a) Vế trái  \(=\dfrac{1.3.5...39}{21.22.23...40}=\dfrac{1.3.5.7...21.23...39}{21.22.23....40}=\dfrac{1.3.5.7...19}{22.24.26...40}\)

               \(=\dfrac{1.3.5.7....19}{2.11.2.12.2.13.2.14.2.15.2.16.2.17.2.18.2.19.2.20}\\ =\dfrac{1.3.5.7.9.....19}{\left(1.3.5.7.9...19\right).2^{20}}=\dfrac{1}{2^{20}}\left(đpcm\right)\)

b) Vế trái

 \(=\dfrac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\\ =\dfrac{1.2.3.4.5.6...\left(2n-1\right).2n}{2.4.6...2n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1.2.3.4...\left(2n-1\right).2n}{2^n.1.2.3.4...n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1}{2^n}.\\ \left(đpcm\right)\)

\(\Rightarrow A=2^{2n}-1=4^n-1=\left(4-1\right)\left(4^{n-1}+4^{n-2}+...+4+1\right)=3\cdot\left(4^{n-1}+4^{n-2}+...+4+1\right)⋮3\forall n\in N\)

Lời giải:

\(M=\frac{1.2.3.4.5.6.7...(2n-1)}{2.4.6...(2n-2).(n+1)(n+2)....2n}=\frac{(2n-1)!}{2.1.2.2.2.3...2(n-1).(n+1).(n+2)...2n}\)

\(=\frac{(2n-1)!}{2^{n-1}.1.2...(n-1).(n+1).(n+2)....2n}=\frac{(2n-1)!}{2^{n-1}.1.2...(n-1).n(n+1)..(2n-1).2}\)

\(=\frac{(2n-1)!}{2^{n-1}.(2n-1)!.2}=\frac{1}{2^{n-1}.2}<\frac{1}{2^{n-1}}\)

Ta có đpcm.

đăng thể hiện mình giỏi hả nhóc, lô ga rít lớp 9 đã hc à, 

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

\(=n^3+2n^2+3n^2+6n-n-2+n^3+2\)

\(=5n^2+5n=5\left(n^2+n\right)⋮5\)

b: \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)

\(=6n^2+30n+n+5-6n^2+3n-10n+5\)

\(=24n+10⋮2\)

d: \(=\left(n+1\right)\left(n^2+2n\right)\)

\(=n\left(n+1\right)\left(n+2\right)⋮6\)

Bài 2:Tìm x biết

(4x+3)3+(5&#x2212;7x)3+(3x&#x2212;8)3=0\" id=\"MathJax-Element-4-Frame\">\\(\\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\\)

M bị phê đá à con

a: TH1: n=1

\(VT=n^2=1;VP=\frac{n\left(n+1\right)\left(2n+1\right)}{6}=\frac{1\cdot\left(1+1\right)\left(2\cdot1+1\right)}{6}=1\)

=>VT=VP

=>Mệnh đề đúng

Giả sử mệnh đề đúng với n=k

Ta cần chứng minh mệnh đề cũng đúng với n=k+1

\(1^2+2^2+\ldots+n^2+\left(n+1\right)^2\)

\(=\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\left(n+1\right)^2\)

\(=\frac{n\left(n+1\right)\left(2n+1\right)+6\left(n+1\right)^2}{6}\)

\(=\frac{\left(n+1\right)\left\lbrack n\left(2n+1\right)+6\left(n+1\right)\right\rbrack}{6}=\frac{\left(n+1\right)\left(2n^2+7n+6\right)}{6}\)

\(=\frac{\left(n+1\right)\left(2n^2+3n+4n+6\right)}{6}=\frac{\left(n+1\right)\left(2n+3\right)\left(n+2\right)}{6}\)

\(=\frac{\left(n+1\right)\left(n+1+1\right)\left\lbrack2\cdot\left(n+1\right)+1\right\rbrack}{6}\)

=>Mệnh đề cũng đúng với n=k+1

=>Mệnh đề đúng với mọi n

b: TH1: n=1

\(VT=1\left(1+1\right)=1\cdot2=2;VP=\frac{n\left(n+1\right)\left(n+2\right)}{3}=\frac{1\left(1+1\right)\left(1+2\right)}{3}=\frac{1\cdot2\cdot3}{3}=2\)

=>VT=VP

=>Mệnh đề đúng

Giả sử mệnh đề đúng với n=k

Ta cần chứng minh mệnh đề cũng đúng với n=k+1

\(1\cdot2+2\cdot3+\cdots+n\left(n+1\right)+\left(n+1\right)\left(n+2\right)\)

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

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

=>Mệnh đề đúng với n=k+1

=>Mệnh đề đúng với mọi n