<=> 2x=x^2-2x-3

<=> \(x^2-4x-3=0\)

Tach ra thanh \(x^2-3x-x-3\)=0 roi dat nhan tu chung ma giai.

Theo bài ra , ta có : 

\(x+\frac{1}{x}=a\)

\(\Rightarrow\left(x+\frac{1}{x}\right)^2=a^2\)

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

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

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

Vậy \(x^2+\frac{1}{x^2}=a^2-2\)

Chúc bạn học tốt =)) 

=>(x-3)*(2x+7)=0

=>(x-3)(2x+7)=(3-3)(2*3-7)=0

vậy x=6 

k cho mình nhé!

c/ Ta có:\(6a-5b=1\)

\(\Rightarrow5b=6a-1\)

Theo đề thì: \(A=4a^2+\left(6a-1\right)^2=40a^2-12a+1\)

\(=\left(\left(2\sqrt{10}a\right)^2-\frac{2.2.\sqrt{10}.3a}{\sqrt{10}}+\frac{9}{10}\right)+\frac{1}{10}\)

\(=\left(2\sqrt{10}a-\frac{3}{\sqrt{10}}\right)^2+\frac{1}{10}\ge\frac{1}{10}\)

còn câu a,b nữa a ơi :((

Ta có: \(S=a^3+b^3+c^3+3a^2+3b^2+3c^2\)

\(=\left(a^3-a\right)+\left(b^3-b\right)+\left(c^3-c\right)+\left(3a^2-3a\right)+\left(3b^2-3b\right)+\left(3c^2-3c\right)+4\left(a+b+c\right)\)

\(=a\left(a+1\right)\left(a-1\right)+b\left(b-1\right)\left(b+1\right)+c\left(c-1\right)\left(c+1\right)+3a\left(a-1\right)+3b\left(b-1\right)+3c\left(c-1\right)+4\left(a+b+c\right)\)

Ta thấy: \(\hept{\begin{cases}a\left(a-1\right)\left(a+1\right)⋮6\\b\left(b-1\right)\left(b+1\right)⋮6\\c\left(c-1\right)\left(c+1\right)⋮6\end{cases}}\)(1)

\(\hept{\begin{cases}3a\left(a-1\right)⋮6\\3b\left(b-1\right)⋮6\\3c\left(c-1\right)⋮6\end{cases}}\)(2)

\(4\left(a+b+c\right)⋮6\)(3)

Từ (1),(2),(3) ta suy ra \(S⋮6\)