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Bài 2 :
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2018}\)
Mà \(2018=a+b+c\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{c-a-b-c}{c\left(a+b+c\right)}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{-\left(a+b\right)}{c\left(a+b+c\right)}\)
\(\Leftrightarrow c\left(a+b\right)\left(a+b+c\right)=-ab\left(a+b\right)\)
\(\Leftrightarrow c\left(a+b\right)\left(a+b+c\right)+ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(a+b\right)\left(b+c\right)=0\)
TH1 : \(a+b=0\Leftrightarrow a=-b\)
\(M=\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2014}}=\frac{1}{-b^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2014}}=\frac{1}{c^{2014}}\)
Mà \(a+b+c=2018\)
\(\Leftrightarrow-b+b+c=2018\)
\(\Leftrightarrow c=2018\)
Khi đó \(M=\frac{1}{2018^{2017}}\)
Các trường hợp còn lại tương tự
Kết quả cuối cùng : \(M=\frac{1}{2018^{2017}}\)
Câu hỏi của nguyễn thị phượng - Toán lớp 9 - Học toán với OnlineMath
Em tham khảo bài 2 ở link này nhé!
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Leftrightarrow\frac{\left(ab+bc+ac\right).\left(a+b+c\right)-abc}{abc.\left(a+b+c\right)}=0\Leftrightarrow\left(ab+bc+ac\right).\left(a+b+c\right)-abc=0\)
\(\Leftrightarrow\left(a+b\right).\left(a+c\right).\left(c+b\right)=0\Leftrightarrow\orbr{\begin{cases}a=-b\\a=-c\end{cases}\text{hoac }c=-b}\)
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 =(
B. \(\frac{x+4}{2015}+1+\frac{x+3}{2016}+1=\frac{x+2}{2017}+1+\frac{x+1}{2018}+1\)
<=> \(\frac{x+2019}{2015}+\frac{x+2019}{2016}=\frac{x+2019}{2017}+\frac{x+2019}{2018}\)
<=>(x+2019).(\(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}>0\)
Vì (\(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}>0\)
=> x+2019>0
=>x>-2019
Ta có:
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(a+b+c\right)=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\)
\(=1+\frac{b+c}{a}+1+\frac{c+a}{b}+1+\frac{a+b}{c}\)
=2017.2018=4070306
=> A= \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)=4070306-3=4070303
Ta có: \(x^2+y^2+z^2\ge xy+yz+zx\)
\(\Rightarrow a^{2018}+b^{2018}+c^{2018}\ge\left(ab\right)^{1009}+\left(bc\right)^{1009}+\left(ca\right)^{1009}\)
Dấu = xảy ra \(\Leftrightarrow a=b=c\)
Mà đẳng thức trên xảy ra dấu =
\(\Leftrightarrow a=b=c\Leftrightarrow P=0\)
Bài kia tí nghĩ nốt, khó v
Sửa đề em nhé: \(\frac{2}{ab}-\frac{1}{c^2}=4\) và tính \(a+b+2c\)
Có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{2}{bc}+\frac{2}{ca}+4=4\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{a}=\frac{-1}{c}\\\frac{1}{b}=\frac{-1}{c}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=-c\\b=-c\end{cases}}\)\(\Leftrightarrow a+b+2c=0\)
a) \(\frac{x+2}{2002}\)+\(\frac{x+5}{1999}\)+\(\frac{x+201}{1803}\)=-3
⇔\(\frac{x+2}{2002}\)+\(\frac{x+5}{1999}\)+\(\frac{x+201}{1803}\)+3=0
⇔\(\frac{x+2}{2002}\)+1+\(\frac{x+5}{1999}\)+1+\(\frac{x+201}{1803}\)+1=0
⇔\(\frac{x+2004}{2002}\)+\(\frac{x+2004}{1999}\)+\(\frac{x+2004}{1803}\)=0
⇔(x+2004)(\(\frac{1}{2002}\)+\(\frac{1}{1999}\)+\(\frac{1}{1803}\))=0
Mà (\(\frac{1}{2002}\)+\(\frac{1}{1999}\)+\(\frac{1}{1803}\))≠0
⇒x+2004=0
⇔x=-2004
Vậy tập nghiệm của phương trình đã cho là:S={-2004}
a, Làm
\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x+5}{2016}+\frac{x+6}{2015}\)
<=>\(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}=\frac{x+2021}{2017}+\frac{x+2021}{2016}+\frac{x+2021}{2015}\)
<=>\(\left(x+2021\right)\left(\frac{1}{2020}+\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
<=> x+2021=0
<=> x=-2021
Kl:......................
b, Làmmmmm
\(\frac{2-x}{2004}-1=\frac{1-x}{2005}-\frac{x}{2006}\)
<=> \(\frac{2006-x}{2004}=\frac{2006-x}{2005}+\frac{2006-x}{2006}\)
<=> \(\left(2006-x\right)\left(\frac{1}{2004}-\frac{1}{2005}-\frac{1}{2006}\right)=0< =>2006-x=0\)
<=> x=2006
Kl:..............