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\(\dfrac{a}{2016}=\dfrac{b}{2017}=\dfrac{c}{2018}=\dfrac{a-b}{-1}=\dfrac{b-c}{-1}=\dfrac{c-a}{2}\)
\(\Rightarrow a-b=b-c=-\dfrac{1}{2}\left(c-a\right)\)
\(\Rightarrow M=4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2=4\left(-\dfrac{1}{2}\left(c-a\right)\right)\left(-\dfrac{1}{2}\left(c-a\right)\right)-\left(c-a\right)^2\)
\(\Rightarrow M=\left(c-a\right)^2-\left(c-a\right)^2=0\)
Đặt a/2016 = b/2017 = c/2018 = k => a=2016k
b=2017k
c=2018k
Ta có (a-c)^3=( 2016k-2018k)^3 = (k(2016-2018))^3 = (k(-2))^3 (1)
Ta lại có 8(a-b)^2*(b-c)= 8(2016k-2017k)^2*(2017k-2018k) = 8(k(2016-2017)^2*(k(2017-2018) = 2^3*(k(-1))^2*(k(-1)) = 2^3*k^2*1*k*(-1) = k^3*(-2)^3 = (k(-2))^3 (2)
Từ (1) và (2) suy ra (a-c0^3 = 8(a-b)^2*(b-c)
Nhớ tick mik nha 
b/ Theo đề bài thì ta có:
\(\left\{{}\begin{matrix}f\left(1\right)=f\left(-1\right)\\f\left(2\right)=f\left(-2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a_4+a_3+a_2+a_1+a_0=a_4-a_3+a_2-a_1+a_0\\16a_4+8a_3+4a_2+2a_1+a_0=16a_4-8a_3+4a_2-2a_1+a_0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a_3+a_1=0\\4a_3+a_1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a_3=0\\a_1=0\end{matrix}\right.\)
Ta có: \(f\left(x\right)-f\left(-x\right)=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0-\left(a_4x^4-a_3x^3+a_2x^2-a_1x+a_0\right)\)
\(=2a_3x^3+2a_1x=0\)
Vậy \(f\left(x\right)=f\left(-x\right)\)với mọi x
a/ Áp dụng tính chất dãy tỷ số bằng nhau ta có:
\(\dfrac{a}{2015}=\dfrac{b}{2016}=\dfrac{c}{2017}=\dfrac{a-b}{-1}=\dfrac{b-c}{-1}=\dfrac{c-a}{2}\)
\(\Rightarrow c-a=-2\left(a-b\right)=-2\left(b-c\right)\)
Thế vào B ta được
\(B=4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2\)
\(=4\left(a-b\right)\left(b-c\right)-\left[-2\left(a-b\right).\left(-2\right).\left(b-c\right)\right]\)
\(=4\left(a-b\right)\left(b-c\right)-4\left(a-b\right)\left(b-c\right)=0\)
Đặt \(\dfrac{a}{2017}=\dfrac{b}{2018}=\dfrac{c}{2019}=k\Rightarrow a=2017k;b=2018k;c=2019k\)
M = 4(2017k - 2018k)(2018k - 2019k) - (2019k - 2017k)2
= 4(-k)(-k) - (2k)2
= 4k2 - 4k2
= 0
b) Tìm min
\(SV=\left|x-2016\right|+\left|x-2017\right|+\left|x-2018\right|\)
\(SV=\left|x-2016\right|+\left|2018-x\right|+\left|x-2017\right|\)
\(SV\ge\left|x-2016+2018-x\right|+\left|x-2017\right|=2+\left|x-2017\right|\ge2\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}2016\le x\le2018\\x=2017\end{matrix}\right.\Leftrightarrow x=2017\)
3) \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=676\)
\(\Rightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=676\)
\(\Rightarrow\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=673\)
a, Đặt \(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5}=k\Rightarrow\left\{{}\begin{matrix}a=3k\\b=4k\\c=5k\end{matrix}\right.\)
Ta có: \(4\left(a-b\right)\left(b-c\right)\)
\(=4\left(3k-4k\right)\left(4k-5k\right)\)
\(=4.\left(-k\right).\left(-k\right)=4k^2\) (1)
\(\left(a-c\right)^2=\left(3k-5k\right)^2=4k^2\) (2)
Từ (1), (2) \(\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\)
\(\Rightarrowđpcm\)
Ta có:\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}=\dfrac{2\left(x+y+x\right)}{x+y+z}=2\)(theo tính chất của DTSBN)
Suy ra:\(\dfrac{1}{x+y+z}=2\)=>x+y+z=\(\dfrac{1}{2}\)
=>y+z=\(\dfrac{1}{2}\)-x
Tương tự, ta có được:
x+z=\(\dfrac{1}{2}-y\)
x+y=\(\dfrac{1}{2}-z\)
Thay các kết quả vừa tìm được, ta có:
\(\dfrac{0,5-x+1}{x}=\dfrac{0,5-y+2}{y}\dfrac{0,5-z-3}{z}=2\)=>\(\dfrac{1,5-x}{x}=\dfrac{2,5-y}{y}=\dfrac{-2,5-z}{z}=2\)
=>x=\(\dfrac{1}{2},y=\dfrac{5}{6},z=\dfrac{-5}{6}\)
Thay x=\(\dfrac{1}{2},y=\dfrac{5}{6},z=\dfrac{-5}{6}\)vào biểu thức A, ta có:
A=2018.\(\dfrac{1}{2}\)+\(\left(\dfrac{5}{6}\right)^{2017}\)+\(\left(\dfrac{-5}{6}\right)^{2017}\)
=>A=1009+\(\left[\left(\dfrac{5}{6}\right)^{2017}+\left(\dfrac{-5}{6}\right)^{2017}\right]\)
=>A=1009+0
=>A=1009
Vậy giá trị của biểu thức A là 1009
Đặt a/2016=b/2017=c/2018=k
=>a=2016k; b=2017k; c=2018k
M=4(a-b)(b-c)(c-a)^2
=4*(2016k-2017k)(2017k-2018k)(2016k-2018k)^2
=4*(-k)*(-k)*(-2k)^2
=4k^2*4k^2=16k^4