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HP
29 tháng 7 2016
bài 1) Đặt \(B=\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\)
Ta có: \(A=B.\left(\frac{p}{m-n}+\frac{m}{n-p}+\frac{n}{p-m}\right)=B.\frac{p}{m-n}+B.\frac{m}{n-p}+B.\frac{n}{p-m}\)
\(B.\frac{p}{m-n}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{p}{m-n}=\frac{m-n}{p}.\frac{p}{m-n}+\frac{n-p}{m}.\frac{p}{m-n}+\frac{p-m}{n}.\frac{p}{m-n}\)
\(=1+\frac{n-p}{m}.\frac{p}{m-n}+\frac{p-m}{n}.\frac{p}{m-n}=1+\frac{p}{m-n}.\left(\frac{n-p}{m}+\frac{p-m}{n}\right)\)
\(=1+\frac{p}{m-n}.\left[\frac{\left(n-p\right).n}{mn}+\frac{\left(p-m\right).m}{mn}\right]=1+\frac{p}{m-n}.\frac{n^2-np+pm-m^2}{mn}\)
\(=1+\frac{p}{m-n}.\frac{\left(m-n\right).\left(p-m-n\right)}{mn}=1+\frac{p.\left(m-n\right).\left(p-m-n\right)}{\left(m-n\right).mn}=1+\frac{p.\left(p-m-n\right)}{mn}\)
\(=1+\frac{p^2-pm-pn}{mn}=1+\frac{p^2-p.\left(m+n\right)}{mn}\)
Vì m+n+p=0=>m+n=-p
\(=>B.\frac{p}{m-n}=1+\frac{p^2-p.\left(-p\right)}{mn}=1+\frac{2p^2}{mn}=1+\frac{2p^3}{mnp}\left(1\right)\)
\(B.\frac{m}{n-p}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{m}{n-p}=\frac{m-n}{p}.\frac{m}{n-p}+\frac{n-p}{m}.\frac{m}{n-p}+\frac{p-m}{n}.\frac{m}{n-p}\)
\(=1+\frac{m-n}{p}.\frac{m}{n-p}+\frac{p-m}{n}.\frac{m}{n-p}=1+\frac{m}{n-p}.\left(\frac{m-n}{p}+\frac{p-m}{n}\right)\)
\(=1+\frac{m}{n-p}.\left[\frac{\left(m-n\right).n}{np}+\frac{\left(p-m\right).p}{np}\right]=1+\frac{m}{n-p}.\frac{mn-n^2+p^2-mp}{np}\)
\(=1+\frac{m}{n-p}.\frac{\left(n-p\right).\left(m-n-p\right)}{np}=1+\frac{m.\left(n-p\right).\left(m-n-p\right)}{\left(n-p\right).np}=1+\frac{m.\left(m-n-p\right)}{np}\)
\(=1+\frac{m^2-mn-mp}{np}=1+\frac{m^2-m\left(n+p\right)}{np}=1+\frac{m^2-m.\left(-m\right)}{np}=1+\frac{2m^2}{np}=1+\frac{2m^3}{mnp}\left(2\right)\) (vì m+n+p=0=>n+p=-m)
\(B.\frac{n}{p-m}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{n}{p-m}=\frac{m-n}{p}.\frac{n}{p-m}+\frac{n-p}{m}.\frac{n}{p-m}+\frac{p-m}{n}.\frac{n}{p-m}\)
\(=1+\frac{m-n}{p}.\frac{n}{p-m}+\frac{n-p}{m}.\frac{n}{p-m}=1+\frac{n}{p-m}.\left(\frac{m-n}{p}+\frac{n-p}{m}\right)\)
\(=1+\frac{n}{p-m}.\left[\frac{\left(m-n\right).m}{pm}+\frac{\left(n-p\right).p}{pm}\right]=1+\frac{n}{p-m}.\frac{m^2-mn+np-p^2}{pm}\)
