\(\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b}{c}\)

\(\Rightarrow\dfrac{b+c}{a}+1=\dfrac{c+a}{b}+1=\dfrac{a+b}{c}+1\)

\(\Rightarrow\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}=\dfrac{a+b+c}{c}\)

Từ \(\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}\Leftrightarrow a\left(a+b+c\right)=b\left(a+b+c\right)\)

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

\(\Rightarrow\left[{}\begin{matrix}a=b\\a+b+c=0\end{matrix}\right.\left(đpcm\right)\)

\(n^5-n=n\left(n^4-1\right)\)

\(=n\left(n^2-1\right)\left(n^2+1\right)\)

\(=n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮6\)(tích của \(3\) số tự nhiên liên tiếp và \(1\) số tự nhiên bất kì)

\(=n\left(n-1\right)\left(n+1\right)\left(n^2-4+5\right)\)

\(=n\left(n-1\right)\left(n+1\right)\left[\left(n-2\right)\left(n+2\right)+5\right]\)

\(=n\left(n-1\right)\left(n+1\right)\left(n-2\right)\left(n+2\right)+5n\left(n-1\right)\left(n+1\right)\)

\(=\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)+5n\left(n-1\right)\left(n+1\right)\)

\(\left\{{}\begin{matrix}\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)⋮5\\5n\left(n-1\right)\left(n+1\right)⋮5\end{matrix}\right.\)(tích \(5\) số tự nhiên liên tiếp và 1 tích có thừa số \(5\))

\(\Rightarrow\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)+5n\left(n-1\right)\left(n+1\right)⋮5\)

Vì \(\left\{{}\begin{matrix}n^5-n⋮6\\n^5-n⋮5\end{matrix}\right.\Leftrightarrow n^5-n⋮30\left(đpcm\right)\)

Mình sẽ làm cách dãy tỉ số bằng nhau ,vì nhân sẽ khá là rối =.=

\(\dfrac{2x+3}{5x+2}=\dfrac{2\left(2x+3\right)}{2\left(5x+2\right)}=\dfrac{4x+6}{10x+4}\)

Hay \(\dfrac{4x+6}{10x+4}=\dfrac{4x+5}{10x+2}=\dfrac{4x+6-4x-5}{10x+4-10x-2}=\dfrac{1}{2}\)

Thay vào ta có:

\(\dfrac{2x+3}{5x+2}=\dfrac{1}{2}\Leftrightarrow5x+2=4x+6\Leftrightarrow5x=4x+4\Leftrightarrow x=4\)

_{\(A=1+2+2^2+2^3+...+2^{10}\)}

_{\(2A=2+2^2+2^3+2^4+...+2^{11}\)}

^{\(2A-A=\left(2+2^2+2^3+2^4+...+2^{11}\right)-\left(1+2+2^2+2^3+...+2^{10}\right)\)}

^{\(A=2^{11}-1< 2^{11}\)}

^{\(B=2.2^2+3.2^3+4.2^4+...+10.2^{10}\)\(2B=2.2^3+3.2^4+4.2^5+...+10.2^{11}\)\(2B-B=\left(2.2^3-3.2^3\right)+\left(3.2^4-4.2^4\right)+...+\left(9.2^{10}-10.2^{10}\right)+10.2^{11}-2.2^2\)\(B=2^3\left(2-3\right)+2^4\left(3-4\right)+...+2^{10}\left(9-10\right)+10.2^{11}-2.2^2\)\(B=-2^3-2^4-....-2^{10}+10.2^{11}-2^3\)}

_{\(B=-\left(2^3+2^4+...+2^{10}\right)+10.2^{11}-2^3\)}

^{\(B=-\left(2^{11}-2^3\right)+10.2^{11}-2^3\)}

_{\(B=-2^{11}+2^3+10.2^{11}-2^3\)}

_{\(B=9.2^{11}\)}

Ta cần so sánh: _\(9.2^{11}\) và _\(2^{14}\)

Hay _\(9\) và _\(2^3\)

_{\(9>8=2^3\Leftrightarrow B>2^{14}\)}

\(M=\dfrac{x}{x+y+z}+\dfrac{y}{x+y+t}+\dfrac{z}{y+z+t}+\dfrac{t}{x+z+t}\)

Ta có:

\(\left\{{}\begin{matrix}\dfrac{x}{x+y+z}>\dfrac{x}{x+y+z+t}\\\dfrac{y}{x+y+t}>\dfrac{y}{x+y+z+t}\\\dfrac{z}{y+z+t}>\dfrac{z}{x+y+z+t}\\\dfrac{t}{x+z+t}>\dfrac{t}{x+y+z+t}\end{matrix}\right.\) Cộng theo \(3\) vế ta có:

