cho a+b+c=2016 và \(\frac{1}{a+b}+\frac{1}{c+a}+\frac{1}{c+a}=\frac{1}{672}\) tính N=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(Cho:a+b+c=2016;\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{2016}\)
Tính:\(S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
Vì \(a+b+c=2016\Rightarrow a=2016-\left(b+c\right);b=2016-\left(a+c\right);c=2016-\left(a+b\right)\)
Ta có:\(S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(S=\frac{2016-\left(b+c\right)}{b+c}+\frac{2016-\left(a+c\right)}{a+c}+\frac{2016-\left(a+b\right)}{a+b}\)
\(S=\frac{2016}{b+c}-1+\frac{2016}{a+c}-1+\frac{2016}{a+b}-1\)
\(S=2016.\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)
\(S=2016.\frac{1}{2016}-3\)
\(S=-2\)
Cho a+b+c=2016 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{90}\)
Tính \(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
Cho 3 số a,b,c thỏa mãn: a+b+c=2016 và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2016}\) Tính : A=(a2016-b2016)(b2016-c2016)(c2016-a2016)
Cho a,b,c thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)và a, b trái dấu.
Tính \(M=\left(a^{15}+b^{15}\right)\left(b^4+c^4\right)\left(c^{2016}+a^{2016}\right)\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{c-a-b-c}{c\left(a+b+c\right)}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{-\left(a+b\right)}{c\left(a+b+c\right)}\)
\(\Leftrightarrow c\left(a+b\right)\left(a+b+c\right)=-ab\left(a+b\right)\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2\right)+ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[c\left(a+c\right)+b\left(a+c\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
TH1 : \(a+b=0\Leftrightarrow a=-b\)
\(M=\left(-b^{15}+b^{15}\right)\left(b^4+c^4\right)\left(c^{2016}+a^{2016}\right)\)
\(M=0\left(b^4+c^4\right)\left(c^{2016}+a^{2016}\right)=0\)
TH2 : \(b+c=0\Leftrightarrow b=-c\)
Đến đây tịt :) bác nào biết giải tiếp giúp Nghị Hồng Vân Anh
đề cho a,b trái dấu rồi nên có một trường hợp thôi nha Trần Thanh Phương, cảm ơn bạn
thực hiện phép tính
A=\(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}+6^{11}}\)
b, cho a+b+c=2016 và \(\frac{1}{a+b}=\frac{1}{b+c}=\frac{1}{c+a}=\frac{1}{7}\)
tính S=\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)
Cho : a + b + c = 2016 với \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{90}\) Tính : S = \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)
Thay các dữ liệu mà đề bài cho vào S,ta có:
\(S=2016.\frac{1}{90}-3=\frac{97}{5}\)
Vậy S=97/5
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 a+b+c=14 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{7}\) . Tính \(A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
ta có: \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{7}\)
\(\Rightarrow14.\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=14.\frac{1}{7}\)
\(\Rightarrow\frac{14}{a+b}+\frac{14}{b+c}+\frac{14}{c+a}=2\)
mà a+b+c =14
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=2\)
\(\Rightarrow\left(\frac{a+b}{a+b}+\frac{c}{a+b}\right)+\left(\frac{a}{b+c}+\frac{b+c}{b+c}\right)+\left(\frac{a+c}{a+c}+\frac{b}{a+c}\right)=2\)
\(\Rightarrow3+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2\)
\(\Rightarrow A=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2-3\)
\(\Rightarrow A=-1\)
CHÚC BN HỌC TỐT!!!!!!
Cho
\(a+b+c=2016\) và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2016}\)
Cmr a hoặc b hoặc c bằng 2016
tìm x y z biết
\(\sqrt{2016.x^2+4}+\sqrt{2017y^2+9}=9-\sqrt{2019z^2+25}\)
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