Cho cac so a,b,c va thoa man\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}=2\)Tinh gia tri bieu thuc \(P=\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\)
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 cac so a,b,c,d thoa man: \(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{d+a+b}=\)
\(\frac{d}{a+b+c}\). Tinh gia tri bieu thuc:
P=\(\frac{a+b}{c+d}=\frac{b+c}{d+a}=\frac{c+d}{b+a}=\frac{d+a}{b+c}\)
Ta có :
\(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{d+a+b}=\frac{d}{a+b+c}\)
\(\Leftrightarrow\)\(\frac{a}{b+c+d}+1=\frac{b}{c+d+a}+1=\frac{c}{d+a+b}+1=\frac{d}{a+b+c}+1\)
\(\Leftrightarrow\)\(\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{c+d+a}=\frac{a+b+c+d}{d+a+b}=\frac{a+b+c+d}{a+b+c}\)
Ta thấy các tử bằng nhau suy ra các mẫu bằng nhau
\(\Rightarrow\)\(b+c+d=c+d+a=d+a+b=a+b+c\)
\(\Rightarrow\)\(a=b=c=d\)
\(\Rightarrow\)\(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{b+a}+\frac{d+a}{b+c}=1+1+1+1=4\)
Đề bị nhầm đúng ko bạn ^^
Cho a,b,c la cac so duong thoa man a+b+c=9.Tim gia tri nho nhat cua bieu thuc:
\(P=a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2}+c^2+\frac{1}{c^2}\)
Ta có:\(P=a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2}+c^2+\frac{1}{c^2}\)
\(\Rightarrow P\ge a^2+b^2+c^2+\frac{9}{a^2+b^2+c^2}\)(bđt cauchy-schwarz)
\(P\ge\frac{a^2+b^2+c^2}{81}+\frac{9}{a^2+b^2+c^2}+\frac{80\left(a^2+b^2+c^2\right)}{81}\)
\(\Rightarrow P\ge\frac{2}{3}+\frac{80\left(a^2+b^2+c^2\right)}{81}\left(AM-GM\right)\)
Sử dụng đánh giá quen thuộc:\(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}=27\)
\(\Rightarrow P\ge\frac{2}{3}+\frac{80\cdot27}{81}=\frac{82}{3}\)
"="<=>a=b=c=3
Gia su a,b,c la cacso thoa man a+b+c=259 va \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}=15\). Khi do gia tri cua bieu thuc \(Q=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)bang
Ta có
\(Q+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\left(\frac{a}{b+c}+\frac{b+c}{b+c}\right)+\left(\frac{b}{a+c}+\frac{a+c}{a+c}\right)+\left(\frac{c}{a+b}+\frac{a+b}{a+b}\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)
\(=259.15\)
\(\Rightarrow Q=259.15-3=3885\)
cho a,b,c la cac so thuc duong thoa man a+b+c=3. tim gia tri nho nhat cua
P=\(\frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}\)
nhận được thông báo thì kéo chuột xuống xem bài giải của t ở phần duyệt bài nhé
Cho a,b,c là cac so thoa man dieu kien \(\frac{2a-b}{a+b}=\frac{b-c+a}{2a-b}=\frac{2}{3}\)
Khi đo gia tri cua bieu thuc \(P=\frac{\left(5b+4a\right)^5}{\left(5b+4c\right)^2.\left(a+3c\right)^3}\)
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 khac 0 thoa man: \(\frac{2a+b+c}{a}\)=\(\frac{2b+c+a}{b}\)=\(\frac{2c+a+b}{c}\)
Tinh gia tri cua bieu thuc: P=\(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
GIUP MINH VOI NHA!
Áp dụng tính chất của dãy tỉ số bằng nhau,ta có:
\(\frac{2a+b+c}{a}=\frac{2b+c+a}{b}=\frac{2c+a+b}{c}=\frac{2a+b+c+2b+c+a+2c+a+b}{a+b+c}=\frac{4\left(a+b+c\right)}{a+b+c}=4\)
\(\Rightarrow\frac{2a+b+c}{a}=4\Rightarrow2a+b+c=4a\Rightarrow b+c=4a-2a=2a\)
\(\frac{2b+c+a}{b}=4\Rightarrow2b+c+a=4b\Rightarrow c+a=4b-2b=2b\)
\(\frac{2c+a+b}{c}=4\Rightarrow2c+a+b=4c\Rightarrow a+b=4c-2c=2c\)
Suy ra \(P=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2c.2a.2b}{abc}=\frac{8abc}{abc}=8\)
Vậy P=8
Cho hỏi tớ sai chỗ nào ạ :>?Góp ý giúp nha?
tinh gia tri cua bieu thuc A=\(\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\))\(\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)\). cho biet a+b+c=0