cho a+b+c=2017 và \(\frac{1}{a+b}\) + \(\frac{1}{b+c}\)+ \(\frac{1}{c+a}\)= \(\frac{1}{10}\).
Tính M= \(\frac{a}{b+c}\)+\(\frac{b}{c+a}\)+\(\frac{c}{a+b}\)
Cho a,b,c thỏa mãn \(a+b+c\ne0\) và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\left(a,b,c\ne0\right)\) . Chứng minh: \(\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{a^{2017}+b^{2017}+c^{2017}}\)
ta có
\(\frac{bc+ac+ab}{abc}=\frac{1}{a+b+c}\)
\(3+\frac{bc\left(b+c\right)+ac\left(b+c\right)+ab\left(a+b\right)}{abc}=0\)
\(\frac{b^2c+bc^2}{abc}>0\)
tương tự các phân thức còn lại suy ra a=b=c
cho a+b+c=2017 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{90}\)
Tính S=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
=> (a+b+c).(1/a+b + 1/b+c +1/c+a) = 2017/90
=> a+b+c/a+b + a+b+c/b+c + a+b+c/c+a = 2017/90
=> 1 + c/a+b + 1 + a/b+c + 1 + b/c+a = 2017/90
=> a/b+c + b/c+a +c/a+b = 2017/90 - 3 = 1747/90
Vậy S = 1747/90
Tk mk nha
Cho a+b+c khác 0;a,b,c khác 0 và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
a Chứng minh \(\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{a^{2107}+b^{2017}+c^{2017}}\)
b Tổng quát bài toán trên
a ) \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{c-\left(a+b+c\right)}{ac+bc+c^2}\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2\right)+ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ab+bc+c^2+ac\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
=> a = - b hoặc b = - c hoặc a = - c
Xét a = - b ta có :
\(\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\left(\frac{1}{-b^{2017}}+\frac{1}{b^{2017}}\right)+\frac{1}{c^{2017}}=\frac{1}{c^{2017}}\) (1)
\(\frac{1}{a^{2017}+b^{2017}+c^{2017}}=\frac{1}{\left(-b^{2017}+b^{2017}\right)+c^{2017}}=\frac{1}{c^{2017}}\) (2)
Từ (1) ; (2) => \(\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{a^{2017}+b^{2017}+c^{2017}}\)
Tới đây bạn xét tiếp 2 TH b = - c và c = - a nữa ta có đpcm nha
b ) TQ :
Nếu a +b +c khác 0; a;b;c khác 0 ; \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\) thì \(\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}\)
Cho a + b + c = 2017 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}=\frac{1}{2017}\)
Tính A = \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
Mk rất gấp, các bn giúp mk vs!!! Mk sẽ tick cho ai trả lời nhanh và đúng nhất
Ta có :
\(A+3=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+3\)
\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)
\(=2017.\frac{1}{2017}=1\)
\(\Rightarrow A=1-3=-2\)
Cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\).Chứng minh \(\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{a^{2017}+b^{2017}+c^{2017}}\)
Có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{a.b}=\frac{-\left(a+b\right)}{c\left(a+b+c\right)}\)
\(\Leftrightarrow\left(a+b\right)c\left(a+b+c\right)=-\left(a+b\right)ab\)
\(\Leftrightarrow\left(a+b\right)\left(ca+cb+c^2+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(c+a\right)\left(c+b\right)=0.\)
Vậy: hoặc a + b = 0 hoặc c + a = 0 hoặc c + b =0.
Vai trò của a, b, c như nhau nên giả sử \(a+b=0\Leftrightarrow a=-b.\)
Khi đó: \(\frac{1}{a^{2007}}+\frac{1}{b^{2007}}+\frac{1}{c^{2007}}=\frac{1}{a^{2007}}+\frac{1}{\left(-a\right)^{2007}}+\frac{1}{c^{2007}}=\frac{1}{c^{2007}}.\)
\(\frac{1}{a^{2007}+b^{2007}+c^{2007}}=\frac{1}{a^{2007}+\left(-a\right)^{2007}+c^{2007}}=\frac{1}{c^{2007}}.\)
Vậy: \(\frac{1}{a^{2007}}+\frac{1}{b^{2007}}+\frac{1}{c^{2007}}=\frac{1}{a^{2007}+b^{2007}+c^{2007}}.\)(đpcm).
cho a, b, c khác 0 và a+b+c khác 0
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\) Chứng minh rằng: a; b; có 2 số đối nhau
Từ đó suy ra \(\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{a^{2017}+b^{2017}+c^{2017}}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{c-\left(a+b+c\right)}{c\left(a+b+c\right)}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{-a-b}{ac+bc+c^2}\)
\(\Leftrightarrow-\left(a+b\right)ab=\left(a+b\right)\left(ac+bc+c^2\right)\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2\right)+\left(a+b\right)ab=0\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
=> a = - b hoặc b = - c hoặc c = - a
Xét a = - b ta có \(\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{-b^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{c^{2017}}\)(1)
\(\frac{1}{a^{2017}+b^{2017}+c^{2017}}=\frac{1}{\left(-b^{2017}+b^{2017}\right)+c^{2017}}=\frac{1}{c^{2017}}\)(2)
Từ (1);(2) \(\Rightarrow\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{a^{2017}+b^{2017}+c^{2017}}\)
Xét tiếp 2 TH b = - c hoặc c = - a nữa ta có đpcm nha
Cho \(a+b+c=\frac{1}{2017}\) và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2017\). Tính \(\left(a^{2009}+b^{2009}+c^{2009}\right)\left(\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}\right)\)
cho a,b,c \(\in\)R thỏa mãn a+b+c=2018 và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2018}\)
tính M=\(\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Leftrightarrow\frac{\left(ab+bc+ac\right).\left(a+b+c\right)-abc}{abc.\left(a+b+c\right)}=0\Leftrightarrow\left(ab+bc+ac\right).\left(a+b+c\right)-abc=0\)
\(\Leftrightarrow\left(a+b\right).\left(a+c\right).\left(c+b\right)=0\Leftrightarrow\orbr{\begin{cases}a=-b\\a=-c\end{cases}\text{hoac }c=-b}\)
thay vào rồi tính (nhớ đưa dấu âm lên tử nha) còn phần phan tích sẽ giải thích sau-bây h bận >:
\(\left(a+b+c\right).\left(ab+ac+bc\right)-abc=0\)
\(\Leftrightarrow a^2c+a^2b+abc+b^2a+b^2c+abc+c^2a+c^2b=0\)
\(\Leftrightarrow\left(abc+a^2c\right)+\left(abc+b^2c\right)+\left(a^2b+ab^2\right)+\left(c^2a+c^2b\right)=0\)
\(\Leftrightarrow ac.\left(a+b\right)+cb.\left(a+b\right)+ab.\left(a+b\right)+c^2.\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right).\left(ac+cb+ab+c^2\right)=0\)
\(\Leftrightarrow\left(a+b\right).\left[c\left(a+c\right)+b.\left(a+c\right)\right]=\left(a+b\right).\left(a+c\right).\left(c+b\right)=0\)
~~ cách này dài dòng >: but t ko nghĩ đc cách nào ngắn hưn =(
cho a + b + c = 2017 thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)=\(\frac{1}{2017}\) tính M = a2017 + b2017 + c2017
Cái đề là \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2017}???\)