cho a^3+b^3+c^3=3abc va a+b+c khac 0 . tinh gia tri bieu thuc \(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}\)
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Ta có: a3+b3+c3=3abc <=> a3+b3+c3-3abc=0
<=>\(a^3+3a^2b+3ab^2+b^3+c^3-3ab\left(a+b\right)-3abc=0\)
<=>\(\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
<=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
<=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
<=>\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Mà a+b+c khác 0
=>\(a^2+b^2+c^2-ab-bc-ca=0\)
<=>\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
<=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
<=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
<=>\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}}a=b=c}\)
=>\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
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\(a,A=\dfrac{5-3}{5+2}=\dfrac{2}{7}\\ b,B=\dfrac{3x-9+2x+6-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ c,C=AB=\dfrac{x-3}{x+2}\cdot\dfrac{2}{x-3}=\dfrac{2}{x+2}\\ C=-\dfrac{1}{3}\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(tm\right)\)
Cho 3 so a,b,c khac 0 thoa man ab/a+b=bc/b+c=ca/c+a
Tinh gia tri cua bieu thuc M=ab+bc+ca/a^2+b^2+c^2
![](https://rs.olm.vn/images/avt/0.png?1311)
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Ta có : \(\frac{a+b-c}{c}=\frac{a+c-b}{b}=\frac{b+c-a}{a}\)
\(\Rightarrow\frac{a+b}{c}-\frac{c}{c}=\frac{a+c}{b}-\frac{b}{b}=\frac{b+c}{a}-\frac{a}{a}\)
\(\frac{a+b}{c}-1=\frac{c+b}{a}-1=\frac{a+c}{b}-1\)
\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}\)
Áp dụng tính chất của dãy tỉ số bằng nhau , ta có
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}=\frac{2a+2b+2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\a+c=2b\end{cases}}\)
Vậy \(P=\left(a+b\right)\left(b+c\right)\left(c+a\right)=2c.2a.2b=8abc\)
mà \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=abc\Rightarrow8abc=abc\Rightarrow abc=0\Rightarrow P=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
- Ta có : \(a^3+b^3+c^3=3abc\)
=> \(a^3+b^3+c^3-3abc=0\)
=> \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
Mà \(a+b+c\ne0\)
=> \(a^2+b^2+c^2-ab-bc-ac=0\)
=> \(\frac{\left(a^2-2ab+b^2\right)+\left(b^2-2ac+c^2\right)+\left(c^2-2ac+a^2\right)}{2}=0\)
=> \(\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{2}=0\)
=> \(a-b=b-c=c-a=0\)
=> \(a=b=c\)
- Thay a = b = c vào biểu thức N ta được :
\(N=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
Vậy giá trị của N = \(\frac{1}{3}\) khi \(a^3+b^3+c^3=3abc\) và \(a+b+c\ne0\)