Cho 2 so thuc duong a,b thoa man a+b<=1.Tim GTNN cua
\(A=\frac{1}{a^3+b^3}+\frac{1}{a^2b}+\frac{1}{ab^2}\)
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 cac so duong a,b,c thoa man : ab+a+b=3
tim GTNN cua bieu thuc C=a^2+b^2
cho 3 so thuc duong a, b, c thoa man 1/a+1/c=2/b. tim GTNN cua (a+b)/(2a-b)+(b+c)(/2c-b)
\(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\Leftrightarrow b=\frac{2ac}{a+c}\)
\(P=\frac{a+b}{2a-b}+\frac{b+c}{2c-b}=\frac{a+\frac{2ac}{a+c}}{2a-\frac{2ac}{a+c}}+\frac{\frac{2ac}{a+c}+c}{2c-\frac{2ac}{a+c}}=\frac{a+3c}{2a}+\frac{3a+c}{2c}=1+\frac{3}{2}\left(\frac{a}{c}+\frac{c}{a}\right)\ge4\)
Dấu "=" xảy ra khi \(a=b=c\)
cho cac so thuc duong thoa man a^100+b^100=a^101+b^101=a ^102+b^102
hay tinh gia trij cuwar bieu thucP= a^2014+b^2015
a^100+b^100=a^101+b^101=a^102+b^102
=>a=1; b=1
P=1^2014+1^2015=2
cho a,b,c la 3 so thuc duong thoa man dieu kien (a+b-c)/c=(b+c-a)/a=(c+a-b)/b tinh gia tri bieu thuc B=(1+b/a)*(1+a/c)*(1+c/b)
(a+b-c)/c+2 =(b+c-a)/c+2 =(c+a-b)/c+2
rồi bạn tự làm tiếp nhé
xét 2 trường hợp
thay vào thôi nhé bạn
Nhớ k cho mình nhé
cho a,b,c la cac so thuc duong thoa man 21ab+2bc+8ac <= 12
khi do gia ti nho nhat cua A=1/a+2/b +3/c
Đặt:⎧⎩⎨⎪⎪⎪⎪⎪⎪a=13xb=45yc=32z{a=13xb=45yc=32z (x,y,z>0)(x,y,z>0)
Khi đó điều kiện đã cho trở thành:3x+5y+7z≤15xyz3x+5y+7z≤15xyz
Áp dụng AM−GMAM−GM ta có:
3x+5y+7z≥15x3y5z7−−−−−−√153x+5y+7z≥15x3y5z715
=>15xyz≥15x3y5z7−−−−−−√15=>x6y5z4≥1.=>15xyz≥15x3y5z715=>x6y5z4≥1.
Ta có:
P=3x+2.54y+3.23z=12(6x+5y+4z)≥12.15x6y5z4−−−−−−√15≥152P=3x+2.54y+3.23z=12(6x+5y+4z)≥12.15x6y5z415≥152 (AM−GM) (AM−GM)
Dấu ′=′′=′ xảy ra <=><=> x=y=z=1x=y=z=1 hay a=13;b=45;c=32
cho 3 so duong thoa man a/(b+c)=b/(a+c)=c/(a+b)
tinh gia tri bieu thuc P=(b+c)/a+(a+c)/b+(a+b)/c
cho a b la cac so nguyen duong thoa man a+20 b+1 chia het cho 21 tim so du trong phep chia cua bieu thuc A=4^a+9^a+a+b cho 21
cho x,y,z la cac so nguyen duong va x+y+z la so le, cac so thuc a,b,c thoa man (a-b)/x=(b-c)/y=(a-c)/z. chung minh rang a=b=c