Đến nội dung

nguyen thi dien

nguyen thi dien

Đăng ký: 25-11-2012
Offline Đăng nhập: 16-06-2014 - 08:43
-----

Trong chủ đề: $\left( \sum kx_k \right) \left( \sum x^2_k...

04-05-2014 - 16:02

minh thu giai trong truong hop n$\geq 4.con cac truong hop khac cac ban tu lam nha.thwo de bai thi ta chi can chung minh trong truong hopx_{n} la day giam.dat y_{n}=x_{n}/n.khi do y_{n} la mot day giam.$$dat \sum_{1}^{n}y_{k}= ln\left ( voi l la so nguyen duong nao do \right ).khi do ta se chung minh $\sum k^{2}y_{k}\geq (\sum ky_{k}\left )( 2n-1 \right )-2ln(n-1)(n+1)/3.bang cach co dinh cac bien va chi cho 2 bien y_{k}va y_{h} thay doi sao cho tong cua chung khong doi$$dat \left ( y_{k}+y_{h} \right )/2=u,\left ( y_{k}-y_{h} \right )/2=v\left ( h\geq k \right )\Rightarrow v\geq 0.khi do dat g\left ( v \right )=k^{2}\left ( u+v \right )+h^{2}(u-v)-(2n-1)\left [ k(u+v)+h(u-v) \right ]$$\Rightarrow g{\left ( v \right )}'= (k-h)(k+h-2n+1)\geq 0.\Rightarrow g(v)\geq g\left ( 0 \right )\Rightarrow ket luan tren$$lai co (k^{2}y_{k}-k^{2}l)^{2}\geq 0\forall k.\Leftrightarrow k^{2}y_{k}^{2}\geq 2lk^{2}y_{k}-k^{2}l^{2}\Rightarrow \sum k^{2}l^{2}\geq 2l(\sum k^{2}y_{k})-l^{2}n\left ( n+1 \right )\left (2n+1 \right )/6\geq 2l(2n-1)\sum ky_{k}-l^{2}n(n+1)(n-3)/6$$$\geq 2l(2n-1)\sum ky_{k}-l^{2}n(n+1)(10n-7)/6$.dat\sum ky_{k}=x.khi do ta co VT\geq [(2n-1)x-2ln(n-1)(n+1)/3]\left [ 2l(2n-1)x-l^{2}n(n+1)(10n-7)/6 \right ]$.$bay gio ta chi can chung minh no\geq 2(2n+1)^{2}x^{3}/\left ( 9(n+1)n \right ).\Leftrightarrow 2(2n+1)^{2}x^{3}/(9n(n+1))-2l(2n-1)^{2}x^{2}+l^{2}n(n+1)(2n-1)(9n-5)x/2+l^{3}n^{2}\left ( n+1 \right )^{2}(n-1)(10n-7)/9\leq 0.dat VT =f\left ( x \right ).f{\left ( x \right )}'=2\left ( 2n+1 \right )^{2}x^{2}/(3n(n+1))-4l(2n-1)^{2}x+l^{2}n(n+!)(2n-1)(9n-5)/2.f\left ( x \right )''=4(2n+1)^{2}x/(3n(n+1))-4l(2n-1)^{2}.do y_{n}la day giam nen ap dung bat dang thuc cheybeychev ta co\sum ky_{k}\leq \left ( \sum k \right ).\left ( \sum y_{k} \right )/n=ln(n+1)/2.\Rightarrow f{}''(x)\leq 0.\Rightarrow f{\left ( x \right )}'\geq f{}'\left ( ln(n+1)/2 \right )\geq 0\forall n\geq 4 \right ) \right )\Rightarrow f\left ( x \right)\leq f\left ( ln(n+1)/2 \right )=0.dpcm.dau = xay ra khi y_{i}=y_{j}\forall i,j=1\overline{,n}.\Leftrightarrow x_{i}/i=x_{j}/j\forall i,j=1\overline{,n}.$neu co gi sai xot thi moi nguoi  bao cho minh qua sdt 01683459471 nha.can on nhieu


Trong chủ đề: $\left( \sum kx_k \right) \left( \sum x^2_k...

