Ta có :

\(\frac{a+b-b-c}{2018-2019}=\frac{a-c}{-1}\)

\(\frac{b+c-c-a}{2019-2020}=\frac{b-a}{-1}\)

\(\frac{b-c}{2018-2020}=\frac{b-c}{-2}\)     

Đặt \(\frac{a-c}{-1}=\frac{b-a}{-1}=\frac{b-c}{-2}=k\left(k\ne0\right)\)

\(\Rightarrow\hept{\begin{cases}\frac{a-c}{-1}=k\\\frac{b-a}{-1}=k\\\frac{b-c}{-2}=k\end{cases}\Rightarrow\hept{\begin{cases}a-c=-k\\b-a=-k\\b-c=k.\left(-2\right)\end{cases}}}\)

\(\Rightarrowđpcm\)

\(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)-3abc+c^3=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-ac-bc+c^2-3ab\right]=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(a;b;c>0\Rightarrow a+b+c>0\)

\(\Rightarrow a^2+b^2+c^2-ab-bc-ac=0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)

\(P=0\)

\(a^3+b^3+c^3=3abc\Leftrightarrow a+b+c=0\)(bổ đề này khá phổ biến ,bạn có thế search gg mk hỏi lười )

sau đó thay vào xem được ko bạn ^_^

Đặt \(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}=k\)

\(\Rightarrow a=2018k\), \(b=2019k\), \(c=2020k\)

Ta có: \(4\left(a-b\right)\left(b-c\right)=4\left(2018k-2019k\right)\left(2019k-2020k\right)\)

                                                 \(=4.\left(-k\right).\left(-k\right)=4k^2=\left(2k\right)^2\)

Ta lại có: \(\left(a-c\right)^2=\left(2018k-2020k\right)^2=\left(-2k\right)^2=\left(2k\right)^2\)

Vậy \(4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\)

Đặt \(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}=k\Rightarrow\hept{\begin{cases}a=2018k\\b=2019k\\c=2020k\end{cases}}\)

Thế vị trí tương ứng ta được :

VT = 4( a - b )( b - c )

       = 4( 2018k - 2019k )( 2019k - 2020k )

       = 4(-k)(-k)

       = 4k²

VP = ( a - c )² 

       = ( 2018k - 2020k )²

       = ( -2k )²

       = 4k²

=> VT = VP

=> đpcm

Đặt \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)\rightarrow\left(x,y,z\right)\)

Khi đó:\(\left(\frac{c}{a-b},\frac{a}{b-c},\frac{b}{c-a}\right)\rightarrow\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\)

Ta có:

\(P\cdot Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)

Mặt khác:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-bc+ac-a^2}{ab}\cdot\frac{c}{a-b}\)

\(=\frac{c\left(a-b\right)\left(c-a-b\right)}{ab\left(a-b\right)}=\frac{c\left(c-a-b\right)}{ab}=\frac{2c^2}{ab}\left(1\right)\)

Tương tự:\(\frac{x+z}{y}=\frac{2a^2}{bc}\left(2\right)\)

\(=\frac{x+y}{z}=\frac{2b^2}{ac}\left(3\right)\)

Từ ( 1 );( 2 );( 3 ) ta có:
\(P\cdot Q=3+\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)

Ta có:\(a+b+c=0\)

\(\Rightarrow\left(a+b\right)^3=-c^3\)

\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

Khi đó:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)

Mách mk nốt 2 bài kia vs

chiju

Đề có sai ko bạn sao lại c-d ?

