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

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

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!

vì \(\frac{a}{b}\)=\(\frac{c}{d}\)=>\(\frac{a^{2017}}{b^{2017}}\) =\(\frac{c^{2017}}{d^{2017}}\) 

áp dụng tính chất dãy tỉ số bằng nhau

=> \(\frac{a^{2017}}{b^{2017}}\) =\(\frac{c^{2017}}{d^{2017}}\)= \(\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}\)=\(\frac{a^{2017}-c^{2017}}{b^{2017}-d^{2017}}\)=\(\frac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}\)(diều phải chứng minh

Từ \(\frac{a}{b}=\frac{c}{d}=k\)

Suy ra a=bk

           c=dk

Ta có

\(\frac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}=\frac{\left(bk\right)^{2017}+b^{2017}}{\left(dk\right)^{2017}+d^{2017}}=\frac{b^{2017}.k^{2017}+b^{2017}}{d^{2017}.k^{2017}+d^{2017}}=\frac{b^{^{2017}}\left(k^{2017}+\right)}{d^{2017}\left(k^{2017}+1\right)}=\frac{b^{2017}}{d^{2017}}\)(1)

Ta có

\(\frac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}=\frac{\left(bk-b\right)^{2017}}{\left(dk-d\right)^{2017}}=\frac{\left(b\left(k-1\right)\right)^{2017}}{\left(d\left(k-1\right)\right)^{2017}}=^{\frac{b^{2017}}{d^{2017}}}\)(2)

Từ (1) và (2)

Ta suy ra

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

từ gt: \(\frac{a}{b}\)=\(\frac{c}{d}\)suy ra ad=bc

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

suy ra \(a^{2017}+b^{2017}.\left(c-d\right)^{2017}=c^{2017}+d^{2017}.\left(a-b\right)^{2017}\)

\(a^{2017}+b^{2017}.c^{2017}-b^{2017}.d^{2017}=c^{2017}+d^{2017}.a^{2017}-d^{2017}.b^{2017}\)

theo mình nghĩ là\(b^{2017}.c^{2017}=d^{2017}.a^{2017}\)

bc=da

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 !!

Là sao?

đề bị bị sai rồi bạn ơi??? !!!

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