Giải:

126 : a dư 25

=> 126 - 25 chia hết cho a

101 chia hết cho a 

U(101)= {1;101} hay a \(\in\) {1;101}

Mà 126 : a dư 25 => a > 25

Vậy a = 101   

Giải:

Đặt \(\dfrac{x}{3}=\dfrac{y}{7}=k\) \(\Rightarrow\) \(\begin{cases}x=3k\\y=7k\end{cases}\)

Ta có:

\(xy=84\Rightarrow3k.7k=84\Rightarrow21k^2=84\)

\(\Rightarrow k^2=\dfrac{84}{21}=4\Leftrightarrow k=\) \(\pm 2\)

Ta có 2 trường hợp:

Trường hợp 1: Nếu \(k=2\) \(\Rightarrow\) \(\begin{cases}x=3.2=6\\y=7.2=14\end{cases}\)

Trường hợp 2: Nếu \(k=-2\) \(\Rightarrow\) \(\begin{cases}x=3.(-2)=-6\\y=7.(-2)=-14\end{cases}\)

Vậy...

Giải:

Ta có: \(VT=a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc\)

\(=\left(a+b+c\right)^3\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)\)

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

\(=VP\) (Đpcm)

Bài 1: Giải:

Đặt \(B=4^{2004}+4^{2003}+...+4^2+4+1\)

\(\Rightarrow4B=4\left(4^{2004}+4^{2003}+...+4^2+4+1\right)\)

\(=4^{2005}+4^{2004}+...+4^3+4^2+4\)

\(\Rightarrow4B-B=\left(4^{2005}+4^{2004}+...+4^2+4\right)-\left(4^{2004}+4^{2003}+...+4+1\right)\)

\(\Rightarrow3B=4^{2005}-1\Rightarrow B=\dfrac{4^{2005}-1}{3}\)

Do đó:

\(A=75.\dfrac{4^{2005}-1}{3}+25=25\left(4^{2005}-1+1\right)\)

\(=25.4^{2005}=25.4.4^{2004}=100.4^{2004}⋮100\) (Đpcm)

Giải:

Ta có:

\(\dfrac{a}{b}< \dfrac{a+n}{b+n}\) \(\Leftrightarrow a\left(b+n\right)< b\left(a+n\right)\)

\(\Leftrightarrow ab+an< ab+bn\Leftrightarrow a< b\) (Vì \(n>0\))

Vậy \(\dfrac{a}{b}< \dfrac{a+n}{b+n}\Leftrightarrow a< b\)

Tương tự ta cũng có:

\(\dfrac{a}{b}>\dfrac{a+n}{b+n}\Leftrightarrow a>b\)

\(\dfrac{a}{b}=\dfrac{a+n}{b+n}\Leftrightarrow a=b\)

Giải:

Ta có:

\(\dfrac{x}{2}+\dfrac{x}{4}+\dfrac{x}{2016}=\dfrac{x}{3}+\dfrac{x}{5}+\dfrac{x}{2017}\)

\(\Leftrightarrow\dfrac{x}{2}+\dfrac{x}{4}+\dfrac{x}{2016}-\dfrac{x}{3}-\dfrac{x}{5}-\dfrac{x}{2017}=0\)

\(\Leftrightarrow x\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{2016}-\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{2017}\right)=0\)

Mà \(\dfrac{1}{2}>\dfrac{1}{3};\dfrac{1}{4}>\dfrac{1}{5};\dfrac{1}{2016}>\dfrac{1}{2017}\)

\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{2016}-\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{2017}\right)\) \(\ne0\)

\(\Leftrightarrow x=0\)