Violympic toán 9
Các câu hỏi tương tự
1. GIải các pt :
a) x^2-2left(sqrt{3}+sqrt{2}right)x+4sqrt{6}0
2. chứng minh rằng các pt sau luôn luôn có nghiệm
a) x^2-2left(m-1right)x-3-m0
b) x^2+left(m+1right)x+m0
c) x^2-left(2m-3right)x+m^2-3m0
d) x^2+2left(m+2right)x-4m-120
e) x^2-left(2m-3right)x+m^2+3m+20
f) x^2-2x-left(m-1right)left(m-3right)0
3. left(a-3right)x^2-2left(a-1right)x+a-50
Tìm a để pt có 2 nghiệm phân biệt
Đọc tiếp
1. GIải các pt :
a) \(x^2-2\left(\sqrt{3}+\sqrt{2}\right)x+4\sqrt{6}=0\)
2. chứng minh rằng các pt sau luôn luôn có nghiệm
a) \(x^2-2\left(m-1\right)x-3-m=0\)
b) \(x^2+\left(m+1\right)x+m=0\)
c) \(x^2-\left(2m-3\right)x+m^2-3m=0\)
d) \(x^2+2\left(m+2\right)x-4m-12=0\)
e) \(x^2-\left(2m-3\right)x+m^2+3m+2=0\)
f) \(x^2-2x-\left(m-1\right)\left(m-3\right)=0\)
3. \(\left(a-3\right)x^2-2\left(a-1\right)x+a-5=0\)
Tìm a để pt có 2 nghiệm phân biệt
2 tháng 3 2020 lúc 23:45
1. a) left{{}begin{matrix}x,y,z0xyz1end{matrix}right.. Tìm max Pfrac{1}{sqrt{x^5-x^2+3xy+6}}+frac{1}{sqrt{y^5-y^2+3yz+6}}+frac{1}{sqrt{z^5-z^2+zx+6}}
b) left{{}begin{matrix}x,y,z0xyz8end{matrix}right.. Min Pfrac{x^2}{sqrt{left(1+x^3right)left(1+y^3right)}}+frac{y^2}{sqrt{left(1+y^3right)left(1+z^3right)}}+frac{z^2}{sqrt{left(1+z^3right)left(1+x^3right)}}
c) x,y,z0. Min Psqrt{frac{x^3}{x^3+left(y+zright)^3}}+sqrt{frac{y^3}{y^3+left(z+xright)^3}}+sqrt{frac{z^3}{z^3+left(x+yright)^3}}
d) a,b,c0;...
Đọc tiếp
1. a) \(\left\{{}\begin{matrix}x,y,z>0\\xyz=1\end{matrix}\right.\). Tìm max \(P=\frac{1}{\sqrt{x^5-x^2+3xy+6}}+\frac{1}{\sqrt{y^5-y^2+3yz+6}}+\frac{1}{\sqrt{z^5-z^2+zx+6}}\)
b) \(\left\{{}\begin{matrix}x,y,z>0\\xyz=8\end{matrix}\right.\). Min \(P=\frac{x^2}{\sqrt{\left(1+x^3\right)\left(1+y^3\right)}}+\frac{y^2}{\sqrt{\left(1+y^3\right)\left(1+z^3\right)}}+\frac{z^2}{\sqrt{\left(1+z^3\right)\left(1+x^3\right)}}\)
c) \(x,y,z>0.\) Min \(P=\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}+\sqrt{\frac{y^3}{y^3+\left(z+x\right)^3}}+\sqrt{\frac{z^3}{z^3+\left(x+y\right)^3}}\)
d) \(a,b,c>0;a^2+b^2+c^2+abc=4.Cmr:2a+b+c\le\frac{9}{2}\)
e) \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=3\end{matrix}\right.\). Cmr: \(\frac{a}{b^3+ab}+\frac{b}{c^3+bc}+\frac{c}{a^3+ca}\ge\frac{3}{2}\)
f) \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca+abc=4\end{matrix}\right.\) Cmr: \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le3\)
g) \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca+abc=2\end{matrix}\right.\) Max : \(Q=\frac{a+1}{a^2+2a+2}+\frac{b+1}{b^2+2b+2}+\frac{c+1}{c^2+2c+2}\)
4 tháng 6 2018 lúc 21:17
B1: giai pt: a, dfrac{left(x+1right)^4}{left(x^2+1right)^2}+dfrac{4x}{x^2+1}6
B2: Tính giá trị của A dfrac{x^2-sqrt{x}}{x+sqrt{x}+1}-dfrac{x^2+sqrt{x}}{x-sqrt{x}+1}+x+1
B3: CMR voi 3 số thực a,b,c tùy ý thì ít nhất 1 trong 3 pt sau phải có nghiệm:
x^2-2ax+2b-10left(1right);x^2-2bx+2c-10left(2right);x^2-2cx+2a-10left(3right)
Đọc tiếp
B1: giai pt: a, \(\dfrac{\left(x+1\right)^4}{\left(x^2+1\right)^2}+\dfrac{4x}{x^2+1}=6\)
B2: Tính giá trị của A= \(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\)
B3: CMR voi 3 số thực a,b,c tùy ý thì ít nhất 1 trong 3 pt sau phải có nghiệm:
\(x^2-2ax+2b-1=0\left(1\right);x^2-2bx+2c-1=0\left(2\right);x^2-2cx+2a-1=0\left(3\right)\)
17 tháng 11 2019 lúc 17:52
1. \(\left\{{}\begin{matrix}x,y,z>0\\xyz=1\end{matrix}\right.\) Cmr: \(\frac{x^2}{\left(x+1\right)^2}+\frac{y^2}{\left(y+1\right)^2}+\frac{z^2}{\left(z+1\right)^2}\ge\frac{3}{4}\)\
2. \(a,b,c>0.\) cmr: \(\Sigma\frac{a^3}{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\le\frac{1}{a+b+c}\)
Cho left{{}begin{matrix}x,y,z0x+2y+3z3end{matrix}right.
