Sửa đề:CM:\(\left(p-m\right)^2=4\left(m-n\right)\left(n-p\right)\)

Ta có:\(\frac{m}{2014}=\frac{n}{2015}=\frac{p}{2016}=\frac{p-m}{2016-2014}=\frac{p-m}{2}=\frac{m-n}{2014-2015}\)=

\(=\frac{m-n}{-1}=\frac{n-p}{2014-2016}=\frac{n-p}{-1}\)

\(\Rightarrow\frac{\left(p-m\right)^2}{4}=\frac{\left(m-n\right).\left(n-p\right)}{\left(-1\right).\left(-1\right)}\)

\(\Rightarrow\frac{\left(p-m\right)^2}{4}=\frac{\left(m-n\right)\left(n-p\right)}{1}\)

\(\Rightarrow\left(p-m\right)^2=4\left(m-n\right)\left(n-p\right)\)

Bổ sung :

➤ Bài 1 :

c/ Tính f(5)

Sửa bài 2 : Tính giá trị của biểu thức M = 4 (a - b) (b - c) = (c - a)².

M~1+1+1=3

N~1

=> M>N

m=n m>n m<n 1 trong 3 chắc chắn đúng ahihi =)))
 

Gọi \(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}=k\Rightarrow a=2014k;b=2015k;c=2016k\left(1\right)\)

Thay (1) vào M ta có :

M=4(2014k-2015k)(2015k-2016k)-(2016k-2014k)²

=>M=4.-k.-k-4k²

=>M=4k²-4k²=0

Vậy M = 0

Đặt:

\(\dfrac{a}{2015}=\dfrac{b}{2016}=\dfrac{c}{2017}=k\Leftrightarrow\left\{{}\begin{matrix}a=2015k\\b=2016k\\c=2017k\end{matrix}\right.\)

Nên \(4\left(a-b\right)\left(b-c\right)=4\left(2015k-2016k\right)\left(2016k-2017k\right)=4.\left(-k\right).\left(-k\right)=4k^2\)\(\left(c-a\right)^2=\left(2017k-2015k\right)^2=4k^2\)

Ta c dpcm

Đặt \(\dfrac{a}{2015}=\dfrac{b}{2016}=\dfrac{c}{2017}\)= k

\(\Rightarrow\) a = 2015 . k

b = 2016 . k

c = 2017 . k

\(\Rightarrow\) 4( a - b ) . ( b - c) = 4( 2015.k - 2016.k) .( 2016.k - 2017.k )

= 4( -k) (-k) = 4k² (1)

( c - a)² =( 2017.k -2015.k)²= (2k)²= 4k²(2)

Từ (1) và ( 2) \(\Rightarrow\)4( a - b).( b - c ) = (c - a )²

Đặt \(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}=k\)

\(\Rightarrow a=2014k;b=2015k;c=2016k\)

\(\Rightarrow4(a-b)(b-c)=4(2014k-2015k)(2015k-2016k)\)

\(\Rightarrow4\cdot k(2014-2015)\cdot k(2015-2016)=4\cdot k\cdot(-1)\cdot k\cdot(-1)=4\cdot k^2\)

\(\Rightarrow(c-a)(c-a)=(c-a)^2=(2016k-2014k)=[k(2016-2014)]^2=(k\cdot2)^2=k^{2\cdot4}\)

Rồi tự suy ra đấy

Bạn Namikaze Minato làm đúng rồi đấy

\(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}=\frac{a-b}{2014-2015}\)

\(=\frac{b-c}{2015-2016}=\frac{c-a}{2016-2014}\)

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

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

do đó: \(\left(a-b\right)\left(b-c\right)=\frac{\left(c-a\right)^2}{4}\)

\(\Rightarrow M=4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2=0\)

Đặt \(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}=k\)

=> \(\hept{\begin{cases}a=2014k\\b=2015k\\c=2016k\end{cases}}\)

Suy ra \(M=4\left(2014k-2015k\right)\left(2015k-2016k\right)-\left(2016k-2014k\right)^2=4k^2-4k^2=0\)

\(1a,\) Ta có: \(\left(2x-6\right)^2\ge0\forall x\Rightarrow\left(2x-6\right)^2+36\ge36\forall x\)

\(\Rightarrow\frac{2016}{\left(2x-6\right)^2+63}\le\frac{2016}{63}=32\)

\(\Rightarrow\left|y+2015\right|+32\le32\)

\(\Rightarrow\left|y+2015\right|\le0\)

\(\Rightarrow\left|y+2015\right|=0\)

\(\Rightarrow y=-2015\)

\(\Rightarrow2x-6=0\Rightarrow x=3\)

Vậy \(x=3;y=-2015\)

b)

Ta có: \(b^2=ac.\)

\(\Rightarrow\frac{a}{b}=\frac{b}{c}.\)

\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{2017b}{2017c}.\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\frac{a}{b}=\frac{b}{c}=\frac{2017b}{2017c}=\frac{a+2017b}{b+2017c}.\)

\(\Rightarrow\frac{a}{b}=\frac{a+2017b}{b+2017c}\)

\(\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{a+2017b}{b+2017c}\right)^2\)

\(\Rightarrow\left(\frac{a}{b}\right)^2=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}.\)

\(\Rightarrow\frac{a}{b}.\frac{a}{b}=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}\)

\(\Rightarrow\frac{a}{b}.\frac{b}{c}=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}.\)

\(\Rightarrow\frac{a}{c}=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}\left(đpcm\right).\)

Chúc bạn học tốt!