\(=1+\frac{n}{p-m}.\frac{\left(p-m\right).\left(n-p-m\right)}{pm}=1+\frac{n.\left(p-m\right).\left(n-p-m\right)}{\left(p-m\right).pm}=1+\frac{n.\left(n-p-m\right)}{pm}\)
\(=1+\frac{n^2-np-mn}{pm}=1+\frac{n^2-n\left(p+m\right)}{pm}=1+\frac{n^2-n.\left(-n\right)}{pm}=1+\frac{2n^2}{pm}=1+\frac{2n^3}{mnp}\left(3\right)\) (vì m+n+p=0=>p+m=-n)
Từ (1),(2),(3) suy ra :
\(A=B.\frac{p}{m-n}+B.\frac{m}{n-p}+B.\frac{n}{p-m}=\left(1+\frac{2p^3}{mnp}\right)+\left(1+\frac{2m^3}{mnp}\right)+\left(1+\frac{2n^3}{mnp}\right)\)
\(=3+\frac{2p^3}{mnp}+\frac{2m^3}{mnp}+\frac{2n^3}{mnp}=3+\frac{2.\left(m^3+n^3+p^3\right)}{mnp}\)
*Tới đây để tính được m3+n3+p3,ta cần CM được bài toán phụ sau:
Đề: Cho m+n+p=0.CMR: \(m^3+n^3+p^3=3mnp\)
Từ m+n+p=0=>m+n=-p
Ta có: \(m^3+n^3+p^3=\left(m+n\right)^3-3m^2n-3mn^2+p^3=-p^3-3mn\left(m+n\right)+p^3\)
\(=-3mn\left(m+n\right)=-3mn.\left(-p\right)=3mnp\)
Vậy ta đã CM được bài toán phụ
*Trở lại bài toán chính: \(A=3+\frac{2.3mnp}{mnp}=3+\frac{6mnp}{mnp}=3+6=9\)
Vậy A=9
HP
29 tháng 7 2016
bài 2)
a)Nhận thấy các thừa số của A đều có dạng tổng quát sau:
\(n^3+1=n^3+1^3=\left(n+1\right)\left(n^2-n+1\right)=\left(n+1\right).\left(n^2-n+\frac{1}{4}+\frac{3}{4}\right)\)
\(=\left(n+1\right).\left(n^2-2.n.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right)=\left(n+1\right).\left[\left(n-\frac{1}{2}\right)^2+\frac{3}{4}\right]=\left(n+1\right).\left[\left(n-0,5\right)^2+0,75\right]\)
\(n^3-1=n^3-1^3=\left(n-1\right)\left(n^2+n+1\right)=\left(n-1\right).\left(n^2+n+\frac{1}{4}+\frac{3}{4}\right)\)
\(=\left(n-1\right).\left(n^2+2.n.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right)=\left(n-1\right).\left[\left(n+\frac{1}{2}\right)^2+\frac{3}{4}\right]=\left(n-1\right).\left[\left(n+0,5\right)^2+0,75\right]\)
suy ra \(\frac{n^3+1}{n^3-1}=\frac{\left(n+1\right).\left[\left(n-0,5\right)^2+0,75\right]}{\left(n-1\right).\left[\left(n+0,5\right)^2+0,75\right]}\)
Do đó: \(\frac{2^3+1}{2^3-1}=\frac{\left(2+1\right).\left[\left(2-0,5\right)^2+0,75\right]}{\left(2-1\right).\left[\left(2+0,5\right)^2+0,75\right]}=\frac{3.\left(1,5^2+0,75\right)}{1.\left(2,5^2+0,75\right)}\)
\(\frac{3^3+1}{3^3-1}=\frac{\left(3+1\right).\left[\left(3-0,5\right)^2+0,75\right]}{\left(3-1\right).\left[\left(3+0,5\right)^2+0,75\right]}=\frac{4.\left(2,5^2+0,75\right)}{2.\left(3,5^2+0,75\right)}\)
...........................