04-05-2014 - 16:00

minh thu giai trong truong hop $n \geq 4$

 

.con cac truong hop khac cac ban tu lam nha.thwo de bai thi ta chi can chung minh trong truong hop $x_{n}$ la day giam.dat $y_{n}=x_{n}/n $ .khi do $y_{n}$ la mot day giam. $

 

\sum_{1}^{n}y_{k}= ln $\left ( voi l la so nguyen duong nao do \right ).khi do ta se chung minh $\sum k^{2}y_{k}\geq (\sum ky_{k}\left )( 2n-1 \right )-2ln(n-1)(n+1)/3.bang cach co dinh cac bien va chi cho 2 bien y_{k}va y_{h} thay doi sao cho tong cua chung khong doi$$dat \left ( y_{k}+y_{h} \right )/2=u,\left ( y_{k}-y_{h} \right )/2=v\left ( h\geq k \right )\Rightarrow v\geq 0.khi do dat g\left ( v \right )=k^{2}\left ( u+v \right )+h^{2}(u-v)-(2n-1)\left [ k(u+v)+h(u-v) \right ]$$\Rightarrow g{\left ( v \right )}'= (k-h)(k+h-2n+1)\geq 0.\Rightarrow g(v)\geq g\left ( 0 \right )\Rightarrow ket luan tren$$lai co (k^{2}y_{k}-k^{2}l)^{2}\geq 0\forall k.\Leftrightarrow k^{2}y_{k}^{2}\geq 2lk^{2}y_{k}-k^{2}l^{2}\Rightarrow \sum k^{2}l^{2}\geq 2l(\sum k^{2}y_{k})-l^{2}n\left ( n+1 \right )\left (2n+1 \right )/6\geq 2l(2n-1)\sum ky_{k}-l^{2}n(n+1)(n-3)/6$$$\geq 2l(2n-1)\sum ky_{k}-l^{2}n(n+1)(10n-7)/6$.dat\sum ky_{k}=x.khi do ta co VT\geq [(2n-1)x-2ln(n-1)(n+1)/3]\left [ 2l(2n-1)x-l^{2}n(n+1)(10n-7)/6 \right ]$.$bay gio ta chi can chung minh no\geq 2(2n+1)^{2}x^{3}/\left ( 9(n+1)n \right ).\Leftrightarrow 2(2n+1)^{2}x^{3}/(9n(n+1))-2l(2n-1)^{2}x^{2}+l^{2}n(n+1)(2n-1)(9n-5)x/2+l^{3}n^{2}\left ( n+1 \right )^{2}(n-1)(10n-7)/9\leq 0.dat VT =f\left ( x \right ).f{\left ( x \right )}'=2\left ( 2n+1 \right )^{2}x^{2}/(3n(n+1))-4l(2n-1)^{2}x+l^{2}n(n+!)(2n-1)(9n-5)/2.f\left ( x \right )''=4(2n+1)^{2}x/(3n(n+1))-4l(2n-1)^{2}.do y_{n}la day giam nen ap dung bat dang thuc cheybeychev ta co\sum ky_{k}\leq \left ( \sum k \right ).\left ( \sum y_{k} \right )/n=ln(n+1)/2.\Rightarrow f{}''(x)\leq 0.\Rightarrow f{\left ( x \right )}'\geq f{}'\left ( ln(n+1)/2 \right )\geq 0\forall n\geq 4 \right ) \right )\Rightarrow f\left ( x \right)\leq f\left ( ln(n+1)/2 \right )=0.dpcm.dau = xay ra khi y_{i}=y_{j}\forall i,j=1\overline{,n}.\Leftrightarrow x_{i}/i=x_{j}/j\forall i,j=1\overline{,n}.$neu co gi sai xot thi moi nguoi  bao cho minh qua sdt 01683459471 nha.can on nhieu


Trong chủ đề: $7x^2+7x=\sqrt{\frac{4x+9}{28}...

20-02-2013 - 21:19

cái này là hệ đối xứng bậc 2 nên dễ dàng giải được

Trong chủ đề: $7x^2+7x=\sqrt{\frac{4x+9}{28}...

20-02-2013 - 21:18

PT $\Leftrightarrow 7(x+\frac{1}{2})^{2}=\sqrt{\frac{1}{7}(x+\frac{1}{2})+\frac{1}{4}}$
Đặt u=$x+\frac{1}{2}$, v=$\sqrt{\frac{1}{7}(x+\frac{1}{2})+\frac{1}{4}}=\sqrt{\frac{1}{7}u+\frac{1}{4}}$
Ta thu được hệ phương trình
$u^{2}-\frac{1}{4}=\frac{1}{7}v$
$v^{2}-\frac{1}{4}=\frac{1}{7}u$

Trong chủ đề: $(m^2-4)x-m-2<o$

28-01-2013 - 20:32

Xét m=2, BPT trở thành -4<0 (Đúng)
Xét m=-2, BPT trở thành 0<0 (sai)
Xét $m\neq \pm 2$, BPT trở thành:
$(m^{2}-4)x< m+2$
$\Leftrightarrow x< \frac{m+2}{m^{2}-4}=\frac{1}{m-2}$
Vậy:với m= 2, bpt thoả mãn với mọi x.
Với $m\neq \pm 2$ Bất phương trình có nghiệm $x< \frac{1}{m-2}$