\(\frac{1+2}{15}=\frac{3b}{23+7a}=\frac{7-3a}{20}\)
=\(\left(1+2a\right).\left(23+7a\right)=3b+15=\frac{7-3a}{20}\)
=\(\left[1+a\left(2+1\right)\right].\left[23+a\left(7+1\right)\right]=3b.15=\frac{7-3a}{20}\)
=[1+a3].[23+a8]=3b.15=7-3a/20
a[23+1(3+8+2)]=3b.15=7-3a/20
a.36=3b.15=7-3a/20
=>\(\frac{15}{a}=\frac{36}{3b}=\frac{7-3a}{20}=>\frac{15}{a}=12b=\frac{7-3a}{20}\)
con lai tu tinh nha
Tìm a, b biết \(\frac{1+2a}{15}=\frac{3b}{23+7a}=\frac{7-3a}{20}\)
1) Rút gọn :
\(B=\frac{\left(a+2b\right)^3-\left(a-2b\right)^3}{\left(2a+b\right)^3-\left(2a-b\right)^3}:\frac{3a^4+7a^2b^2+3b^4}{4a^4+7a^2b^2+3b^4}\)
Cho 2a-b=7.Tìm GTBT \(P=\frac{5a-b}{3a+7}-\frac{3b-2a}{2b-7}\)
cho a,b,c>0 thỏa mãn a+b+c=2016
Tìm GTNN P=\(\frac{2a+3b+3c+1}{2015+a}+\frac{3a+2b+3c}{2016+b}+\frac{3a+3b+2c-1}{2017+c}\)
Cho a,b,c>0 thỏa mãn a+b+c=2016
Tìm GTNN P=\(\frac{2a+3b+3c-1}{2015+a}+\frac{3a+2b+3c}{2016+b}+\frac{3a+3b+2c+1}{2017+c}\)
\(B=\frac{3a-b}{2a+7}+\frac{3b-a}{2b-7}vsa-b=7\left(a\ne-\frac{7}{2};b\ne\frac{7}{2}\right)\)
Cho a;b;c là các số dương thay đổi thỏa mãn \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=2017\)
TÍNH \(MaxP=\frac{1}{2a+3b+3c}+\frac{1}{3a+2b+3c}+\frac{1}{3a+3b+2c}\)
Cho a, b, c >0. Chứng minh:
a)\(\frac{1}{2a+3b+3c}\) +\(\frac{1}{2b+3c+3a}\) +\(\frac{1}{2c+3a+3b}\) \(\le\) \(\frac{1}{4}\) (\(\frac{1}{a+b}\) +\(\frac{1}{b+c}\) +\(\frac{1}{c+a}\) )
b)\(\frac{1}{a+2b+3c}\) +\(\frac{1}{b+2c+3a}\) +\(\frac{1}{c+2a+3b}\) \(\le\) \(\frac{1}{2}\) (\(\frac{1}{a+2c}\) +\(\frac{1}{b+2a}\) +\(\frac{1}{c+2b}\) )
cm voi moi so duong a b c thi
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\left(1+\sqrt{2}+\sqrt{3}\right)\left(\frac{1}{a+\sqrt{2b}+\sqrt{3a}}+\frac{1}{b+\sqrt{2c}+\sqrt{3a}}+\frac{1}{c+\sqrt{2a}+\sqrt{3b}}\right)\)
