Ta có:

\(M=1^2+2^2+3^2+...+99^2+100^2\)

\(=1+2\left(1+1\right)+...+99\left(98+1\right)+100\left(99+1\right)\)

\(=1+1.2+2+...+98.99+99+99.100+\) \(100\)

\(=\left(1.2+2.3+...+99.100\right)\) \(+\left(1+2+...+99+100\right)\)

\(=333300+5050=338050\)

\(\Rightarrow\) Tổng quát: \(M=\dfrac{\left(n-1\right)n\left(n+1\right)}{3}+\dfrac{n\left(n+1\right)}{2}\)

b) Giải:

Đặt \(A=n^3+3n^2-n-3\) ta có

\(A=n^3+3n^2-n-3=n^2\left(n+3\right)-\left(n+3\right)\)

\(=\left(n^2-1\right)\left(n+3\right)=\left(n+1\right)\left(n-1\right)\left(n+3\right)\)

Thay \(n=2k+1\left(k\in Z\right)\) ta được:

\(A=\left(2k+2\right)2k\left(2k+4\right)=\) \(2\left(k+1\right).2k.2\left(k+2\right)\)

\(=8\left(k+1\right)k\left(k+2\right)\)

Mà \(\left(k+1\right)k\left(k+2\right)\) là tích của \(3\) số tự nhiên nhiên tiếp nên chia hết cho \(6\) \(\Rightarrow A⋮8.6=48\)

Vậy \(n^3+3n^2-n-3\) \(⋮48\forall x\in Z;x\) lẻ (Đpcm)

a) Ta có:

\(5\sqrt{a}-4b\sqrt{25a^3}+5a\sqrt{16ab^2}-2\sqrt{9a}\)

\(=5\sqrt{a}-4b.5a\sqrt{a}+5a.4b\sqrt{a}-2.3\sqrt{a}\)

\(=5\sqrt{a}-20ab\sqrt{a}+20ab\sqrt{a}-6\sqrt{a}\) \(=-\sqrt{a}\)

b) Ta có:

\(5a\sqrt{64ab^3}-\sqrt{3}.\sqrt{12a^3b^3}+2ab\sqrt{9ab}\) \(-5b\sqrt{81a^3b}\)

\(=5a.8b\sqrt{ab}-\sqrt{3.12a^3b^3}+2ab.3\sqrt{ab}\) \(-5b.9a\sqrt{ab}\)

\(=40ab\sqrt{ab}-6ab\sqrt{ab}+6ab\sqrt{ab}-45ab\)\(\sqrt{ab}\)

\(=-5ab\sqrt{ab}\)

Giải:

Áp dụng BĐT AM - GM ta có:

\(x+y\ge2\sqrt{xy}\)

\(y+z\ge2\sqrt{yz}\)

\(x+z\ge2\sqrt{xz}\)

Cộng theo vế các BĐT trên ta có:

\(x+y+y+z+z+x\) \(\ge2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}\)

\(\Leftrightarrow2\left(x+y+z\right)\ge\) \(2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)\)

\(\Leftrightarrow x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)

d) Giải:

Ta có: \(\left\{{}\begin{matrix}2222\equiv-4\left(\text{mod }7\right)\\5555\equiv4\left(\text{mod }7\right)\end{matrix}\right.\)

\(\Rightarrow2222^{5555}+5555^{2222}\equiv\left(-4\right)^{5555}\) \(+4^{2222}\)

\(\equiv-4+4=0\left(\text{mod }7\right)\)

Mà \(\left(-4\right)^{5555}+4^{2222}=\left(-4\right)^{2222}\left(4^{3333}-1\right)\) \(⋮4^3-1=63⋮7\)

Vậy \(2222^{5555}+5555^{2222}⋮7\)

Giải:

Ta biết rằng \(\left\{{}\begin{matrix}\left|A\right|\ge A\Leftrightarrow A\ge0\\\left|A\right|\ge0\Leftrightarrow A=0\end{matrix}\right.\)

Ta có: \(D=\left|x-\dfrac{1}{2}\right|+\left|x-\dfrac{5}{3}\right|+\left|x-3\right|\)

\(=\left|x-\dfrac{1}{2}\right|+\left|x-\dfrac{5}{3}\right|+\left|3-x\right|\) \(\ge x-\dfrac{1}{2}+0+3-x=\dfrac{5}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}\ge0\\x-\dfrac{5}{3}=0\\3-x\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x=\dfrac{5}{3}\\x\le3\end{matrix}\right.\) \(\Leftrightarrow x=\dfrac{5}{3}\)

Vậy \(MIN_D=\dfrac{5}{2}\Leftrightarrow x=\dfrac{5}{3}\)