\(M>\dfrac{x}{x+y+z+t}+\dfrac{y}{x+y+z+t}+\dfrac{z}{x+y+z+t}+\dfrac{t}{x+y+z+t}=1\)

Lại có:

\(\left\{{}\begin{matrix}\dfrac{x}{x+y+z}< \dfrac{x+t}{x+y+z+t}\\\dfrac{y}{x+y+t}< \dfrac{y+z}{x+y+z+t}\\\dfrac{z}{y+z+t}< \dfrac{z+x}{x+y+z+t}\\\dfrac{t}{x+z+t}< \dfrac{t+y}{x+y+z+t}\end{matrix}\right.\)Cộng theo \(3\) vế ta có:

\(M< \dfrac{x+t}{x+y+z+t}+\dfrac{y+z}{x+y+z+t}+\dfrac{z+x}{x+y+z+t}+\dfrac{t+y}{x+y+z+t}=2\)Như vậy \(1< M< 2\Leftrightarrow M\notin N\left(đpcm\right)\)

\(\dfrac{x+y-z}{z}=\dfrac{y+z-x}{x}=\dfrac{x-y+z}{y}\)

\(\Rightarrow\dfrac{x+y-z}{z}+2=\dfrac{y+z-x}{x}+2=\dfrac{x-y+z}{y}+2\)

\(\Rightarrow\dfrac{x+y-z}{z}+\dfrac{2z}{z}=\dfrac{y+z-x}{x}+\dfrac{2x}{x}=\dfrac{x-y+z}{y}+\dfrac{2y}{y}\)

\(\Rightarrow\dfrac{x+y-z+2z}{z}=\dfrac{y+z-x+2x}{x}=\dfrac{x-y+z+2y}{y}\)

\(\Rightarrow\dfrac{x+y+z}{z}=\dfrac{y+z+x}{x}=\dfrac{x+z+y}{y}\)

Điều này xảy ra khi và chỉ khi: \(\left[{}\begin{matrix}x+y+z=0\\x=y=z\end{matrix}\right.\)

\(\circledast\)Với \(x+y+z=0\Leftrightarrow\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\)

Thay vào \(A\) ta có: \(A=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{x}{z}\right)=\left(\dfrac{x+y}{y}\right)\left(\dfrac{y+z}{z}\right)\left(\dfrac{z+x}{z}\right)=\dfrac{-z.-x.-y}{xyz}=\dfrac{-xyz}{xyz}=-1\)

\(\circledast\) Với \(x=y=z\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{y}=1\\\dfrac{y}{z}=1\\\dfrac{x}{z}=1\end{matrix}\right.\)

Thay vào \(A\) ta có:

\(A=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

Ta gọi tổng đó là A

Ta có:

A= 3^0+3^1+3^2+....+3^11

A=(3⁰+3¹+3²+3³) +(3⁴+3⁵+3⁶+3⁷)+(3⁸+3⁹+3¹⁰+3¹¹)

A=40+3⁴(3⁰+3¹+3²+3³)+3⁸(3⁰+3¹+3²+3³)

A=40+3⁴.40+3⁸.40

A=40(1+3⁴+38) chia hết cho 40

=>A chia hết cho 40

tick nha

Xét thừa số tổng quát:

\(\dfrac{1}{t^3}< \dfrac{1}{t^3-t}=\dfrac{1}{t\left(t^2-1\right)}=\dfrac{1}{t\left(t+1\right)\left(t-1\right)}=\dfrac{1}{\left(t-1\right)t\left(t+1\right)}\)

Thay vào bài toán:

\(\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+...+\dfrac{1}{2017^3}< \dfrac{1}{\left(2-1\right)2\left(2+1\right)}+\dfrac{1}{\left(3-1\right)3\left(3+1\right)}+\dfrac{1}{\left(4-1\right)4\left(4+1\right)}+....+\dfrac{1}{\left(2017-1\right)2017\left(2017+1\right)}=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{2016.2017.2018}\)

Đặt:\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{2016.2017.2018}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{2016.2017}-\dfrac{1}{2017.2018}\right)=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2017.2018}\right)=\dfrac{1}{1.2.2}-\dfrac{1}{2.2017.2018}=\dfrac{1}{4}-\dfrac{1}{2.2017.2018}< \dfrac{1}{4}\left(đpcm\right)\)