Cho biết \(\frac{a-b}{3}=\frac{b+c}{6}=\frac{c-a}{7}\)
Hãy tính \(P=c+8\left(a+b\right)-2020\)
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XO
25 tháng 8 2020
Ta có :\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=36\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=36\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=12\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
\(\Rightarrow\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}=\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\)
=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{ab}-\frac{2}{bc}-\frac{2}{ca}=0\)
=> \(\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\left(\frac{1}{b^2}-\frac{2}{bc}+\frac{1}{c^2}\right)+\left(\frac{1}{c^2}-\frac{2}{ac}+\frac{1}{a^2}\right)=0\)
=> \(\left(\frac{1}{a}-\frac{1}{b}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2+\left(\frac{1}{c}-\frac{1}{a}\right)^2=0\)
=> \(\hept{\begin{cases}\frac{1}{a}-\frac{1}{b}=0\\\frac{1}{b}-\frac{1}{c}=0\\\frac{1}{c}-\frac{1}{a}=0\end{cases}}\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)
Khi đó \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\Leftrightarrow3\frac{1}{a}=6\Rightarrow\frac{1}{a}=2\Leftrightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=2\)
Khi đó Đặt P = \(\left(\frac{1}{a}-3\right)^{2020}+\left(\frac{1}{b}-3\right)^{2020}+\left(\frac{1}{c}-3\right)^{2020}\)
= (2 - 3)2020 + (2 - 3)2020 + (2 - 3)2020
= 1 + 1 + 1 = 3
Vậy P = 3
17 tháng 12 2020
Méo bt trẩu là gì à =))
Bảo ezzz thì chỉ hộ cách làm ko bt thì đừng cư xử như 1 đứa trẻ trâu=))
30 tháng 1 2020
Đặt \(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=2018k\\b=2019k\\c=2020k\end{matrix}\right.\)
\(\Rightarrow\left(a-c\right)^3=\left(2018k-2020k\right)^3=\left(-2k\right)^3=-8k^3\) (1)
\(8\left(a-b\right)^2.\left(b-c\right)=8\left(2018k-2019k\right)^2.\left(2019k-2020k\right)=8k^2\left(-k\right)=8\left(-k\right)^3=-8k^3\left(2\right)\)
Từ (1) và (2) ⇒ \(\left(a-c\right)^3=8\left(a-b\right)^2.\left(b-c\right)\left(đpcm\right)\)
G
14 tháng 3 2019
\(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)-3abc+c^3=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-ac-bc+c^2-3ab\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(a;b;c>0\Rightarrow a+b+c>0\)
\(\Rightarrow a^2+b^2+c^2-ab-bc-ac=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)
\(P=0\)
9 tháng 6 2018
Đặt A=\(\left(\frac{-a}{2}+\frac{b}{3}+\frac{c}{6}\right)^3+\left(\frac{a}{3}+\frac{b}{6}-\frac{c}{2}\right)^3+\left(\frac{a}{6}-\frac{b}{2}+\frac{c}{3}\right)^3\)
\(=\left(\frac{-3a+2b+c}{6}\right)^3+\left(\frac{2a+b-3c}{6}\right)^3+\left(\frac{a-3b+2c}{6}\right)^3\)
\(=\left(\frac{-3a+2b+c+2a+b-3c+a-3b+2c}{6}\right)^3-\frac{\left(-a+3b-2c\right)\left(3a-2b-c\right)\left(-2a-b+3c\right)}{72}\)
(Hằng đẳng thức)
\(=0-\frac{\left(-a+3b-2c\right)\left(3a-2b-c\right)\left(-2a-b+3c\right)}{72}\)
\(\Rightarrow\frac{\left(a-3b+2c\right)\left(-3a+2b+c\right)\left(2a+b-3c\right)}{72}=\frac{1}{8}\)
\(\Leftrightarrow\left(a-3b+2c\right)\left(2a+b-3c\right)\left(-3a+2b+c\right)=9\)(đpcm).
26 tháng 12 2020
Xét \(A=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\)
\(=a.\frac{a}{b+c}+b.\frac{b}{c+a}+c.\frac{c}{a+b}\)
\(=a.\left(\frac{a}{b+c}+1-1\right)+b.\left(\frac{b}{c+a}+1-1\right)+c.\left(\frac{c}{a+b}+1-1\right)\)
\(=a.\frac{a+b+c}{b+c}-a+b.\frac{a+b+c}{c+a}-b+c.\frac{a+b+c}{a+b}-c\)
\(=\left(a+b+c\right).\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-\left(a+b+c\right)\)
\(=\left(a+b+c\right).2020-\left(a+b+c\right)\)
\(\Rightarrow P=\frac{A}{a+b+c}=\frac{\left(a+b+c\right).2019}{a+b+c}=2019\)
Vậy...
HM
17 tháng 7 2016
Từ a/b=c/d =>a/c=b/d
Đặt a /c =b /d =k =>a =ck, b= dk
=>a2020/b2020 =(ck)2020/(dk)2020 = c2020 . k2020/ d2020 .k2020 = c2020/d2020
(a-c)2020/ (b-d)2020 = (ck-c)2020/ (dk-d)2020 =[ c.(k-1)]2020/ [ d.(k-1)]2020 =c2020.(k-1)2020 / d2020. (k-1)2020 = c2020/ d2020
=> a2020/ b2020 = (a-c)2020 / (b-d)2020 (vì đều bằng c2020/d2020)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a-b}{3}=\frac{b+c}{6}=\frac{c-a}{7}=\frac{a-b+b+c+c-a}{3+6+7}=\frac{2c}{16}=\frac{c}{8}\)
mà \(\frac{b+c}{6}=\frac{c-a}{7}=\frac{\left(b+c\right)-\left(c-a\right)}{6-7}=\frac{b+c-c+a}{-1}=-\left(a+b\right)\)
\(\Rightarrow\frac{c}{8}=-\left(a+b\right)\)\(\Rightarrow c=-8\left(a+b\right)\)
Ta có: \(P=c+8\left(a+b\right)-2020=-8\left(a+b\right)+8\left(a+b\right)-2020=-2020\)
Ta có :\(\frac{a-b}{3}=\frac{b+c}{6}=\frac{c-a}{7}=\frac{a-b+b+c-c+a}{3+6-7}=\frac{2a}{2}=a\)(1)(dãy tỉ số bằng nhau)
\(\frac{a-b}{3}=\frac{b+c}{6}=\frac{c-a}{7}=\frac{a-b-b-c+c-a}{3-6+7}=\frac{-2b}{4}=-\frac{b}{2}\)(2)(dãy tỉ số bằng nhau)
\(\frac{a-b}{3}=\frac{b+c}{6}=\frac{c-a}{7}=\frac{a-b+b+c+c-a}{3+6+7}=\frac{2c}{16}=\frac{c}{8}\)(3)(dãy tỉ số bằng nhau)
Từ (1)(2)(3) => \(\frac{a}{1}=\frac{-b}{2}=\frac{c}{8}\)
Đựt \(\frac{a}{1}=\frac{-b}{2}=\frac{c}{8}=k\Rightarrow\hept{\begin{cases}a=k\\b=-2k\\c=8k\end{cases}}\)
Khi đó P = c + 8(a + b) - 2020 = 8k + 8(k - 2k) - 2020 = 8k - 8k - 2020 = -2020
Vậy P = -2020