Biết
1)\(\frac{a+b}{a-b}=\frac{c+a}{c-a}\).CM: \(a^2=b\cdot c\)
2)CMR:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}=\frac{1}{1002}+...+\frac{1}{2002}\)
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Câu hỏi của Cristiano Ronaldo - Toán lớp 7 - Học toán với OnlineMath
\(1-\frac{1}{2}+\frac{1}{3}-...+\frac{1}{2001}-\frac{1}{2002}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2001}\right)\)\(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)\)
= \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}+\frac{1}{2002}\right)\)\(-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2002}\right)\)\(-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1001}\right)\)
\(=\frac{1}{1002}+\frac{1}{1003}+\frac{1}{1004}+...+\frac{1}{2002}\)
Ta có \(VT=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2001}-\frac{1}{2002}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2001}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}+\frac{1}{2002}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}+\frac{1}{2002}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1001}\right)\)
\(=\frac{1}{1002}+...\frac{1}{2002}=VP\)
Vậy...
b)Ta có: \(a^{2000}+b^{2000}=a^{2001}+b^{2001}\)
\(\Rightarrow a^{2001}+b^{2001}\)\(-a^{2000}-b^{2000}=0\)
\(\Rightarrow a^{2000}\left(a-1\right)+b^{2000}\left(b-1\right)=0\)(1)
và \(a^{2001}+b^{2001}=a^{2002}+b^{2002}\)
\(\Rightarrow a^{2002}+b^{2002}\)\(-a^{2001}-b^{2001}=0\)
\(\Rightarrow a^{2001}\left(a-1\right)+b^{2001}\left(b-1\right)=0\)(2)
Lấy (2) - (1), ta được: \(a^{2000}\left(a-1\right)^2+b^{2000}\left(b-1\right)^2=0\)(3)
Mà \(a^{2000}\left(a-1\right)^2\ge0\forall a\)và \(b^{2000}\left(b-1\right)^2\ge0\forall b\)
nên (3) xảy ra\(\Leftrightarrow\hept{\begin{cases}a^{2000}\left(a-1\right)^2=0\\b^{2000}\left(b-1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1hoaca=0\\b=1hoacb=0\end{cases}}\)
Mà a,b dương nên a = 1 và b = 1
a) Áp dụng BĐT Svac - xơ:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=9\)
(Dấu "="\(\Leftrightarrow a=b=c=\frac{1}{3}\))
Theo mình thì câu 2 là :
a/ b+c + b/c+a + c/a+b =1
suy ra (a+b+c) * (a/ b+c + b/c+a + c/a+b ) = a+b+c
suy ra a*(a+b+c)/(b+c) + b*(a+b+c)/(c+a) + c*(a+b+c)/(a+b) = a+b+c
suy ra a^2+a*(b+c)/b+c +b^2 +b*(c+a)/ c+a +c^2+c*(a+b)/a+b =a=b+c
suy ra a^2/(b+c) +a +b^2/(c+a) +b +c^2/(a+b) +c =a+b+c
suy ra a^2/(b+c) +b^2/(c+a) +c^2/(a+b) =a+b+c -a-b-c
suy ra a^2/(b+c) +b^2/(c+a) +c^2/(a+b) = 0
A = -4/5x(1/2+1/3+1/4)= -4/5x1 = -4/5
B = 6/19 x ( 3/4+4/3+-1/2)= 6/19x 19 = 6
C = 2002/2003x(3/4+5/6-19/12)=2003/2002x0=0