Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
Ta dùng bđt Cô si nhé :)
Do x > 1 nên x + 1 > 0. Từ đó ta có:
\(A=4\left(x+1\right)+\frac{25}{x+1}-4\)
Áp dụng bđt Cosi ta có : \(4\left(x+1\right)+\frac{25}{x+1}\ge2\sqrt{\frac{4\left(x+1\right).25}{\left(x+1\right)}}=20\Rightarrow A\ge20-4=16\)
Vậy GTNN của A là 16, khi x = 1,5.
cho 0<x>1. Tim GTNN cua \(A=\frac{x}{1-x}+\frac{5}{x}\)
cho x,y,z>0 tm xy+xz+yz=1. tim gtnn cua
S=\(\frac{1}{4x^2-yz+2}+\frac{1}{4y^2-xz+2}+\frac{1}{4z^2-xy+2}\)
Cho x >1. Tìm GTNN A = \(4x+\frac{25}{x-1}\)
Tim GTNN cua
\(A=x^2-\frac{1}{3}x+\frac{1}{27x}+2016...\)
ch 0<x<1. tim GTNN cua A=\(\frac{x}{1-x}+\frac{5}{x}\)
1.Tim a de \(N=\frac{6}{M}\in Z\)biet \(M=\frac{a+2\sqrt{a}+1}{\sqrt{a}}\)
2.tim nghiem nguyen to cua pt:\(x^2-2y^3=1\)
3.Cho a, b,c la 3 canh cua tam giac.CM:\(a^2\left(b+c\right)+b^2\left(c+a\right)+c^2\left(a+b\right)\subseteq a^3+b^3+c^3\)
4.ve do thi ham so \(y=\sqrt{x^2-4x+4}-\sqrt{x^2+4x+4}\)dung do thi. tim GTLN,GTNN.
cho x,y,z la cac so thuc thoa x+y+z=0, x+1>0, y+1>0, z+1>0. tim GTLN cua P=\(\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+4}\)
cho x,y,z,t la cac so duong. tim GTNN cua A=\(\frac{x-t}{t+y}+\frac{t-y}{y+z}+\frac{y-z}{z+x}+\frac{z-x}{x+t}\)
Tim GTLN ,GTNN cua bieu thuc :A=\(\frac{x+1}{x^2+x+1}\)
tim gtnn va gtln cua
a)\(\frac{x^2+1}{x^2-x+1}\)
b)\(\frac{5y^2-3xy}{x^2-3xy+4y^2}\)
c)Cho \(x^2+2xy-x^2y-y+7=0\) .Tim gtnn va gtln cua \(x^2+6xy+12y^2\)
