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

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

\(\Rightarrowđpcm\)

Cứu mình với 9:00 sáng nay mình nộp bài rùi

bạn ơi bạn có câu trả lời chưa, cho mik xin vs

b) trước hết ta cần chứng minh nếu x+y+z=0 thì x^3+y^3+z^3=3xyz

ta có x+y+z=0==> x=-(y+z) 

             <=> \(x^3=-\left(y^3+z^3+3yz\left(y+z\right)\right)\)

           <=> \(x^3+y^3+z^3=-3yz\left(y+z\right)\)

      <=> \(x^3+y^3+z^3=3xyz\)( cì y+z=-x)

 áp dụng vào bài ta có \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

 do đó M=\(\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc\cdot\frac{3}{abc}=3\)

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

TH1: Nếu a+b+c \(\ne0\)

áp dụng tính chất của dãy tỉ số bằng nhau ta có:

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

mà \(\frac{a+b-c}{c}+1=\frac{b+c-a}{a}+1=\frac{c+a-b}{b}+1=2\)

\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=2\)

Vậy \(B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\left(\frac{a+b}{a}\right)\left(\frac{a+c}{c}\right)\left(\frac{b+c}{b}\right)=8\)

TH2 : Nếu a+b+c = 0

áp dụng tính chất của dãy tỉ số bằng nhau ta có :

        \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c+b+c-a+c+a-b}{a+b+c}=0\)

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

\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=1\)

vậy \(B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\left(\frac{a+b}{a}\right)\left(\frac{a+c}{c}\right)\left(\frac{b+c}{b}\right)=1\)

\(\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{c+a-b}{b}+2\)

\(\Leftrightarrow\frac{a+b+c}{c}=\frac{a+b+c}{b}=\frac{a+b+c}{a}\)

TH1: a+b+c=0 

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

TH2: a+b+c khác 0

 \(\Rightarrow a=b=c\Rightarrow B=\left(1+\frac{a}{a}\right).\left(1+\frac{a}{a}\right).\left(1+\frac{a}{a}\right)=2^3=8\)

a)

⇒ \(\frac{11x-1}{4}=\frac{10}{4}\)

⇒ 11x - 1 = 10

11x = 10 + 1 = 11

x = 11 : 11 = 1

b)

\(\left[{}\begin{matrix}3x-6=0\\\frac{x}{9}-\frac{1}{3}=0\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}3x=0+6\\\frac{x}{9}=0+\frac{1}{3}\end{matrix}\right.\)⇒ \(\left[{}\begin{matrix}3x=6\\\frac{x}{9}=\frac{1}{3}\end{matrix}\right.\)⇒ \(\left[{}\begin{matrix}x=6:3\\\frac{x}{9}=\frac{3}{9}\end{matrix}\right.\)⇒\(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy x = 2 hoặc x = 3

c)

\(M=c\left(\frac{5}{7}+\frac{7}{14}-\frac{17}{14}\right)\)

\(M=c\left(\frac{10}{14}+\frac{7}{14}-\frac{17}{14}\right)\)

\(M=\left(\frac{2018}{2019}-\frac{2019}{2020}\right).0\)

M = 0

d)

\(N=\frac{-7}{13}+2-\frac{19}{13}+\frac{2020}{2018}.\frac{2018}{202}\)

\(N=\left(\frac{-7}{13}-\frac{19}{13}\right)+2+10\)

N = \(-2+2+10\)

N = 10

Tí ăn xong giải tiếp

Câu 3a này cái cuối là 1/2018.2020 mới đúng chứ