\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=\dfrac{1}{a+b+c}\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(a+b+c\right)=abc\)

\(\Leftrightarrow a^2b+ab^2+abc+abc+b^2c+bc^2+a^2c+abc+ac^2-abc=0\)

\(\Leftrightarrow ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc\right)+ac\left(a+c\right)=0\)

\(\Leftrightarrow\left(a+c\right)\left(ab+b^2+bc+ac\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)

\(\circledast Với:a=-b\) , ta có :

\(P=\left(-b+b\right)\left(b^3+c^3\right)\left(c^5+a^5\right)=0\)

\(\circledast Với:b=-c\) , ta có :

\(P=\left(a+b\right)\left(b^3-b^3\right)\left(c^5+a^5\right)=0\)

\(\circledast Với:c=-a\) , ta có :

\(P=\left(a+b\right)\left(b^3+c^3\right)\left(-a^5+a^5\right)=0\)

KL..............

\(x\left(x+7\right)=0\)

\(x=0;-7\)

từ từ gửi hết cho 

\(\left(x+12\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+12=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-12\\x=3\end{cases}}}\)

\(\left(-x+5\right)\left(3-x\right)\)thiếu nha bn 

\(x\left(2+x\right)\left(7-x\right)=0\)

\(x=0;-2;7\)

\(\left(x-1\right)\left(x+2\right)\left(-x-3\right)=0\)

\(x=1;-2;3\)

a)        \(a+\frac{1}{a}=3\)

\(\Leftrightarrow\)\(\left(a+\frac{1}{a}\right)^2=9\)

\(\Leftrightarrow\)\(a^2+2+\frac{1}{a^2}=9\)

\(\Leftrightarrow\)\(a^2+\frac{1}{a^2}=7\)

  Ta có:      \(\left(a+\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}\right)=3.7\)

\(\Leftrightarrow\)\(a^3+\frac{1}{a}+a+\frac{1}{a^3}=21\)

\(\Leftrightarrow\)\(a^3+\frac{1}{a^3}=21-3=18\)

Ta lại có:    \(\left(a^2+\frac{1}{a^2}\right)\left(a^3+\frac{1}{a^3}\right)=7.18\)

\(\Leftrightarrow\)\(a^5+\frac{1}{a}+a+\frac{1}{a^5}=126\)

\(\Leftrightarrow\)\(a^5+\frac{1}{a^5}=126-3=123\)

Tham khảo ở đây : /hoi-dap/question/77428.html

Xin lỗi mình nhập bị nhầm. Này là toán 8 ạ

1 là 15

2 là 452

3 là 7258

nha nhớ nghe

a/ BĐT sai, cho \(a=b=c=2\) là thấy

b/ \(VT=\frac{a^4}{a^2+2ab}+\frac{b^4}{b^2+2bc}+\frac{c^4}{c^2+2ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a+b+c\right)^2}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)}{\left(a+b+c\right)^2}\)

\(VT\ge\frac{\left(a^2+b^2+c^2\right)\left(a+b+c\right)^2}{3\left(a+b+c\right)^2}=\frac{1}{3}\left(a^2+b^2+c^2\right)\)

Dấu "=" xảy ra khi \(a=b=c\)

c/ Tiếp tục sai nữa, vế phải là \(\frac{3}{2}\) chứ ko phải \(2\), và hy vọng rằng a;b;c dương

\(VT=\frac{a^2}{abc.b+a}+\frac{b^2}{abc.c+b}+\frac{c^2}{abc.a+c}\ge\frac{\left(a+b+c\right)^2}{abc\left(a+b+c\right)+a+b+c}\)

\(VT\ge\frac{9}{3abc+3}\ge\frac{9}{\frac{3\left(a+b+c\right)^3}{27}+3}=\frac{9}{\frac{3.3^3}{27}+3}=\frac{9}{6}=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Ta có:

\(a^3+b^3+b^3\ge3ab^2\) ; \(b^3+c^3+c^3\ge3bc^2\) ; \(c^3+a^3+a^3\ge3ca^2\)

Cộng vế với vế \(\Rightarrow a^3+b^3+c^3\ge ab^2+bc^2+ca^2\)

\(\frac{a^5}{b^2}+\frac{b^5}{c^2}+\frac{c^5}{a^2}=\frac{a^6}{ab^2}+\frac{b^6}{bc^2}+\frac{c^6}{ca^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{ab^2+bc^2+ca^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{a^3+b^3+c^3}=a^3+b^3+c^3\)