Ta có: \(B=4^{2017}+4^{2016}+...+4^2+4^1+4^0\)

\(\Leftrightarrow4\cdot B=4^{2018}+4^{2017}+...+4^3+4^2+4^1\)

\(\Leftrightarrow3\cdot B=4^{2018}-1\)

\(\Leftrightarrow A=165\cdot\dfrac{4^{2018}-1}{3}+55\)

\(\Leftrightarrow A=4^{2018}\)

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/ Ta có

\(200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)\)

\(=1+2\left(1-\frac{1}{3}\right)+2\left(1-\frac{1}{4}\right)+...+2\left(1-\frac{1}{100}\right)\)

\(=1+2\left(\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\right)\)

\(=2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\)

Thế lại bài toán ta được:

\(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\)

\(=\frac{2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}=2\)

b/ Ta có: 

A - B\(=\frac{-21}{10^{2016}}+\frac{12}{10^{2016}}+\frac{21}{10^{2017}}-\frac{12}{10^{2017}}\)

\(=\frac{9}{10^{2017}}-\frac{9}{10^{2016}}< 0\)

Vậy A < B

cảm ơn bạn

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{97\cdot99}-\frac{5}{4}\cdot\frac{13}{99}+\frac{5}{99}\cdot\frac{1}{4}\)

\(A=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right)-\frac{13}{4}\cdot\frac{5}{99}+\frac{5}{99}\cdot\frac{1}{4}\)

\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{5}{99}\cdot\left(\frac{13}{4}-\frac{1}{4}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)-\frac{5}{99}\cdot3\)

\(A=\frac{1}{2}\cdot\frac{32}{99}-\frac{5}{33}\)

\(A=\frac{16}{99}-\frac{5}{33}=\frac{1}{99}\)

3/\(7a+b=0\Rightarrow b=-7a\)

\(f\left(x\right)=ax^2-7ax+c\).Ta có: \(f\left(10\right)=100a-70a+c=30a+c\)

\(f\left(-3\right)=30a+c\).Nhân theo vế ta có đpcm:

\(f\left(10\right).f\left(-3\right)=\left(30a+c\right)^2\ge0\) (đúng)

Câu 2 tự quy đồng rồi so sánh:v

Ta có :

\(M=2^{2017}-\left(2^{2016}+2^{2017}+...............+2+1\right)\)

Đặt :

\(A=2^{2016}+2^{2015}+................+2+1\)

\(\Leftrightarrow2A=2^{2017}+2^{2016}+2^{2015}+............+2^2+2\)

\(\Leftrightarrow2A-A=\left(2^{2017}+2^{2016}+........+2\right)-\left(2^{2016}+2^{2015}+..........+1\right)\)

\(\Leftrightarrow A=2^{2017}-1\)

\(\Leftrightarrow M=2^{2017}-A\)

\(\Leftrightarrow M=2^{2017}-\left(2^{2017}-1\right)\)

\(\Leftrightarrow M=2^{2017}-2^{2017}+1\)

\(\Leftrightarrow M=0+1=1\)

\(M=2^{2017}-\left(2^{2016}+2^{2015}+...+2^1+2^0\right)\)
Đặt :

\(S=2^{2016}+2^{2015}+...+2^1+2^0\)

\(\Rightarrow S=2^0+2^1+...+2^{2015}+2^{2016}\)

\(\Rightarrow2S=2\left(2^0+2^1+...+2^{2015}+2^{2016}\right)\)

\(\Rightarrow2S=2^1+2^2+...+2^{2016}+2^{2017}\)

\(\Rightarrow2S-S=\left(2^1+2^2+...+2^{2016}+2^{2017}\right)-\left(2^0+2^1+...+2^{2015}+2^{2016}\right)\)

\(\Rightarrow S=2^{2017}-1\)

Thay S vào M ta có:

\(M=2^{2017}-\left(2^{2017}-1\right)\)

\(M=2^{2017}-2^{2017}+1=1\)