![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có : \(a^2+b^2+c^2+42=2a+8b+10c\)
\(\Leftrightarrow\) \(a^2+b^2+c^2+42-2a-8b-10c=0\)
\(\Leftrightarrow\) \(\left(a^2-2a+1\right)+\left(b^2-8b+16\right)+\left(c^2-10c+25\right)=0\)
\(\Leftrightarrow\) \(\left(a-1\right)^2+\left(b-4\right)^2+\left(c-5\right)^2=0\)
mà \(\left\{{}\begin{matrix}\left(a-1\right)^2\ge0\forall a\\\left(b-4\right)^2\ge0\forall b\\\left(c-5\right)^2\ge0\forall c\end{matrix}\right.\)
\(\Rightarrow\) \(\left(a-1\right)^2+\left(b-4\right)^2+\left(c-5\right)^2=0\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}\left(a-1\right)^2=0\\\left(b-4\right)^2=0\\\left(c-5\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a-1=0\\b-4=0\\c-5=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=1\\b=4\\c=5\end{matrix}\right.\)
khi đó \(a+b+c=1+4+5=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu hỏi của Hattory Heiji - Toán lớp 8 - Học toán với OnlineMath
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
2:
a: =>a^2+2ab+b^2-2a^2-2b^2<=0
=>-(a^2-2ab+b^2)<=0
=>(a-b)^2>=0(luôn đúng)
b; =>a^2+b^2+c^2+2ab+2ac+2bc-3a^2-3b^2-3c^2<=0
=>-(2a^2+2b^2+2c^2-2ab-2ac-2bc)<=0
=>(a-b)^2+(b-c)^2+(a-c)^2>=0(luôn đúng)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\)Ta có: \(a+b+c=0 \Rightarrow b+c=-a \Rightarrow (b+c)^2=(-a)^2 \Leftrightarrow b^2+c^2+2bc=a^2 \Leftrightarrow a^2-b^2-c^2=2bc\)
Tương tự: \(b^2-c^2-a^2=2ca;c^2-a^2-b^2=2ab\)
\(P=...=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)
----
Bổ đề \(a+b+c=0 \Leftrightarrow a^3+b^3+c^3\)
Ở đây ta c/m chiều thuận:
Với \(a+b+c=0 \Leftrightarrow a+b=-c \Rightarrow (a+b)^3=(-c)^3 \Leftrightarrow a^3+b^3+3ab(a+b)=-c^3 \Leftrightarrow a^3+b^3+c^3=3abc(QED)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}\ge\dfrac{4}{a+2b+c}\ge\dfrac{4}{\dfrac{a^2+1}{2}+b^2+1+\dfrac{c^2+1}{2}}=\dfrac{8}{b^2+7}\)
Tương tự
\(\dfrac{1}{a+b}+\dfrac{1}{a+c}\ge\dfrac{8}{a^2+7}\)
\(\dfrac{1}{b+c}+\dfrac{1}{a+c}\ge\dfrac{8}{c^2+7}\)
Cộng vế:
\(2\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge\dfrac{8}{a^2+7}+\dfrac{8}{b^2+7}+\dfrac{8}{c^2+7}\)
\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{4}{a^2+7}+\dfrac{4}{b^2+7}+\dfrac{4}{c^2+7}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
a2 + b2 + c2 + 42=2a +8b +10c
\(\Rightarrow a^2+b^2+c^2+42-2a-8b-10c=0\)
\(\Rightarrow\left(a^2-2a+1\right)+\left(b^2-8b+16\right)+\left(c^2-10c+25\right)=0\)
\(\Rightarrow\left(a-1\right)^2+\left(b-4\right)^2+\left(c-5\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}a-1=0\\b-4=0\\c-5=0\end{cases}}\Rightarrow\hept{\begin{cases}a=1\\b=4\\c=5\end{cases}}\)
Khi đó \(a+b+c=1+4+5=10\)
cho x<0 thỏa mãn \(\frac{1}{x^2+9x+20}\)+\(\frac{1}{x^2+11x+30}\)+\(\frac{1}{x^2+13x+42}\)=\(\frac{1}{18}\) tìm x=?
mn giải giúp mk với