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\(2a^{2020}+2b^{2020}+2c^{2020}-2\left(ab\right)^{1010}-2\left(bc\right)^{1010}-2\left(ca\right)^{1010}=0\)
\(\Leftrightarrow\left(a^{1010}-b^{1010}\right)^2+\left(b^{1010}-c^{1010}\right)^2+\left(c^{1010}-a^{1010}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^{1010}-b^{1010}=0\\b^{1010}-c^{1010}=0\\c^{1010}-a^{1010}=0\end{matrix}\right.\)
\(\Rightarrow\left|a\right|=\left|b\right|=\left|c\right|\)
Nếu đề không cho a;b;c dương thì không tính được cụ thể giá trị A
Nếu a;b;c dương thì \(a=b=c\Rightarrow A=0\)
Nhân 2 vế của 2 ĐT đề bài ta có
\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)=\frac{47}{10}\)
<=> \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\left(\frac{a}{a+b}+\frac{b}{a+b}\right)+\left(\frac{b}{b+c}+\frac{c}{b+c}\right)+\left(\frac{c}{a+c}+\frac{a}{a+c}\right)=\frac{47}{10}\)
=>\(P=\frac{17}{10}\)
Vậy \(P=\frac{17}{10}\)
Ta có: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+a+b+c=2+2018\)
\(\Leftrightarrow\frac{a+ab+bc}{b+c}+\frac{b+bc+ab}{c+a}+\frac{c+ac+bc}{a+b}=2020\)
\(\Leftrightarrow a\left(\frac{1+b+c}{b+c}\right)+b\left(\frac{1+a+c}{a+c}\right)+c\left(\frac{1+a+b}{a+b}\right)=2020\left(1\right)\)
Vì \(a+b+c=2018\Rightarrow\hept{\begin{cases}a+b=2018-c\\b+c=2018-a\\c+a=2018-b\end{cases}\left(2\right)}\)
Thay (2) vào (1) ta được:
\(a\left(\frac{2019-a}{b+c}\right)+b\left(\frac{2019-b}{a+c}\right)+c\left(\frac{2019-c}{a+b}\right)=2020\)
\(\Leftrightarrow\frac{2019a-a^2}{b+c}+\frac{2019b-b^2}{a+c}+\frac{2019c-c^2}{a+b}=2020\)
\(\Leftrightarrow\frac{2019a}{b+c}-\frac{a^2}{b+c}+\frac{2019b}{a+c}-\frac{b^2}{a+c}+\frac{2019c}{a+b}-\frac{c^2}{a+b}=2020\)
\(\Leftrightarrow2019\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-\left(\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)=2020\)
\(\Leftrightarrow4038-\left(\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)=2020\)( vì \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=2\))
\(\Leftrightarrow\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=2018\)
\(\Leftrightarrow\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+1=2019\)
cho a^3 +b^3+c^3=3abc và a+b+c khác 0 tính giá trị của biểu thức M=a^2020+b^2020+c^2020/(a+b+c)^2020
Ta có : a3 + b3 + c3 = 3abc
=> (a + b)(a2 - ab + b2) + c3 - 3abc = 0
=> (a + b)3 - 3ab(a + b) + c3 - 3abc = 0
=> [(a + b)3 + c3] - [(3ab(a + b) + 3abc] = 0
=> (a + b + c)(a2 + b2 + 2ab - ac - bc + c2) - 3ab(a + b + c) = 0
=> (a + b + c)(a2 + b2 + c2 - ab - ac - bc) = 0
=> a2 + b2 + c2 - ab- ac - bc = 0
=> 2(a2 + b2 + c2 - ab- ac - bc) = 0
=> 2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc = 0
=> (a2 - 2ab + b2) + (b2 - 2bc + c2) + (a2 - 2ac + c2) = 0
=> (a - b)2 + (b - c)2 + (a - c)2 = 0
=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow a=b=c\)
Khi đó M = \(\frac{a^{2020}+b^{2020}+c^{2020}}{\left(a+b+c\right)^{2020}}=\frac{3.c^{2020}}{\left(3c\right)^{2020}}+\frac{3c^{2020}}{3^{2020}.c^{2020}}=\frac{1}{3^{2019}}\)
\(a^{2020}+b^{2020}+c^{2020}=a^{1010}b^{1010}+b^{1010}c^{1010}+c^{1010}a^{1010}\)
\(\Leftrightarrow a^{2020}+b^{2020}+c^{2020}-a^{1010}b^{1010}-b^{1010}c^{1010}-c^{1010}a^{1010}=0\)
\(\Leftrightarrow2a^{2020}+2b^{2020}+2c^{2020}-2a^{1010}b^{1010}-2b^{1010}c^{1010}-2a^{1010}c^{1010}=0\)
\(\Leftrightarrow\left(a^{2020}-2a^{1010}b^{1010}+b^{2020}\right)+\left(b^{2020}-2b^{1010}c^{1010}+c^{2020}\right)+\left(c^{2020}-2a^{1010}c^{1010}+a^{2020}\right)=0\)
\(\Leftrightarrow\left(a^{1010}-b^{1010}\right)^2+\left(b^{1010}-c^{1010}\right)^2+\left(c^{1010}-a^{1010}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(a^{1010}-b^{1010}\right)=0\\b^{1010}-c^{1010}=0\\c^{1010}-a^{1010}=0\end{cases}}\Leftrightarrow a^{1010}=b^{1010}=c^{1010}\Leftrightarrow\pm a=\pm b=\pm c\)
Rồi thay :> Còn thay kiểu nào thì mình cũng hong biết :">