Vì \(n\in Z^+\)nên\(n\left(n+1\right)\left(n+2\right)>n^3\Rightarrow\sqrt[3]{n\left(n+1\right)\left(n+2\right)}>n\)

\(\Rightarrow\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}>n\)(1)

Lại có:\(n^2+2n+1>n^2+2n\Rightarrow\left(n+1\right)^2>n\left(n+2\right)\Rightarrow\left(n+1\right)^3>n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow n+1>\sqrt[3]{n\left(n+1\right)\left(n+2\right)}\\ \Rightarrow\sqrt[3]{n^3+3n^2+3n+1}>\sqrt[3]{n^3+3n^2+2n}\)

\(\Rightarrow\sqrt[3]{n^3+3n^2+2n+n+1}>\sqrt[3]{n^3+3n^2+2n+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)

\(\Rightarrow\sqrt[3]{\left(n+1\right)^3}>\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)

Tương tự \(\Rightarrow n+1>\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)(2)

Từ (1) và (2) suy ra:

\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}< n+1\)

\(n\in Z^+\)nên n² < n² + 2n < n² + 2n + 1 <=> n² < n(n + 2) < (n + 1)² => n³ < n(n + 1)(n + 2) < (n + 1)³ 

=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)}< n+1\)

=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)}< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n}\)\(< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n+1}\)\(=\sqrt[3]{\left(n+1\right)\left(n^2+2n+1\right)}=\sqrt[3]{\left(n+1\right)\left(n+1\right)^2}=n+1\)

=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n}\)

\(< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}}< n+1\)

Tiếp tục như vậy,ta có đpcm.

Sorry ! n² < n(n + 2) nên n³ < n(n + 1)(n + 2) (vì n < n + 1)

Ta có: \(\dfrac{2}{\left(n-1\right)n\left(n+1\right)}=\dfrac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\)

??? Cái gì đây, đây là câu hỏi hay câu trả lời ???

rảnh ghê ta

Bài làm mà mấy thánh cứ vào phốt thế

c: \(\left(n-2\right)^2-\left(n+3\right)\left(n-3\right)=4\left(n-1\right)\)

\(\Leftrightarrow n^2-4n+4-n^2+9=4n-4\)

=>-4n+13=4n-4

=>-8n=-17

hay n=17/8

a: \(\left(n-2\right)\left(n+2\right)+6\left(n-1\right)=\left(n+1\right)^2\)

\(\Leftrightarrow n^2-4+6n-6=n^2+2n+1\)

=>6n-10=2n+1

=>4n=11

hay n=11/4

d: \(2\left(3-x\right)-3\left(x-1\right)=4\left(x-3\right)\)

=>6-2x-3x+3=4x-12

=>-5x+9=4x-12

=>-9x=-21

hay x=7/3

\(1,\)

\(a,\) Sửa: \(A=10^n+72n-1⋮81\)

Với \(n=1\Leftrightarrow A=10+72-1=81⋮81\)

Giả sử \(n=k\Leftrightarrow A=10^k+72k-1⋮81\)

Với \(n=k+1\Leftrightarrow A=10^{k+1}+72\left(k+1\right)-1\)

\(A=10^k\cdot10+72k+72-1\\ A=10\left(10^k+72k-1\right)-648k+81\\ A=10\left(10^k+72k-1\right)-81\left(8k-1\right)\)

Ta có \(10^k+72k-1⋮81;81\left(8k-1\right)⋮81\)

Theo pp quy nạp 

\(\Rightarrow A⋮81\)

\(b,B=2002^n-138n-1⋮207\)

Với \(n=1\Leftrightarrow B=2002-138-1=1863⋮207\)

Giả sử \(n=k\Leftrightarrow B=2002^k-138k-1⋮207\)

Với \(n=k+1\Leftrightarrow B=2002^{k+1}-138\left(k+1\right)-1\)

\(B=2002\cdot2002^k-138k-138-1\\ B=2002\left(2002^k-138k-1\right)+276138k+1863\\ B=2002\left(2002^k-138k-1\right)+207\left(1334k+1\right)\)

Vì \(2002^k-138k-1⋮207;207\left(1334k+1\right)⋮207\)

Nên theo pp quy nạp \(B⋮207,\forall n\)

\(2,\)

\(a,\) Sửa đề: CMR: \(1\cdot2+2\cdot3+...+n\left(n+1\right)=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)

Đặt \(S_n=1\cdot2+2\cdot3+...+n\left(n+1\right)\)

Với \(n=1\Leftrightarrow S_1=1\cdot2=\dfrac{1\cdot2\cdot3}{3}=2\)

Giả sử \(n=k\Leftrightarrow S_k=1\cdot2+2\cdot3+...+k\left(k+1\right)=\dfrac{k\left(k+1\right)\left(k+2\right)}{3}\)

Với \(n=k+1\)

