xin loi ban minh khong biet

Bài 1:

\(x^2-x+1=x^2-x+\frac{1}{4}+\frac{3}{4}\)

\(=\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}\)

\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" khi \(x=\frac{1}{2}\)

Vậy \(Min=\frac{3}{4}\) khi \(x=\frac{1}{2}\)

Bài 2:

\(x^2+10x+2041=x^2+10x+25+2016\)

\(=\left(x^2+10x+25\right)+2016\)

\(=\left(x+5\right)^2+2016\ge2016\)

Dấu "=" khi \(x=-5\)

Vậy \(Min=2016\) khi \(x=-5\)

nhìn là bit tu lam

Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)

\(=\left(x+y\right)^3=1^3=1\)

Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)

\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)

Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)

\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)

\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)

\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)

\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)

Đặt \(A=\dfrac{x^2-10x+25}{x^2-5}\)

ĐK : \(x^2-5\ne0\\ \Leftrightarrow\left\{{}\begin{matrix}x\ne\sqrt{5}\\x\ne-\sqrt{5}\end{matrix}\right.\)

\(A=0\\ \Leftrightarrow\dfrac{x^2-10x+25}{x^2-5}=0\\ \Leftrightarrow x^2-10x+25=0\\ \Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\left(TM\right)\)

Vậy x =5 thì A =0