cho a+b+c=2014 va \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{5}{1007}\)
tinh gtbt S=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
Cho a+b+c=2014 và \(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{1}{2014}\).Tính S=\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(S+3=\left(1+\frac{a}{b+c}\right)+\left(1+\frac{b}{a+c}\right)+\left(1+\frac{c}{a+b}\right)\)
\(S+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)
\(S+3=\frac{2014.1}{2014}=1\Rightarrow S=1-3=-2\)
cho cac so a,b,c va thoa man \(\frac{ab}{a+b}=\frac{1}{3},\frac{bc}{b+c}=\frac{1}{4},\frac{ca}{c+a}=\frac{1}{5}\)Tinh gia tri bieu thuc P=\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Thêm đk \(a,b,c\ne0\)
Ta có: \(\frac{ab}{a+b}=\frac{1}{3}\Rightarrow\frac{a+b}{ab}=3\)
\(\frac{bc}{b+c}=\frac{1}{4}\Rightarrow\frac{bc}{b+c}=4\)
\(\frac{ca}{c+a}=\frac{1}{5}\Rightarrow\frac{c+a}{ca}=5\)
\(\Rightarrow\frac{a+b}{ab}+\frac{b+c}{bc}+\frac{c+a}{ca}=12\)
\(\Leftrightarrow\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}=12\)
\(\Leftrightarrow2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=12\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)
cho a +b+c=2014 vÀ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2014\)
Tinh M = \(\frac{1}{a^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}\)
a, cho day ti so bang nhau : \(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
tinh gia tri bieu thuc M: \(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
b,cho x= \(1+\frac{1}{2013}+\frac{1}{2013^2}+\frac{1}{2013^3}+....+\frac{1}{2013^{2013}}\)
tinh gia tri bieu thuc: S= (2012x+\(\frac{1}{2013^{2013}}\)) : 2013^2014
Cho a, b, c là 3 số thực khác 0
\(\frac{a+b-2017c}{c}=\frac{b+c-2017a}{a}=\frac{c+a-2017b}{b}\)
Tính GTBT: B = \((1+\frac{b}{a}^a)\times(1\times\frac{a}{c})\times1+\frac{b}{c}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a+b-2017c}{c}=\frac{b+c-2017a}{a}=\frac{c+a-2017b}{b}\)
\(=\frac{a+b-2017c+b+c-2017a+c+a-2017b}{a+b+c}=\frac{-2015\left(a+b+c\right)}{a+b+c}=-2015\)
Do đó :
\(\frac{a+b-2017c}{c}=-2015\)\(\Leftrightarrow\)\(a+b=2c\) \(\left(1\right)\)
\(\frac{b+c-2017a}{a}=-2015\)\(\Leftrightarrow\)\(b+c=2a\) \(\left(2\right)\)
\(\frac{c+a-2017b}{b}=-2015\)\(\Leftrightarrow\)\(c+a=2b\) \(\left(3\right)\)
Thay (1), (2) và (3) vào \(B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}\) ta được :
\(B=\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=\frac{8abc}{abc}=8\)
Vậy \(B=8\)
Chúc bạn học tốt ~
Cho a+b+c=0,tính GTBT:
\(\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}+\frac{1}{a^2+b^2-c^2}\)
Câu hỏi của Hattory Heiji - Toán lớp 8 - Học toán với OnlineMath
Em tham khảo link trên!
Bài 1 cho x,y,z>2014 và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{1007}\)
chứng minh rằng \(\sqrt{x+y+z}\ge\sqrt{x-2014}+\sqrt{y-2014}+\sqrt{z-2014}\)
Bài 2
cho a,b,c>0. chứng minh rằng
\(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}\ge\frac{4}{ab+bc+ca}\)
Bài 2 : đã cm bên kia
Bài 1: :|
we had điều này:
\(2=\frac{2014}{x}+\frac{2014}{y}+\frac{2014}{z}\)
\(\Leftrightarrow\frac{x-2014}{x}+\frac{y-2014}{y}+\frac{z-204}{z}=1\)
Xòng! bunyakovsky
P/s : Bệnh lười kinh niên tái phát nên ít khi ol sorry :<
Cho a,b,c khác 0 thỏa mãn \(\frac{ab}{a+b}\)=\(\frac{bc}{b+c}\)=\(\frac{ca}{c+a}\)
Tính M = \(\frac{\left(ab+bc+ca\right)^{1007}}{a^{2014}+b^{2014}+c^{2014}}\)