Cần cm \(S_{k+1}=1\cdot2+2\cdot3+...+k\left(k+1\right)+\left(k+1\right)\left(k+2\right)=\dfrac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{3}\)

Thật vậy, ta có:

\(\Leftrightarrow S_{k+1}=S_k+\left(k+1\right)\left(k+2\right)\\ \Leftrightarrow S_{k+1}=\dfrac{k\left(k+1\right)\left(k+2\right)}{3}+\left(k+1\right)\left(k+2\right)\\ \Leftrightarrow S_{k+1}=\dfrac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{3}\)

Theo pp quy nạp ta có đpcm

\(b,\) Với \(n=0\Leftrightarrow0^3=\left[\dfrac{0\left(0+1\right)}{2}\right]^2=0\)

Giả sử \(n=k\Leftrightarrow1^3+2^3+...+k^3=\left[\dfrac{k\left(k+1\right)}{2}\right]^2\)

Với \(n=k+1\)

Cần cm \(1^3+2^3+...+k^3+\left(k+1\right)^3=\left[\dfrac{\left(k+1\right)\left(k+2\right)}{2}\right]^2\)

Thật vậy, ta có

\(1^3+2^3+...+k^3+\left(k+1\right)^3\\ =\left[\dfrac{k\left(k+1\right)}{2}\right]^2+\left(k+1\right)^3\\ =\dfrac{k^2\left(k+1\right)^2+4\left(k+1\right)^3}{4}=\dfrac{\left(k+1\right)^2\left(k^2+4k+4\right)}{4}\\ =\dfrac{\left(k+1\right)^2\left(k+2\right)^2}{4}=\left[\dfrac{\left(k+1\right)\left(k+2\right)}{2}\right]^2\)

Theo pp quy nạp ta được đpcm

\(b,lim\dfrac{\left(n^2+1\right)\left(n-10\right)^2}{\left(n+1\right)\left(3n-3\right)^3}\)

\(=lim\dfrac{\left(1+\dfrac{1}{n^2}\right)\left(\dfrac{1}{n}-\dfrac{10}{n^2}\right)^2}{\left(1+\dfrac{1}{n}\right)\left(\dfrac{3}{n^2}-\dfrac{3}{n^3}\right)}=0\)

\(a,lim\dfrac{4n^5-3n^2}{\left(3n^2-2\right)\left(1-4n^3\right)}\)

\(=lim\dfrac{4-\dfrac{3}{n^3}}{\left(3-\dfrac{2}{n^2}\right)\left(\dfrac{1}{n^3}-4\right)}\)

\(=\dfrac{4-0}{\left(3-0\right)\left(0-4\right)}=\dfrac{4}{-12}=-\dfrac{1}{3}\)

\(\lim\dfrac{\left(n^2+1\right)\left(n-10\right)^2}{\left(n+1\right)\left(3n-3\right)^3}=\lim\dfrac{\left(1+\dfrac{1}{n^2}\right)\left(1-\dfrac{10}{n}\right)^2}{\left(1+\dfrac{1}{n}\right)\left(3-\dfrac{3}{n}\right)^3}=\dfrac{1.1^2}{1.3}=\dfrac{1}{3}\)

Bài 2:Tìm x biết

\\(\\left(4x+3\\right)^3+\\left(5-7x\\right)^3+\\left(3x-8\\right)^3=0\\)

\\(\\Leftrightarrow\\left[\\left(4x\\right)^3+3.\\left(4x\\right)^2.3+3.4x.3^2+3^3\\right]+\\left[5^3-3.5^2.7x+3.5.\\left(7x\\right)^2-\\left(7x\\right)^3\\right]+\\left[\\left(3x\\right)^3-3.\\left(3x\\right)^2.8+3.3x.8^2-8^3\\right]=0\\)

\\(\\Leftrightarrow64x^3+144x^2+108x+27+125-525x+735x^2-343x^3+27x^3-216x^2+576x-512=0\\)

\\(\\Leftrightarrow-252x^3+663x^2+159x-360=0\\)

\\(\\Leftrightarrow3\\left(-84x^3+221x^2+53x-120\\right)=0\\)

Bài 2: Đặt \(4x+3=a;5-7x=b;3x-8=c\Rightarrow a+b+c=0\)

Kết hợp với đề bài ta có \(\left\{{}\begin{matrix}a^3+b^3+c^3=0\\a+b+c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^3+b^3+c^3-3abc+3abc=0\\a+b+c=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc=0\left(1\right)\\a+b+c=0\left(2\right)\end{matrix}\right.\)

Thay (2) vào (1) suy ra \(3abc=0\Leftrightarrow a=0\text{hoặc }b=0\text{hoặc }c=0\)

+) a = 0 suy ra \(x=-\frac{3}{4}\)

+) b = 0 suy ra \(x=\frac{5}{7}\)

+) c = 0 suy ra \(x=\frac{8}{3}\)

Vậy...