Bài 1:
b, \(x^2+y^2-2x+4y+5\)
\(=x^2-x-x+1+y^2+2y+2y+4\)
\(=\left(x^2-x\right)-\left(x-1\right)+\left(y^2+2y\right)+\left(2y+4\right)\)
\(=x.\left(x-1\right)-\left(x-1\right)+y.\left(y+2\right)+2.\left(y+2\right)\)
\(=\left(x-1\right)^2+\left(y+2\right)^2\)
Chúc bạn học tốt!!!
Bài 7b:
Ta có: \(a^2+b^2+c^2=ab+ac+bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2ac+2bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
Mà \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(a-c\right)^2\ge0\end{matrix}\right.\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\Leftrightarrow a=b=c\)
\(\Rightarrowđpcm\)
Bài 1b:
\(x^2+y^2-2x+4y+5\)
\(=x^2-2x+1+y^2+4y+4\)
\(=\left(x-1\right)^2+\left(y+2\right)^2\)
Câu hỏi của Trần Thị Minh Hậu - Toán lớp 8 - Học toán với OnlineMath
\(a^2+b^2+c^2=ab+ac+bc\Leftrightarrow2a^2+2b^2+2c^2=2ab+2ac+2bc\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\)
\(\Leftrightarrow a=b=c\)
