1. Cho a,b la 2 so duong thoa a+b<=1.chung minh rang \(6b+\frac{1}{3a}+\frac{4}{b}\ge11\).
2. cho a,b,c la cac so nguyen duong sao cho (a-b).(a-c).(b-c)=a+b+c
a. chung minh rang a+b+c chia het cho 2
b. Tim gia tri nho nhat cua M=a+b+c
cho a,b,c, la cac so duong thoa man a+b+c=1, c/m :1/a+1/b+1/c>=9
Số học sinh nữ là:
40x3/8 = 15(học sinh)
Số học sinh nam là:
40-15=25(học sinh)
Cho a,b,c la ba so thuc duong thoa man (a^2).(b+c)=(b^2).(a+c)=20172018.tinh (c^2).(a+b)
cho 3 so a,b,c la so nguyen . Trong do co 1 so nguyen am , 1 so nguyen duong va 1 so bang 0 , thoa man IaI=b^2.(b-c) . Hoi a,b,c thuoc loai so nao
Cho a, b, c, d la cac so nguyen duong thoa man a2+b2=b2+d2. CM a+b+c+d la hop so
cho cac so nguyen duong a,b,c thoa man :(a,b,c)=1;ab=c(a+b).Cm a+b la so chinh phuong
cho a,b,c la cac so duong thoa man (1/a+1/b+1/c)>=(a+b+c), chung minh a+b+c>=3abc
cho a,b,c la cac so duong thoa man (1/a+1/b+1/c)>=(a+b+c), chung minh a+b+c>=3abc
https://www.facebook.com/OnThiDaiHocKhoiA/posts/508217699295984
cho a,b,c la ba so duong thoa man a+b+c=1 CMR:c+ab/a+b + a+bc/b+c + b+ac/a+c \(\ge\) 2
Áp dụng BĐT AM-GM ta có:
\(VT=\dfrac{c+ab}{a+b}+\dfrac{a+bc}{b+c}+\dfrac{b+ac}{a+c}\)
\(=\dfrac{c\left(a+b+c\right)+ab}{a+b}+\dfrac{a\left(a+b+c\right)+bc}{b+c}+\dfrac{b\left(a+b+c\right)+ac}{a+c}\)
\(=\dfrac{ac+bc+c^2+ab}{a+b}+\dfrac{a^2+ab+ac+bc}{b+c}+\dfrac{ab+b^2+bc+ac}{a+c}\)
\(=\dfrac{\left(b+c\right)\left(c+a\right)}{a+b}+\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}\)
\(\ge2\left(a+b+c\right)=2\left(a+b+c=1\right)\)
Khi \(a=b=c=\dfrac{1}{3}\)
cho a,b la so thuc duong thoa man:a^2+2ab+2b^2-2b=8. cm 0<a+b<=3
a^2 + 2ab + 2b^2 - 2b= 8
<=> (a^2 + 2ab + b^2) + (b^2 - 2b + 1)=9
<=>(a + b)^2 + (b - 1)^2=9
Vì (b - 1)^2 >=0 nên (a + b)^2 =< 9
=> a + b =< 3.
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