đặt a/2014=b/2015=c/2016=k

=>a=2014k;b=2015k;c=2016k

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

=4.k(2014-2015).k92015-2016)=4.k.(-1).k.(-1)=4.k^2(1)

=>(c-a)(c-a)=(c-a)^2=(2016k-2014k)(2016k-2014k)=[k(2016-2014)]^2=[k.2]^2=k^2.4(2)

từ (1)và (2)=>4(a-b)(b-c) = (c-a).(c-a)

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

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

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 \(\left(1\right)\left(2\right)\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)

Câu hỏi của NGUYỄN DOÃN ANH THÁI - Toán lớp 9 - Học toán với OnlineMath làm tương tự chỗ cuối thay a+b+c=2015 là dc

giả sử :c^2>a^2>b^2 khi đó ta có :

\(\frac{b^2+c^2}{a^2+3}+\frac{c^2-a^2}{b^2+4^2}+\frac{a^2-b^2}{c^2+5}\le\frac{b^2+c^2}{b^2+3}+\frac{c^2-a^2}{b^2+3}+\frac{a^2-b^2}{b^2+3}=\frac{2c^2}{b^2+3}\le\frac{2}{3}.c^2\)

Như vậy ta có :\(a^2+b^2+c^2\le\frac{2}{3}.c^2\). Điều này xảy ra khi a=b=c

                 chuc bn hk tốt!