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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
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) \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=676\)
\(\Rightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=676\)
\(\Rightarrow\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=673\)
\(\dfrac{a}{2016}=\dfrac{b}{2017}=\dfrac{c}{2018}=\dfrac{a-b}{-1}=\dfrac{b-c}{-1}=\dfrac{c-a}{2}\)
\(\Rightarrow a-b=b-c=-\dfrac{1}{2}\left(c-a\right)\)
\(\Rightarrow M=4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2=4\left(-\dfrac{1}{2}\left(c-a\right)\right)\left(-\dfrac{1}{2}\left(c-a\right)\right)-\left(c-a\right)^2\)
\(\Rightarrow M=\left(c-a\right)^2-\left(c-a\right)^2=0\)
\(\dfrac{a+b+c}{b+c-2016+a+c+2015+a+b+1}=\dfrac{a+b+c}{2016}\)
Sau đó rút gọn đi rồi tính nha
STUDY WELL 
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a+b+c}{b+c-2016+a+c+2015+a+b+1}=\dfrac{a+b+c}{2016}\\ \Rightarrow\dfrac{a+b+c}{2a+2b+2c}=\dfrac{a+b+c}{2016}\\ \Rightarrow\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2016}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{2015}=\dfrac{b}{2016}=\dfrac{c}{2017}=\dfrac{a-b}{2015-2016}=\dfrac{b-c}{2016-2017}=\dfrac{c-a}{2015-2017}\\ \Rightarrow\dfrac{a-b}{-1}=\dfrac{b-c}{-1}=\dfrac{c-a}{-2}\\\dfrac{a-b}{-1}=\dfrac{b-c}{-1}=\dfrac{c-a}{-2}=k\\ \Rightarrow a-b=-k;b-c=-k ;c-a=-2k\\ 4\left(a-b\right)\left(b-c\right)=4\left(-k\right)\left(-k\right)=4k^2\\ \left(c-a\right)^2=\left(-2k\right)^2=4k^2\\ \Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\left(ĐPCM\right)\)
a) Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\\\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\end{matrix}\right.\Rightarrowđpcm\)
b) \(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}=\dfrac{x-3}{2015}+\dfrac{x-4}{2014}\)
\(\Rightarrow\left(\dfrac{x-1}{2017}-1\right)+\left(\dfrac{x-2}{2016}-1\right)=\left(\dfrac{x-3}{2015}-1\right)+\left(\dfrac{x-4}{2014}-1\right)\)\(\Rightarrow\dfrac{x-2018}{2017}+\dfrac{x-2018}{2016}=\dfrac{x-2018}{2015}+\dfrac{x-2018}{2014}\)
\(\Rightarrow\left(x-2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
vì \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\) nên \(x-2018=0\Leftrightarrow x=2018\)
Câu 1:
\(\frac{a^{2016}+b^{2016}}{c^{2016}+d^{2016}}=\frac{a^{2016}-b^{2016}}{c^{2016}-d^{2016}}\)
\(\Rightarrow (a^{2016}+b^{2016})(c^{2016}-d^{2016})=(a^{2016}-b^{2016})(c^{2016}+d^{2016})\)
\(\Leftrightarrow 2(bc)^{2016}=2(ad)^{2016}\Rightarrow (bc)^{2016}=(ad)^{2016}\)
\(\Rightarrow (\frac{a}{b})^{2016}=(\frac{c}{d})^{2016}\)
\(\Rightarrow \frac{a}{b}=\pm \frac{c}{d}\) (đpcm)
Câu 2:
Nếu $a+b+c+d=0$ thì: \(\left\{\begin{matrix} a+b=-(c+d)\\ b+c=-(d+a)\\ c+d=-(a+b)\\ d+a=-(b+c)\end{matrix}\right.\)
\(\Rightarrow M=(-1)+(-1)+(-1)+(-1)=-4\)
Nếu $a+b+c+d\neq 0$
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{5(a+b+c+d)}{a+b+c+d}=5\)
\(\Rightarrow \left\{\begin{matrix} 2a+b+c+d=5a\\ a+2b+c+d=5b\\ a+b+2c+d=5c\\ a+b+c+2d=5d\end{matrix}\right.\) \(\Rightarrow \left\{\begin{matrix} b+c+d=3a(1)\\ a+c+d=3b(2)\\ a+b+d=3c(3)\\ a+b+c=3d(4)\end{matrix}\right.\)
Từ \((1);(2)\Rightarrow b+a+2(c+d)=3(a+b)\Rightarrow c+d=a+b\)
\(\Rightarrow \frac{a+b}{c+d}=1\)
Tương tự: \(\frac{b+c}{d+a}=\frac{c+d}{a+b}=\frac{d+a}{b+c}=1\)
\(\Rightarrow M=1+1+1+1=4\)
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
\(B=4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2\)
\(=4\left(a-b\right)\left(b-c\right)-\left[-2\left(a-b\right).\left(-2\right).\left(b-c\right)\right]\)
\(=4\left(a-b\right)\left(b-c\right)-4\left(a-b\right)\left(b-c\right)=0\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{a+b+c}=1\Leftrightarrow a=b=c\)
Ta có: \(a-b=b-c=c-a=0\)
\(M=0\)