Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=t\)

\(\Rightarrow a=2016t,b=2017t,c=2018t\)

Ta có: \(4\left(a-b\right)\left(b-c\right)=4\left(2016t-2017t\right)\left(2017t-2018t\right)=4.\left(-t\right).\left(-t\right)=4t^2\)

\(\left(c-a\right)^2=\left(2018t-2016t\right)^2=\left(2t\right)^2=4t^2\)

Vậy \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

a/2016 = b/2017 = c/2018 = (a-b) / (2016-2017) = (b-c) / (2017-2018) = (c-a) / (2018-1026)

                                          = (a-b) / (-1) = (b-c) / ( -1) = (c-a) / 2

Vì (a-b) / (-1) = (b-c) / ( -1) = (c-a) / 2 nên (a-b) / (-1) . (b-c) / (-1) =[ (c-a) / 2 ]²

                                                      => (a-b)(b-c) / (-1).(-1) = (c-a)² /  2²

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

                                                       => (a-b)(b-c) =(c-a)² / 4

                                                       => 4(a-b)(b-c)= (c-a)²

C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)

c=\(\frac{1}{1}-\frac{1}{10}\)

c=\(\frac{9}{10}\)

còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!

bài này dễ vào TH 0,5 điểm trong bài thi

nghe có vẻ khó nhưng chú ý 1 chút là có thể làm được

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^{2016}}{c^{2016}}=\frac{b^{2016}}{d^{2016}}\)\(\Rightarrow\left(\frac{a^{2016}}{c^{2016}}\right)^{2017}=\left(\frac{b^{2016}}{d^{2016}}\right)^{2017}\)

áp dụng t/c dãy t/s = nhau

\(\Rightarrow\left(\frac{a^{2016}}{c^{2016}}\right)^{2017}=\left(\frac{b^{2016}}{d^{2016}}\right)^{2017}=\)\(\frac{\left(a^{2016}+b^{2016}\right)^{2017}}{\left(c^{2016}+d^{2016}\right)^{2017}}\)

biến đổi tiếp cái kia tương tự rồi suy ra chúng = nhau nhé

Bài 1:

ta có: \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)

\(B=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)

\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)

\(B=\frac{1}{2^2}-\frac{1}{100^2}\)

\(B=\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)

\(\Rightarrow B< \frac{1}{4}\)

Bài 2:

ta có: \(B=\frac{2015+2016+2017}{2016+2017+2018}\)

\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)

mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)

\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)

\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)

\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)

\(\Rightarrow A>B\)

Học tốt nhé bn !!

a, \(M=\frac{3}{2}\cdot\frac{4}{3}\cdot\cdot\cdot\cdot\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{3.4...2019}{2.3...2018}=\frac{2019}{2}\)

b, c cùng 1 câu phải k

ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(=1+\frac{1}{2}+...+\frac{1}{2018}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)

\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}=B\)

\(\Rightarrow\frac{A}{B}=1\Rightarrow\left(\frac{A}{B}\right)^{2018}=1^{2018}=1\)

A,\(M=\frac{3}{2}\cdot\frac{4}{3}....\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{4\cdot3...2019}{2\cdot3...2018}=\frac{2019}{2}\)

NHA

HỌC TỐT

bạn giải thử để tham khảo đi

Nè Phong Đãng - Trang của Phong Đãng - Học toán với OnlineMath: 

Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=k\left(k\in R\right)\)    

\(\Rightarrow a=2016k,b=2017k,c=2018k\)

Thay vào biểu thức: \(4\left(a-b\right)\left(b-c\right)\), có: 

\(=4\left(2016k-2017k\right)\left(2017k-2018k\right)\)

\(=4\cdot\left(-1\right)\cdot k\cdot\left(-1\right)\cdot k=4k^2\)                                                                         (1)

Làm tương tự như vậy: \(\left(c-a\right)^2=\left(2018k-2016k\right)^2=\left(2k\right)^2=4k^2\)(2)

Từ (1)(2) suy ra: \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)

rất đúng,To Kill A Mockingbird

A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)

\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)

\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)

\(=\frac{3}{5}+\frac{2}{5}=1\)

b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)

\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)

\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)

\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)

 \(=\frac{1}{3.2}-\frac{5.2}{7.3}\)

\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)

\(=\frac{7}{42}-\frac{20}{42}\)

\(=-\frac{13}{42}\)

cs ng làm đung r

đag định lm

a) \(a^2+b^2+c^2=ab+bc+ac\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(c-a\right)^2+\left(b-c\right)^2=0\)

Ta có : \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(c-a\right)^2\ge0\\\left(b-c\right)^2\ge0\end{cases}}\)

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

\(\Leftrightarrow a=b=c\)

a. \(a^2+b^2+c^2=ab+bc+ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ab-2ca=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\left(đpcm\right)\)

c) \(\frac{x+5}{2015}+\frac{x+4}{2016}+\frac{x+3}{2017}+\frac{x+2}{2018}=\frac{x+2015}{5}+\frac{x+2016}{4}+\frac{x+2017}{3}+\frac{x+2018}{2}\)

Ta có VT + 4 = VP + 4

VT + 4 = \(\left(\frac{x+5}{2015}+1\right)+\left(\frac{x+4}{2016}+1\right)+\left(\frac{x+3}{2017}+1\right)+\left(\frac{x+2}{2018}+1\right)\) 

\(=\frac{x+2020}{2015}+\frac{x+2020}{2016}+\frac{x+2020}{2017}+\frac{x+2020}{2018}\)

\(=\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)\)

VP + 4 = \(\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)+\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}\right)\)

\(=\frac{x+2020}{5}+\frac{x+2020}{4}+\frac{x+2020}{3}+\frac{x+2020}{2}\)

\(=\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\right)\)

Khi đó \(\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)=\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\right)\)

=> \(\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

Vì \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)

=> x + 2020 = 0

=> x = -2020