cho abc = 2020. Tính \(T=\frac{2020a}{2020+2020a+ab}+\frac{2020b}{2020+2020b+bc}+\frac{2020c}{2020+2020c+ca}\)
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Cho a,b,c là các số dương thỏa mãn abc=1. Chứng minh rằng:
\(\frac{1}{a+2020}+\frac{1}{b+2020}+\frac{1}{c+2020}\le\frac{1}{2020a+1}+\frac{1}{2020b+1}+\frac{1}{2020c+1}\)
giúp mk vs mk đang cần gấp
Cho các số dương a,b,c thỏa mãn a+b+c=2020
Tìm Max của biểu thức \(P=\frac{a}{a+\sqrt{2020a+bc}}+\frac{b}{b+\sqrt{2020b+ca}}+\frac{c}{c+\sqrt{2020c+ab}}\)
\(\sqrt{a\left(a+b+c\right)+bc}=\sqrt{\left(a+b\right)\left(c+a\right)}\ge\sqrt{\left(\sqrt{ac}+\sqrt{ab}\right)^2}=\sqrt{ac}+\sqrt{ab}\)
\(\Rightarrow\frac{a}{a+\sqrt{2020a+bc}}\le\frac{a}{a+\sqrt{ac}+\sqrt{ab}}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)
Tương tự: \(\frac{b}{b+\sqrt{2020b+ca}}\le\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\) ; \(\frac{c}{c+\sqrt{2020c+ab}}\le\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)
Cộng vế với vế: \(P\le1\)
Dấu "=" xảy ra khi \(a=b=c=...\)
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Cho abc=2020. Rút gọn A=\(\frac{2020a}{ab+2020a+2020}+\frac{b}{bc+b+2020}+\frac{c}{ac+c+1}\)
thay 2020 = abc vào biểu thức A ta được :
\(A=\frac{2020a}{ab+2020a+2020}+\frac{b}{bc+b+2020}+\frac{c}{ac+c+1}\)
\(\Rightarrow A=\frac{abc.a}{ab+abc.a+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)
\(\Rightarrow A=\frac{abc.a}{ab\left(1+ac+c\right)}+\frac{b}{b\left(c+1+ac\right)}+\frac{c}{ac+c+1}\)
\(\Rightarrow A=\frac{ac}{ac+c+1}+\frac{1}{ac+c+1}+\frac{c}{ac+c+1}\)
\(\Rightarrow A=\frac{ac+1+c}{ac+c+1}=1\)
VẬy A=1
Cho các phân số : ab/a+2020b =3/2 , bc/b+2020c = 4/3 ,ac/c+2020a = -12/5
Rút gọn phân số : T= abc/ab+bc+ca
tính A=( 2020a^2+2021 bc/a^2+2bc)+(2020b^2+2021ac/b^2+2ac)+(2020c^2+2021ab/c^2+2ab) biết 1/a +1/b+1/c=0
Cho \(a,b,c\ge0\)Thỏa mãn: a + b + c = 1010
Chứng minh: \(\sqrt{2020a+\frac{\left(b-c\right)^2}{2}}+\sqrt{2020b+\frac{\left(c-a\right)^2}{2}}+\sqrt{2020c+\frac{\left(a-b\right)^2}{2}}\le2020\sqrt{2}\)
\(\sqrt{2020a+\frac{\left(b-c\right)^2}{2}}\le\sqrt{2020a+\frac{\left(b+c\right)^2}{2}}=\sqrt{2020a+\frac{\left(1010-a\right)^2}{2}}\)
\(=\sqrt{\frac{1}{2}\left(a^2+2020a+1010^2\right)}=\frac{1}{\sqrt{2}}\left(a+1010\right)\)
=> \(VT\le\frac{1}{\sqrt{2}}\left(a+b+c+3.1010\right)=2020\sqrt{2}\)
Dấu "=" xảy ra khi a=1010;b=0;c=0 và các hoán vị
Cho \(\frac{a}{c}\)= \(\frac{b}{d}\)
Chứng minh :a) \(\frac{a+2020b}{a-2020b}\) = \(\frac{e+2020d}{e-2020d}\)
b) \(\frac{2020\left(a+c\right)}{2020a}\)= \(\frac{b+d}{b}\)
c) 2a+3c(b+d)=(a+c)(2b+3d)
Giải giúp em nha
Cho \(\frac{a}{c}\)= \(\frac{b}{d}\)
Chứng minh :a) \(\frac{a+2020b}{a-2020b}\) = \(\frac{e+2020d}{e-2020d}\)
b) \(\frac{2020\left(a+c\right)}{2020a}\)= \(\frac{b+d}{b}\)
c) 2a+3c(b+d)=(a+c)(2b+3d)
Giải giúp em nha
a) Áp dụng dãy tỉ số bằng nhau:
\(\frac{a}{c}=\frac{b}{d}=\frac{2020b}{2020d}=\frac{a+2020b}{c+2020d}=\frac{a-2020b}{c-2020d}\)
=> \(\frac{a+2020b}{c+2020d}=\frac{a-2020b}{c-2020d}\)
=> \(\frac{a+2020b}{a-2020b}=\frac{c+2020d}{c-2020d}\)
b) \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\)
Áp dụng dãy tỉ số bằng nhau:
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
=> \(\frac{a}{b}=\frac{a+c}{b+d}\Rightarrow\frac{a}{a+c}=\frac{b}{b+d}\)
=> \(\frac{2020a}{2020\left(a+c\right)}=\frac{b}{b+d}\)
=> \(\frac{2020\left(a+c\right)}{2020a}=\frac{b+d}{b}\)
c) \(2a+3c\left(b+d\right)=\left(a+c\right)\left(2b+3d\right)\)
Câu c sai đề.
Cho a/b=c/d. CMR:
\(\frac{2020a+2019b}{2020a-2019b}=\frac{2020c+2019d}{20120c-2019d}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{2020a}{2020c}=\frac{2019b}{2019d}=\frac{2020a+2019b}{2020c+2019d}=\frac{2020a-2019b}{2020c-2019d}\)
\(\Rightarrow\frac{2020a+2019b}{2020a-2019b}=\frac{2020c+2019d}{2020c-2019d}\)