Tìm MaxP biết:
Pdfrac{88y^3-x^3}{2xy+16y^2}+dfrac{297z^3-8y^3}{6yz+36z^2}+dfrac{11x^3-27z^3}{3xz+4x^2}
Đặt 2ya, 3zb Rightarrow x+a+b3
Rightarrow Pdfrac{11a^3-x^3}{ax+4a^2}+dfrac{11b^3-a^3}{ab+4b^2}+dfrac{11x^3-b^3}{bx+4x^2}
Ta chứng minh bđt sau:
dfrac{11a^3-x^3}{ax+4a^2}le3a-xLeftrightarrow11a^3-x^3leleft(3a-xright)left(ax+4a^2right)Leftrightarrow11a^3-x^3le12a^3+3a^2x-ax^2-4a^2xLeftrightarrow a^3-a^2x-ax^2+x^3ge0Leftrightarrow a^2le...
Đọc tiếp
Cho \(\left\{{}\begin{matrix}x,y,z>0\\x+2y+3z=3\end{matrix}\right.\)
Tìm MaxP biết:
\(P=\dfrac{88y^3-x^3}{2xy+16y^2}+\dfrac{297z^3-8y^3}{6yz+36z^2}+\dfrac{11x^3-27z^3}{3xz+4x^2}\)
Đặt 2y=a, 3z=b \(\Rightarrow x+a+b=3\)
\(\Rightarrow P=\dfrac{11a^3-x^3}{ax+4a^2}+\dfrac{11b^3-a^3}{ab+4b^2}+\dfrac{11x^3-b^3}{bx+4x^2}\)
Ta chứng minh bđt sau:
\(\dfrac{11a^3-x^3}{ax+4a^2}\le3a-x\Leftrightarrow11a^3-x^3\le\left(3a-x\right)\left(ax+4a^2\right)\Leftrightarrow11a^3-x^3\le12a^3+3a^2x-ax^2-4a^2x\Leftrightarrow a^3-a^2x-ax^2+x^3\ge0\Leftrightarrow a^2\left(a-x\right)-x^2\left(a-x\right)\ge0\Leftrightarrow\left(a-x\right)^2\left(a+x\right)\ge0\left(luondung\right)\)tương tự:
\(\dfrac{11x^3-b^3}{bx+4x^2}\le3x-b,\dfrac{11b^3-a^3}{ab+4b^2}\le3b-a\)
\(\Rightarrow P\le3\left(x+a+b\right)-\left(a+b+x\right)=2\left(a+b+x\right)=2.3=6\)
\(MaxP=6\Leftrightarrow x=1,y=\dfrac{1}{2},z=\dfrac{1}{3}\)
11 tháng 2 2020 lúc 21:04
1. left{{}begin{matrix}a,b,c0a+b+c1end{matrix}right.. Cmr: frac{ab}{sqrt{left(1-cright)^2left(1+cright)}}+frac{bc}{sqrt{left(1-aright)^2left(1+aright)}}+frac{ca}{sqrt{left(1-bright)^3left(1+bright)}}lefrac{3sqrt{2}}{8}
2. left{{}begin{matrix}a,b,c0a+b+cle1end{matrix}right.. Cmr: frac{1}{a^2+b^2+c^2}+frac{1}{ableft(a+bright)}+frac{1}{bcleft(b+cright)}+frac{1}{acleft(a+cright)}gefrac{87}{2}
3. left{{}begin{matrix}a,b,c0ab+bc+ca2abcend{matrix}right.. Cmr: frac{1}{aleft(2a-1right)^2}+frac{1}{bleft...
