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Ta có S m-n = (√2 + 1)m /(√2 + 1)n + (√2 - 1)m /(√2 - 1)n = (√2 + 1)m (√2 - 1)n + (√2 - 1)m (√2 + 1)n
Từ đó
S m+n + S m-n = (√2 + 1)m+n + (√2 - 1)m+n +(√2 + 1)m (√2 - 1)n + (√2 - 1)m (√2 + 1)n
= (√2 + 1)m [(√2 + 1)n + (√2 -1)n] + (√2 - 1)m [(√2 - 1)n + (√2 + 1)n]
= [(√2 + 1)n + (√2 - 1)n] [(√2 + 1)m + (√2 - 1)m]
= S m .S n
sorry mk ko bít!!! ^^
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Link : https://123doc.org/document/3369350-ung-dung-cua-dinh-ly-viet.htm
Trang 2 nhé :33
\(B1,1,S_{3n}+3S_n=\left(2-\sqrt{3}\right)^{3n}+\left(2+\sqrt{3}\right)^{3n}+3\left[\left(2-\sqrt{3}\right)^n+\left(2+\sqrt{3}\right)^n\right]\)
\(=\left[\left(2-\sqrt{3}\right)^n\right]^3+\left[\left(2+\sqrt{3}\right)^n\right]^3\)
\(+3\left[\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)\right]^n\left[\left(2-\sqrt{3}\right)^n+\left(2+\sqrt{3}\right)^n\right]\)
Ta có hằng đẳng thức \(a^3+b^3+3ab\left(a+b\right)=\left(a+b\right)^3\)
Ở đây với \(a=\left(2-\sqrt{3}\right)^n\)và \(b=\left(2+\sqrt{3}\right)^n\)
Nên \(S_{3n}+3S_n=\left[\left(2-\sqrt{3}\right)^n+\left(2+\sqrt{3}\right)^n\right]^3=S_n^3\)
\(2,S_3=\left(2-\sqrt{3}\right)^3+\left(2+\sqrt{3}\right)^3\)
\(=\left(2-\sqrt{3}+2+\sqrt{3}\right)\left(2-\sqrt{3}-\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)+2+\sqrt{3}\right)\)
\(=4\left[4-\left(4-3\right)\right]\)
\(=12\)
Ta có \(S_4=\left(2-\sqrt{3}\right)^4+\left(2+\sqrt{3}\right)^4\)
\(=\left[\left(2-\sqrt{3}\right)^2\right]^2+\left[\left(2+\sqrt{3}\right)^2\right]^2\)
\(=\left(7-4\sqrt{3}\right)^2+\left(7+4\sqrt{3}\right)^2\)
\(=97-56\sqrt{3}+97+56\sqrt{3}\)
\(=194\)
\(B2,F=x^4+6x^3+13x^2+12x+12\)(Bài này cẩn thận dấu "=")
\(=\left(x^4+6x^3+9x^2\right)+4x^2+12x+12\)
\(=\left(x^2+3x\right)^2+4\left(x^2+3x\right)+4+8\)
\(=\left(x^2+3x+2\right)^2+8\ge8\)
Dấu "=" tại \(x^2+3x+2=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
D = S35 + S60 +S100 với Sn = 1 - 2 + 3 - 4 + .... + ( -1 ) n -1 .n ( n\(\in\)N* )
Giúp mk nhanh nhé !
Với \(n\)lẻ: \(n=2k-1\)
\(S_n=1-2+3-...+\left(-1\right)^{n-1}n=1+\left(3-2\right)+...+\left[\left(-1\right)^{n-1}n-\left(-1\right)^{n-2}\left(n-1\right)\right]\)
\(=1+1+...+1=k\)
Với \(n\)chẵn: \(n=2k\)
\(S_n=1-2+3-...+\left(-1\right)^{n-1}n=\left(1-2\right)+\left(3-4\right)+...+\left[\left(-1\right)^{n-1}n-\left(-1\right)^{n-2}\left(n-1\right)\right]\)
\(=-1-1-...-1=-k\)
Áp dụng:
\(D=S_{35}+S_{60}+S_{100}=18-30-50=-62\)
a) \(u_n=\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}}\)
\(=\sqrt{\frac{\left[n\left(n+1\right)\right]^2+2n^2+2n+1}{\left[n\left(n+1\right)\right]^2}}=\sqrt{\frac{\left[n\left(n+1\right)\right]^2+2n\left(n+1\right)+1}{\left[n\left(n+1\right)\right]^2}}\)
\(=\sqrt{\frac{\left[n\left(n+1\right)+1\right]^2}{\left[n\left(n+1\right)\right]^2}}=\frac{n\left(n+1\right)+1}{n\left(n+1\right)}\in Q\)
b) \(u_n=\frac{n\left(n+1\right)+1}{n\left(n+1\right)}=1+\frac{1}{n\left(n+1\right)}=1+\frac{1}{n}-\frac{1}{n+1}\)
Vậy \(S_{2021}=u_1+u_2+...+u_{2021}=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2021}-\frac{1}{2022}\)
\(=2022-\frac{1}{2022}=\frac{2022^2-1}{2022}\)
\(\text{Δ}=\left(m+3\right)^2-4m^2\)
\(=m^2+6m+9-4m^2=-3m^2+6m+9\)
\(=-3\left(m^2-2m-3\right)=-3\left(m-3\right)\left(m+1\right)\)
Để phương trình có hai nghiệm phân biệt thì (m-3)(m+1)<0
=>-1<m<3
b:\(\Leftrightarrow x1+x2+2\sqrt{x_1x_2}=5\)
\(\Leftrightarrow m+3+2\sqrt{m^2}=5\)
=>2|m|=5-m-3=2-m
TH1: m>=0
=>2m=2-m
=>3m=2
=>m=2/3(nhận)
TH2: m<0
=>-2m=2-m
=>-2m+m=2
=>m=-2(loại)
c: P(x1)=P(x2)
=>\(x_1^3+a\cdot x_1^2+b=x_2^3+a\cdot x_2^2+b\)
=>\(\left(x_1-x_2\right)\left(x_1^2+x_1x_2+x_2^2\right)+a\left(x_1-x_2\right)\left(x_1+x_2\right)=0\)
=>(x1-x2)(x1^2+x1x2+x2^2+ax1+ax2)=0
=>x=0 và a=0
=>\(\left\{{}\begin{matrix}a=0\\b\in R\end{matrix}\right.\)