\(S=1-\frac{1}{4}+1-\frac{1}{9}+1-\frac{1}{16}+...+1-\frac{1}{n^2}\)

\(S=n-1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< n-1\)

\(S=n-1-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)

\(>n-1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\right)\)

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

\(=n-1-\left(1-\frac{1}{n}\right)\)

\(=n-2+\frac{1}{n}>n-2\)

\(\Rightarrow n-2< S< n-1\)

ta có đpcm. 

1)đề bài  \(\Leftrightarrow\frac{7^x.7^2+7^x.7+7^x.1}{57}=\frac{5^{2x}.1+5^{2x}.5+5^{2x}.5^3}{131}\)

\(\Leftrightarrow\frac{7^x\left(49+7+1\right)}{57}=\frac{5^{2x}\left(1+5+125\right)}{131}\)

\(\Leftrightarrow7^x=5^{2x}\Leftrightarrow\frac{7^x}{25^x}=1\Leftrightarrow\left(\frac{7}{25}\right)^x=\left(\frac{7}{25}\right)^0\)

=> x=1

1)=>x=0 nhầm

\(S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n+1\right)}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)

\(=1-\frac{1}{n+1}=\frac{n}{n+1}\)

\(A=3^0+3^1+3^2+...+3^{2018}\)

\(3A=3^1+3^2+3^3+...+3^{2018}+3^{2019}\)

\(\Rightarrow3A-A=\left(3^1+3^2+...+3^{2019}\right)-\left(3^0+3^1+...+3^{2018}\right)\)

\(2A=3^{2019}-3^0=3^{2019}-1\)

1) \(A=\frac{\left|x-2016\right|+2017}{\left|x-2016\right|+2018}=\frac{\left|x-2016\right|+2018-1}{\left|x-2016\right|+2018}=1-\frac{1}{\left|x-2016\right|+2018}\)

\(A\)nhỏ nhất nên \(\frac{1}{\left|x-2016\right|+2018}\)lớn nhất nên \(\left|x-2016\right|+2018\)dương nhỏ nhất. 

mà \(\left|x-2016\right|+2018\ge2018\)

Dấu \(=\)khi \(x=2016\).

Vậy \(minA=1-\frac{1}{2018}=\frac{2017}{2018}\)đạt tại \(x=2016\).

2) \(x-2xy+y=0\)

\(\Leftrightarrow x\left(1-2y\right)+\frac{1}{2}-y-\frac{1}{2}=0\)

\(\Leftrightarrow\left(2x+1\right)\left(1-2y\right)=1=1.1=\left(-1\right).\left(-1\right)\)

Từ đây xét 2 trường hợp nha. Ra kết quả cuối cùng là: \(\left(x,y\right)\in\left\{\left(0,0\right),\left(1,1\right)\right\}\).

\(\frac{27^{n+2}}{3^{2n+4}}=3^4\)

=> 27^{n + 2} : 3^{2n + 4} = 3⁴

=> (3³)^{n + 2} : 3^{2n + 4} = 3⁴

=> 3^{3n + 6} : 3^{2n + 4} = 3⁴

=> 3^{3n + 6 - 2n - 4} = 3⁴

=> 3^{n + 2} = 3⁴

=> n + 2 = 4

=> n = 2

Vậy n = 2 là giá trị cần tìm

\(\frac{a}{b}=\frac{9}{2}-\frac{1}{2}.\frac{4}{9}=\frac{77}{18}\)

\(\Rightarrow a=77,b=18\)

\(a-2b=77-2.18=41\)

3x² + 4x + 100 = 0

=> 3(x² + 4/3x + 100/3) = 0

=> 3(x² + 2.2/3x + 4/9 + 296/9) = 0

=> x² + 2.2/3x + 4/9 + 296/9 = 0

=> (x + 2/3)² + 296/9 = 0 (vô lý vì (x + 2/3)² + 296/9 \(\ge\frac{296}{9}\)> 0 \(\forall x\))

=> Đa thức vô ngiệm