\(\left\{{}\begin{matrix}a+b+c=1\\a^2+b^2+c^2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a+b+c\right)^2=1\\a^2+b^2+c^2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2+c^2+2ab+2bc+2ca=1\\a^2+b^2+c^2=2\end{matrix}\right.\)

trừ 2 vế ta được:
\(2ab+2bc+2ca=-1\)

\(\Leftrightarrow ab+bc+ca=-\dfrac{1}{2}\)

\(x^3+2x-3=0\)

\(\Leftrightarrow x^3-x+3x-3=0\)

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

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

\(\Leftrightarrow\left(x-1\right)\left[x\left(x+1\right)+3\right]=0\)

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

\(\Leftrightarrow\left(x-1\right)\left[\left(x^2+x+\dfrac{1}{4}\right)+3-\dfrac{1}{4}\right]=0\) 

\(\Leftrightarrow\left(x-1\right)\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\right]=0\) (*)

Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\forall x\) (**)

Từ (*);(**) => \(x-1=0\) =>x=1

Vậy x=1

Xét tứ giác ABCD có:

\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)

\(\Leftrightarrow\widehat{A}+\widehat{B}+80^o+100^o=360^o\)

=> \(\widehat{A}+\widehat{B}=180^o\)

Và \(\widehat{A}-\widehat{B}=20^o\)

=>\(\widehat{A}=\dfrac{180^o+20^o}{2}=100^0\)

=>\(\widehat{B}=100^o-20^o=80^o\) 

x= 6,5344468891

(-c)(-b)=bc 

\(A=x^2-x+\dfrac{1}{2}=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}\\ =\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)

Dấu ''='' xảy ra khi : \(\left(x-\dfrac{1}{2}\right)^2=0< =>x=\dfrac{1}{2}\)

Vậy GTNN của `A` là : `1/4<=>x=1/2`