cho a>0,b>0 thoa man a+b lon hon hoac bang 2 .tim max cua M =1/a+b^2 +1/b+a^2
cho ba so a,b,c thoa man 0be honhoac bang a be hon hoac bang b+1 be hon hoac bang c+2 va a+b+c=1 tim gia tri nho nhat cua c
cau 1 ,cho m,n thuoc n va p la so nguyen to thoa man p/m-1=m+n/p
CMRp^2=n+2
cau 2,cho a,b,c thoa man a+b+c=0 CMRab+bc+ca be hon hoac bang 0
cau 3,bay gio la 4 gio 10 phut hoi sau it nhat bao lau thi hai kim dong ho nam doi dien nhau tren mot duong thang
cau 4,so 2^100 viet trong he thap phan tao thanh 1 so hoi so do co bao nhieu chu so
cau 5,cho a,b,c la so do 3 canh cua mot tam giac vuong voi c la so do canh huyen CMRa^2n+b^2n be hon hoac bang c^2n (n la so tu nhien lon hon 0 )
1. cho 4 stn a lon hon hoac bang b, b lon hon hoac bang c, clon hon hoac bang d.
CM:(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)
2. CM: co the tim dc 1 stn k sao cho: (1997^k)-1 chia het cho 10^4
3. tong cac chu so cua 1 so chinh phuong co the bang 1995 dc k?
4. tong cua 1995 stn khac 0 dung bang 1995. Hoi UCLN cua chung la bao nhieu?
đề này sai bét .ngồi đến năm sau cũng trả giải được
Tim x biet:
A. (x-3/4).(3x+1/2)lon hon hoac bang 0
B. (2x+1).(4x+3)be hon hoac bang 0
cho a,b,c lon hon hoac bang 1.Tim gtln cua P = (a+1)(b+1)(c+1)/abc+1
Cho a,b > 0 thoa man a +b ≥ 2. Tim GTLN cua \(M=\dfrac{1}{a+b^2}+\dfrac{1}{b+a^2}\)
Ta có : \(M=\dfrac{1}{a+b^2}+\dfrac{1}{b+a^2}=\dfrac{a+1}{\left(a+b^2\right)\left(a+1\right)}+\dfrac{b+1}{\left(b+1\right)\left(b+a^2\right)}\le\dfrac{a+1}{\left(a+b\right)^2}+\dfrac{b+1}{\left(a+b\right)^2}=\dfrac{1}{a+b}+\dfrac{2}{\left(a+b\right)^2}\le\dfrac{1}{2}+\dfrac{2}{4}=1\)đẳng thức xả ra khi và chỉ khi a=b=1. Do đó GTLN của M là 1.
cho x,y>0 thoa man x lon hon hoac bang 2y.TTim min M=x^2+y^2/xy
cmr : a^2/b^2 +b^2/c^2+c^2/a^2 luon luon lon hon hoac bang c/b+ b/a+ a/c voi a,b,c lon hon 0
so tu nhien x thoa man dieu kien 0.(x-3)=0.so x bang
(A)0
(B)3
(C)so tu nhien bat ki
(D) so tu nhien bat ki lon hon hoac bang 3