Sửa đề : Cần chứng minh \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)

Đặt :\(\frac{a}{2017}=\frac{b}{2018}=\frac{c}{2019}=k\)

\(\Rightarrow\hept{\begin{cases}a=2017k\\b=2018k\\c=2019k\end{cases}}\)

Khi đó :

\(4\left(a-b\right)\left(b-c\right)=4\left(2017k-2018k\right)\left(208k-2019k\right)\)

\(=4\cdot\left(-k\right)\cdot\left(-k\right)=4k^2\)

\(\left(c-a\right)^2=\left(2019k-2017k\right)^2=\left(2k\right)^2=4k^2\)

Do đó : \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\) (đpcm)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)  hinh nhu theo co dieu kien a,b,c  ko dong thoi = 0

<=> \(\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)

<=>  \(\frac{a+b}{ab}=\frac{c-a-b-c}{c\left(a+b+c\right)}\)

<=> \(\left(a+b\right)\left(ac+bc+c^2\right)=-ab\left(a+b\right)\)

<=> \(\left(a+b\right)\left(ac+bc+c^2\right)+ab\left(a+b\right)=0\)

<=> \(\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)

<=> \(\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)

<=> a+b=0 hoac a+c=0 hoac b+c=0

do khi luy thua a,b,c len cach so mu le la 27,41,2019 thi a,b,c ko doi dau nen \(a^{27}+b^{27}=0.hoac.b^{41}+c^{41}=0.hoac.c^{2019}+a^{2019}=0\)

P = 0 

Vay P = 0 

Study well

Ta có : \(\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}-\frac{1}{a}\Rightarrow\frac{b+c}{bc}=\frac{a-a-b-c}{a^2+ab+ac}\)

\(\Leftrightarrow\frac{b+c}{bc}=\frac{-b-c}{a^2+ab+ac}\Leftrightarrow\left(b+c\right)\left(a^2+ab+ac\right)=-\left(b+c\right)bc\)

\(\left(b+c\right)\left(a^2+ab+ac\right)+\left(b+c\right)bc=0\)

\(\Rightarrow\left(b+c\right)\left(a^2+ab+ac+bc\right)=0\)

\(\Leftrightarrow\left(b+c\right)[\left(a+b\right)a+c\left(a+b\right)]=0\)

\(\Leftrightarrow\left(b+c\right)\left(a+b\right)\left(a+c\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}b=-c\\\orbr{\begin{cases}a=-b\\c=-a\end{cases}}\end{cases}\Leftrightarrow\orbr{\begin{cases}b^{41}+c^{41}=0\\\orbr{\begin{cases}a^{27}+b^{27}=0\\c^{2019}+a^{2019}=0\end{cases}}\end{cases}}}\)\(\Leftrightarrow\orbr{\begin{cases}b=-c\\\orbr{\begin{cases}a=-b\\c=-a\end{cases}}\end{cases}\Leftrightarrow\orbr{\begin{cases}b^{41}+c^{41}=0\\\orbr{\begin{cases}a^{27}+b^{27}=0\\a^{2019}+c^{2019}=0\end{cases}}\end{cases}}}\)

\(\Rightarrow\)b= -c hoặc   a= -b hoặc c= -a           

\(\Rightarrow\)\(b^{41}+c^{41}=0\)hoặc     \(a^{27}+b^{27}=0\)hoặc    \(c^{2019}+a^{2019}=0\)

Trong 3 trường hợp trên , thì P luôn bằng 0 do có một thừa số bằng 0 

                               Vậy P =0

Đặt \(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=2018k\\b=2019k\\c=2020k\end{matrix}\right.\)

\(\Rightarrow\left(a-c\right)^3=\left(2018k-2020k\right)^3=\left(-2k\right)^3=-8k^3\) (1)

\(8\left(a-b\right)^2.\left(b-c\right)=8\left(2018k-2019k\right)^2.\left(2019k-2020k\right)=8k^2\left(-k\right)=8\left(-k\right)^3=-8k^3\left(2\right)\)

Từ (1) và (2) ⇒ \(\left(a-c\right)^3=8\left(a-b\right)^2.\left(b-c\right)\left(đpcm\right)\)

mn giúp mk vs

chiều mk nộp rùi

Méo bt trẩu là gì à =))

Bảo ezzz thì chỉ hộ cách làm ko bt thì đừng cư xử như 1 đứa trẻ trâu=))