>x-5 =>A<0<=>x+3>0 và x-5<0
<=>-3<x<5
Vậy -3<x<5 thì A<0
Tính giá trị của biểu thức: \(\left(-2\right)\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2010}\right)\)
tính giá trị của biểu thức:
\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2017^2}\right)\)
\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2017^2}\right)\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{2016.2018}{2017^2}\)
\(=\frac{2.3^2.4^2.5^2...2016^2.2017.2018}{2^2.3^2.4^2.5^2...2017^2}\)
\(=\frac{2018}{2.2017}=\frac{1009}{2017}\)
mk mới học lớp 6 thôi
tk mk nha mk đang âm điểm nè hu hu
tính giá trị của biểu thức:
A=\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2017^2}\right)\)
Tính giá trị biểu thức: A=\(\frac{\left(1+17\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right)....\left(1+\frac{17}{19}\right)}{\left(1+19\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right)...\left(1+\frac{19}{17}\right).}\)
Tính giá trị biểu thức
\(A=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)\left(1-\frac{3}{7}\right)...\left(1-1\frac{2}{7}\right)\left(1-1\frac{3}{7}\right)\)
\(\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right)\left(1-1\frac{1}{7}\right)...\left(1-1\frac{3}{7}\right)\)
\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-1\frac{1}{7}\right)...\left(1-1\frac{3}{7}\right)\left(1-1\right)\)
\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-1\frac{3}{7}\right).0\)
\(=0\)
Trong dãy nhất định có \(\left[1-\frac{7}{7}\right]=0\)nên tích dãy trên là 0
tính giá trị của biểu thức M=\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{16}.\left(1+2+3+...+16\right)\)
Tính giá trị của biểu thức:
\(A=\frac{1}{2}\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{2017\cdot2019}\right)\)
A=\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{2017.2019}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1.3+1}{1.3}\right).\left(\frac{2.4+1}{2.4}\right).\left(\frac{3.5+1}{3.5}\right)..........\left(\frac{2017.2019+1}{2017.2019}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.............\frac{4072324}{2017.2019}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...................\frac{2018^2}{2017.2019}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{\left(2.3.4..........2018\right).\left(2.3.4............2018\right)}{\left(1.2.3............2017\right).\left(3.4.5..........2019\right)}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2018.2}{1.2019}\right)=\frac{2018.2}{2.2019}=\frac{2018}{2019}\)
Vậy \(A=\frac{2018}{2019}\)
Chúc bn học tốt
\(A:\frac{1}{2}=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{2017.2019+1}{2017.2019}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}......\frac{2018^2}{2017.2019}\)
\(=\frac{2.2.3.3.4.4.....2018.2018}{1.3.2.4.3.5....2017.2019}\)
\(=\frac{2.3.4.....2018}{1.2.3.4.....2017}.\frac{2.3.4....2018}{3.4.5.....2019}\)
\(=2018.\frac{2}{2019}\)
\(=\frac{4036}{2019}\)
\(\Rightarrow A=\frac{4036}{2019}.\frac{1}{2}\)
\(A=\frac{2018}{2019}\)