\(\frac{10^3+1}{10^3-1}=\frac{\left(10+1\right).\left[\left(10-0,5\right)^2+0,75\right]}{\left(10-1\right).\left[\left(10+0,5\right)^2+0,75\right]}=\frac{11.\left(9,5^2+0,75\right)}{9.\left(10,5^2+0,75\right)}\)
\(=>A=\frac{3\left(1,5^2+0,75\right).4\left(2,5^2+0,75\right)........11.\left(9,5^2+0,75\right)}{1\left(2,5^2+0,75\right).2.\left(3,5^2+0,75\right)........9\left(10,5^2+0,75\right)}=\frac{3.4........11}{1.2......9}.\frac{1,5^2+0,75}{10,5^2+0,75}\)
\(=\frac{10.11}{2}.\frac{1}{37}=\frac{2036}{37}\)
Vậy A=2036/37
b) có thể ở chỗ 1+1/4 bn nhầm,phải là \(1^4+\frac{1}{4}\) ,mà chắc cũng chẳng sao,vì 14=1 mà
Nhận thấy các thừa số của B có dạng tổng quát:
\(n^4+\frac{1}{4}=n^4+n^2+\frac{1}{4}-n^2=\left(n^2\right)^2+2.n^2.\frac{1}{2}+\frac{1}{4}-n^2=\left(n^2+\frac{1}{2}\right)^2-n^2\)
\(=\left(n^2+\frac{1}{2}-n\right)\left(n^2+\frac{1}{2}+n\right)\)
\(B=\frac{\left(1^2+\frac{1}{2}-1\right).\left(1^2+\frac{1}{2}+1\right).\left(3^2+\frac{1}{2}+3\right).\left(3^2+\frac{1}{2}-3\right)..........\left(9^2+\frac{1}{2}-9\right).\left(9^2+\frac{1}{2}+9\right)}{\left(2^2+\frac{1}{2}-2\right).\left(2^2+\frac{1}{2}+2\right).\left(4^2+\frac{1}{2}-4\right).\left(4^2+\frac{1}{2}+4\right)......\left(10^2+\frac{1}{2}-10\right).\left(10^2+\frac{1}{2}+10\right)}\)
Mặt khác,ta cũng có: \(\left(a+1\right)^2-\left(a+1\right)+\frac{1}{2}=a^2+2a+1-a-1+\frac{1}{2}=a^2+a+\frac{1}{2}\)
Suy ra \(B=\frac{1^2+\frac{1}{2}-1}{10^2+\frac{1}{2}+10}=\frac{1}{221}\)
Vậy B=1/221
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
26 tháng 9
a; A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\) + ... + \(\dfrac{1}{\left(2n\right)^2}\)
A = \(\dfrac{1}{2^2}\).(\(\dfrac{1}{1^2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{n^2}\))
A = \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + ... + \(\dfrac{1}{n.n}\))
Vì \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\); \(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\); ...; \(\dfrac{1}{n.n}\) < \(\dfrac{1}{\left(n-1\right)n}\)
nên A < \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + ... + \(\dfrac{1}{\left(n-1\right)n}\))
A < \(\dfrac{1}{4.}\)(1 + \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{n-1}\) - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{4}\).(1 + 1 - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{4}\).(2 - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{2}\) - \(\dfrac{1}{4n}\) < \(\dfrac{1}{2}\) (đpcm)
PT
1
13 tháng 11 2017
nâng cao và phát triển toán bài 128 b gần giống câu này nha mình mới học được hết ptđttnt thôi
TP
3 tháng 8 2017
Ta có :
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
Ta có:
\(\frac{1}{\sqrt{x}+\sqrt{x-1}}=\frac{\sqrt{x}-\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}=\sqrt{x}-\sqrt{x-1}\)
Do đó:
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
\(\Leftrightarrow A=\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+...+\sqrt{n-1}+\sqrt{n}\)
\(\Leftrightarrow A=\sqrt{n}-1\left(dpcm\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 11 2019
\(\frac{n^3-1}{n^3+1}=\frac{\left(n-1\right)\left(n^2+n+1\right)}{\left(n+1\right)\left(n^2-n+1\right)}=\frac{\left(n-1\right)\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(n+1\right)\left(n^2-n+1\right)}\)
\(\Rightarrow A=\frac{1\left(3^2-3+1\right)}{3\left(2^2-2+1\right)}.\frac{2.\left(4^2-4+1\right)}{4.\left(3^2-3+1\right)}.\frac{3\left(5^2-5+1\right)}{5.\left(4^2-4+1\right)}...\frac{\left(n-1\right)\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(n+1\right)\left(n^2-n+1\right)}\)
\(=\frac{1.2.\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(2^2-2+1\right).n\left(n+1\right)}=\frac{2\left(n^2+n+1\right)}{3\left(n^2+n\right)}\)
\(F=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....+\frac{1}{3^n}\)
\(\Rightarrow\)\(3F=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{n-1}}\)
\(\Rightarrow\)\(3F-F=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{n-1}}\right)-\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^n}\right)\)
\(\Rightarrow\)\(2F=\frac{1}{3}-\frac{1}{3^n}\)\(=\frac{3^{n-1}-1}{3^n}\)
\(\Rightarrow\)\(F=\frac{3^{n-1}-1}{2.3^n}\)
P/S: tham khảo nha. mk ko chắc làm đúng
\(F=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^n}\)
\(\Rightarrow3F=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{n-1}}\)
\(\Rightarrow3F-F=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{n-1}}-\frac{1}{3^2}-\frac{1}{3^3}-\frac{1}{3^4}-...-\frac{1}{3^n}\)
\(\Rightarrow2F=\frac{1}{3}-\frac{1}{3^n}\)
\(\Rightarrow F=\frac{\frac{1}{3}-\frac{1}{3^n}}{2}\)