Đọc tiếp
1. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=1\end{matrix}\right.\). Cmr: \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}+\frac{bc}{\sqrt{\left(1-a\right)^2\left(1+a\right)}}+\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\le\frac{3\sqrt{2}}{8}\)
2. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c\le1\end{matrix}\right.\). Cmr: \(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab\left(a+b\right)}+\frac{1}{bc\left(b+c\right)}+\frac{1}{ac\left(a+c\right)}\ge\frac{87}{2}\)
3. \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca=2abc\end{matrix}\right.\). Cmr: \(\frac{1}{a\left(2a-1\right)^2}+\frac{1}{b\left(2b-1\right)^2}+\frac{1}{c\left(2c-1\right)^2}\ge\frac{1}{2}\)
4. \(\left\{{}\begin{matrix}x,y,z>0\\x+y+z=2015\end{matrix}\right.\). Tìm min \(A=\frac{x^4+y^4}{x^3+y^3}+\frac{y^4+z^4}{y^3+z^3}+\frac{z^4+x^4}{z^2+x^2}\)
Mn giúp mk với ạ! Thanks nhiều
23 tháng 2 2020 lúc 11:23
1. a) Tìm nin N*, n2008 sao cho 2^{2008}+2^{2012}+2^{2013}+2^{2014}+2^{2016}+2^n là số chính phương
b) tìm x,y 0 thỏa mãn x^2+y^22left(x+yright)left(sqrt{x}+sqrt{y}-2right)
2. a) left{{}begin{matrix}age0a+bge1end{matrix}right.. Min Afrac{8a^2+b}{4a}+b^2
b) left{{}begin{matrix}a,bge0left(a-bright)^2a+b+2end{matrix}right.. Cmr: left(1+frac{a^3}{left(b+1right)^3}right)left(1+frac{b^3}{left(b+1right)^3}right)le9
c) x,y0;left(x+sqrt{1+x^2}right)left(y+sqrt{1+y^2}right)2020. Min P x + y
d) x,y,...
Đọc tiếp
1. a) Tìm \(n\in N\)*, \(n>2008\) sao cho \(2^{2008}+2^{2012}+2^{2013}+2^{2014}+2^{2016}+2^n\) là số chính phương
b) tìm x,y > 0 thỏa mãn \(x^2+y^2=2\left(x+y\right)\left(\sqrt{x}+\sqrt{y}-2\right)\)
2. a) \(\left\{{}\begin{matrix}a\ge0\\a+b\ge1\end{matrix}\right.\). Min \(A=\frac{8a^2+b}{4a}+b^2\)
b) \(\left\{{}\begin{matrix}a,b\ge0\\\left(a-b\right)^2=a+b+2\end{matrix}\right.\). Cmr: \(\left(1+\frac{a^3}{\left(b+1\right)^3}\right)\left(1+\frac{b^3}{\left(b+1\right)^3}\right)\le9\)
c) \(x,y>0;\left(x+\sqrt{1+x^2}\right)\left(y+\sqrt{1+y^2}\right)=2020\). Min P = x + y
d) \(x,y,z>0;\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}=6\). Min \(P=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
e) \(\left\{{}\begin{matrix}x,y,z>0\\x+y+z+4xyz=4\end{matrix}\right.\) Cmr: \(\left(1+xy+\frac{y}{z}\right)\left(1+yz+\frac{z}{x}\right)\left(1+zx+\frac{x}{y}\right)\ge27\)
f) \(\left\{{}\begin{matrix}x,y,z\ge1\\3x^2+4y^2+5z^2=52\end{matrix}\right.\). Min P = x + y + z
g) \(x,y>0\). Min \(P=\frac{2}{\sqrt{\left(2x+y\right)^3+1}-1}+\frac{2}{\sqrt{\left(x+2y\right)^3+1}-1}+\frac{\left(2x+y\right)\left(x+2y\right)}{4}-\frac{8}{3\left(x+y\right)}\)
Chứng minh các pt sau luôn có nghiệm :
a, \(x^2-\left(m+1\right)x+m=0\)
b, \(x^2-2\left(m+1\right)+2m+1=0\)
c, \(x^2+\left(m+3\right)x+m+1\) = 0
d, \(x^2-3x+1-m^2=0\)
17 tháng 8 2019 lúc 22:20
Cho các số dương a,b,c tm a+2b+3c=1. Chứng minh rằng ít nhất 1 trong 3 pt sau có nghiệm:
\(4x^2-4\left(2a+1\right)x+4a^2+192abc+1=0\left(1\right)\);\(4x^2-4\left(2b+1\right)x+4b^2+96abc+1=0\left(2\right)\)
Đây ạNguyễn Việt Lâm