Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt: \(\frac{a}{2013}=\frac{b}{2012}=\frac{c}{2011}=k\Rightarrow\hept{\begin{cases}a=2013k\\b=2012k\\c=2011k\end{cases}}\)
\(P=\frac{\left(a-c\right)^4}{\left(a-b\right)^2\left(b-c\right)^2}=\frac{\left(2013k-2011k\right)^4}{\left(2013k-2012k\right)^2\left(2012k-2011k\right)^2}=\frac{16k^4}{k^4}=16\)
Đặt \(\hept{\begin{cases}a-b=x\\b-c=y\\c-a=z\end{cases}}\)
Thế vào bài toán trở thành
Cho: \(\frac{x+z}{xz}+\frac{x+y}{xy}+\frac{y+z}{yz}=2013\left(1\right)\)
Tính \(M=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Từ (1) ta có
\(\left(1\right)\Leftrightarrow\frac{xy+yz+zx+yz+xy+zx}{xyz}=2013\)
\(\Leftrightarrow\frac{2\left(xy+yz+zx\right)}{xyz}=2013\)
\(\Leftrightarrow\frac{xy+yz+zx}{xyz}=\frac{2013}{2}\)
Ta lại có
\(M=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{xy+yz+zx}{xyz}=\frac{2013}{2}\)
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-a\right)\left(b-c\right)}+\frac{a-b}{\left(c-b\right)\left(c-a\right)}\)
\(=\frac{\left(a-c\right)-\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(b-a\right)-\left(b-c\right)}{\left(b-a\right)\left(b-c\right)}+\frac{\left(c-b\right)-\left(c-a\right)}{\left(c-b\right)\left(c-a\right)}\)
\(=\frac{1}{a-b}-\frac{1}{a-c}+\frac{1}{b-c}-\frac{1}{b-a}+\frac{1}{c-a}-\frac{1}{c-b}\)
\(=2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)=2013\)
\(\Rightarrow M=\frac{2013}{2}\)
Xét \(a+b+c=0\) thì \(\hept{\begin{cases}a+2b=c\\b+2c=a\\c+2a=b\end{cases}}\)\(\Rightarrow P=\frac{\left(2a+b\right)\left(2b+c\right)\left(2c+a\right)}{abc}=1\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(a+b+c=\frac{a+2b-c}{c}=\frac{b+2c-a}{a}+\frac{c+2a-b}{b}=\frac{a+2b-c+b+2c-a+c+2a-b}{a+b+c}=\frac{2a+2b+2c}{a+b+c}=2\)
\(\Rightarrow\hept{\begin{cases}a+2b=3c\\b+2c=3a\\c+2a=3b\end{cases}}\)\(\Rightarrow P=\frac{3a.3b.3c}{abc}=27\)
Có a+2b-c/c=b+2c-a/a=c+2a-b/b
suy ra a+2b-c/c=b+2c-a/a=c+2a-b/b=a+2b-c+b+2c-a+c+2a-b/a+b+c=2a+2b+2c/a+b+c=2
suy ra a+2b-c=2c suy ra a+2b=3c
b+2c-a=2a suy ra b+2c=3a
c+2a-b=2b suy ra c+2a=3b
Có P=(2+a/b)(2+b/c)(2+c/a)=(2b+a/b)(2c+b/c)(2a+c/a)=(3c/b)(3a/c)(3b/a)=27abc/abc=27
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-a\right)\left(b-c\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=2013\)
<=>\(\frac{\left(b-a\right)-\left(c-a\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(c-b\right)-\left(a-b\right)}{\left(b-c\right)\left(b-a\right)}+\frac{\left(a-c\right)-\left(b-c\right)}{\left(c-a\right)\left(c-b\right)}=2013\)
<=>\(\frac{1}{c-a}+\frac{1}{a-b}+\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{b-c}+\frac{1}{c-a}=2013\)
<=>\(2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)=2013\)
<=>\(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}=\frac{2013}{2}=1006,5\)
1) \(\frac{a}{b}=\frac{c}{d}=\frac{a-c}{b-d}\)
-->\(\frac{a}{b}=\frac{a-c}{b-d}\left(đpcm\right)\)
2) ta có \(\frac{a}{b}=\frac{c}{d}\)
đặt a=kb và c=kd
\(\frac{a+b}{a-b}=\frac{kb+b}{kb-b}=\frac{b\left(k+1\right)}{b\left(k-1\right)}=\frac{k+1}{k-1}\left(1\right)\)
\(\frac{c+d}{c-d}=\frac{kd+d}{kd-d}=\frac{d\left(k+1\right)}{d\left(k-1\right)}=\frac{k+1}{k-1}\left(2\right)\)
từ (1) và (2) --> \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\left(đpcm\right)\)
Ta có \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)(dãy tỉ số bằng nhau)
=> a = b = c
Khi đó \(P=\left(1+\frac{2a}{b}\right)\left(1+\frac{2b}{c}\right)\left(1+\frac{2c}{a}\right)=\left(1+\frac{2b}{b}\right)\left(1+\frac{2c}{c}\right)\left(1+\frac{2a}{a}\right)\)
= (1 + 2)(1 + 2)(1 + 2) = 3.3.3 = 27
Vậy P = 27
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\) ( do a + b + c khác 0 )
\(\Rightarrow\hept{\begin{cases}\frac{a}{b}=1\\\frac{b}{c}=1\\\frac{c}{a}=1\end{cases}}\Rightarrow a=b=c\)
Thế vào P ta được :
\(P=\left(1+\frac{2b}{b}\right)\left(1+\frac{2c}{c}\right)\left(1+\frac{2a}{a}\right)=\left(1+2\right)\left(1+2\right)\left(1+2\right)=27\)
a) Áp dụng dãy tỉ số bằng nhau.
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{a+b+c}=1\)
=> a=b=c
b) \(S=\frac{a^5.b^7.c^{2013}}{a.b^8.c^{2016}}=\frac{a^4}{b.c^3}=\frac{a^4}{a.a^3}=\frac{a^4}{a^4}=1\)