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\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{c-a-b-c}{c\left(a+b+c\right)}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{-\left(a+b\right)}{c\left(a+b+c\right)}\)
\(\Leftrightarrow c\left(a+b\right)\left(a+b+c\right)=-ab\left(a+b\right)\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2\right)+ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[c\left(a+c\right)+b\left(a+c\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
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
\(A=\frac{2016a}{ab+2016a+2016}+\frac{b}{bc+b+2016}+\frac{c}{ac+c+1}\)
\(A=\frac{2016a}{ab+2016a+abc}+\frac{b}{bc+b+2016}+\frac{bc}{abc+bc+b}\)
\(A=\frac{2016a}{a\left(b+2016+bc\right)}+\frac{b}{bc+b+2016}+\frac{bc}{2016+bc+b}\)
\(A=\frac{2016}{b+2016+bc}+\frac{b}{bc+b+2016}+\frac{bc}{2016+bc+b}\)
\(A=\frac{2016+b+bc}{2016+b+bc}=1\)
Thay : 2016 = abc
ta có :
\(A=\frac{a^2bc}{ab+a^2bc+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)
\(A=\frac{a^2bc}{ab\left(1+ac+c\right)}+\frac{b}{b\left(c+1+ac\right)}+\frac{c}{ac+c+1}\)
\(A=\frac{ac}{ac+c+1}+\frac{1}{ac+c+1}+\frac{c}{ac+c+1}\)
\(A=\frac{ac+c+1}{ac+c+1}\)
\(A=1\)
vậy \(A=\frac{2016.a}{ab+2016.a+2016}+\frac{b}{bc+b+2016}+\frac{c}{ac+c+1}=1\)
Chúc bạn học tốt !
tìm x y z biết
\(\sqrt{2016.x^2+4}+\sqrt{2017y^2+9}=9-\sqrt{2019z^2+25}\)
đăng bài này nè
Ta có : \(a+b+c=2016\Rightarrow\frac{1}{a+b+c}=\frac{1}{2016}\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{c}-\frac{1}{a+b+c}\right)=0\)
\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left(a+b\right)\left[\frac{c^2+ac+bc+ab}{abc\left(a+b+c\right)}\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(c^2+ac+bc+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}a+b=0\\b+c=0\\c+a=0\end{array}\right.\)
- Nếu a + b = 0 => c = 2016 (1)
- Nếu b + c = 0 => a = 2016 (2)
- Nếu a + c = 0 => b = 2016 (3)
Từ (1) , (2) và (3) ta có điều phải chứng minh.
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:\(S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(S=\frac{2016-\left(b+c\right)}{b+c}+\frac{2016-\left(a+c\right)}{a+c}+\frac{2016-\left(a+b\right)}{a+b}\)
\(S=\frac{2016}{b+c}-1+\frac{2016}{a+c}-1+\frac{2016}{a+b}-1\)
\(S=2016.\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)
\(S=2016.\frac{1}{2016}-3\)
\(S=-2\)