Cho a, b, c, thỏa mãn:
\(\frac{a}{2014}\)=\(\frac{b}{2015}\)=\(\frac{c}{2016}\)
Chứng minh rằng:
4 . ( a - b ) . ( b - c ) = ( c - a )2
cho 3 sơn thức a,b,c thỏa mãn
\(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}\)
chứng minh rằng: 4(a-b) (b-c) = (c-a)2
giải hẳn ra
\(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}=\frac{a-b}{2014-2015}=\frac{b-c}{2015-2016}=\frac{c-a}{2016-2014}\)
=\(\frac{a-b}{-1}=\frac{b-c}{-1}=\frac{c-a}{2}\)=>\(\frac{\left(a-b\right)\left(b-c\right)}{\left(-1\right)\left(-1\right)}=\frac{\left(c-a\right)^2}{2^2}=\frac{\left(a-b\right)\left(b-c\right)}{1}=\frac{\left(c-a\right)^2}{4}\Leftrightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
Cho 3 số a,b,c thỏa mãn : \(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}\). Tính M=\(4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2\)
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
Cho ba số thực a, b và c thỏa mãn \(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}.\)
CMR: 4(a-b)(b-c)= (c-a)2
Cho ba số thực a,b và c thỏa mãn a/2014=b/2015=c/2016
Chứng minh rằng : 4(a-b)(b-c) = (c-a)^2
đặt \(\frac{a}{2014}\)=\(\frac{b}{2015}\)=\(\frac{c}{2016}\)= K
---> a = 2014k, b=2015k , c=2016k
về trái : 4. ( 2014k-2015k). (2015k-2016k)=4. (-1k).(-1k)=4k2
Về phai: (2016k-2014k)2=(2k)2=4k2
---> ve trai = ve phai----> dpcm
Cho a,b,c thỏa mãn
a/2014 = b/2015 = c/2016. Chứng minh rằng:
4(a-b)(b-c) = (c-a)2
Đặt : \(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}=k\)
\(\Rightarrow\frac{a}{2014}=k\Rightarrow a=2014k\)
\(\Rightarrow\frac{b}{2015}=k\Rightarrow b=2015k\)
\(\Rightarrow\frac{c}{2016}=k\Rightarrow c=2016k\)
Ta có : \(4\left(a-b\right)\left(b-c\right)=4\left(2014k-2015k\right)\left(2015k-2016k\right)\)
\(=4k\left(2014-2015\right).k\left(2015-2016\right)=4k.\left(-1\right).k.\left(-1\right)=4.k^2\)( 1 )
\(\Rightarrow\left(c-a\right)^2=\left(2016k-2014k\right)\left(2016k-2014k\right)=\left[\left(2016k-2014k\right)^2\right]=\left[k\left(2016-2014\right)\right]=\left(k^2\right)^2=k^{2.4}\)( 2 )
Từ \(\left(1\right)\left(2\right)\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
Cho a,b,c là 3 số thỏa mãn: \(\frac{a}{2015}=\frac{b}{2016}=\frac{c}{2017}\)
Chứng minh: \(4\left(a-b\right).\left(b-c\right)=\left(c-a\right)^2\)
cho a,b,c thỏa mãn \(\frac{a}{2012}=\frac{b}{2013}=\frac{c}{2014}\)
chứng minh rằng 4(a-b)(b-c)=(c-a)2
Cho ba số a, b, c thỏa mãn
\(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}\)
tính giá trị của biểu thức:
\(M=4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2\)
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 \(M=4\left(2014k-2015k\right)\left(2015k-2016k\right)-\left(2016k-2014k\right)^2=4k^2-4k^2=0\)
Cho a,b,c thỏa mãn a/2014=b/2015=c/2016
Chứng minh: 4(a-b)(b-c)=(c-a)^2
Đặt \(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}=k\Rightarrow a=2014k;b=2015k;c=2016k\)
=>\(4\left(a-b\right)\left(b-c\right)=4\left(2014k-2015k\right)\left(2015k-2016k\right)=4\left(-1k\right)\left(-1k\right)=4k^2\)
\(\left(c-a\right)^2=\left(2016k-2014k\right)^2=\left(2k\right)^2=4k^2\)
=>đpcm