Bài 1:
a, \(\left(x-y\right)^2=x^2+y^2+2xy-4xy=\left(x+y\right)^2-4xy\)
Thay \(x+y=3,xy=-4\), ta có:
\(\left(x-y\right)^2=3^2-4.\left(-4\right)=25\)
b, \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)
Thay \(x+y=3,xy=-4\),ta có:
\(x^3+y^3=3^3-3.\left(-4\right).3=63\)
c, Giải \(\left\{{}\begin{matrix}x+y=3\\xy=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^3-y^3=65\\x^3-y^3=-65\end{matrix}\right.\)
Bài 1:
\(a,\left(x-y\right)^2=\left(x+y\right)^2-4xy=3^2-4.\left(-4\right)=25\)
\(b,x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=3\left[\left(x+y\right)^2-3xy\right]\)
\(=3\left(3^2-3.\left(-4\right)\right)=63\)
\(c,\)\(x+y=3\Rightarrow x=3-y\)
Thay vào xy = -4 ,có :
\(\left(3-y\right)y=-4\Leftrightarrow-y^2+3y+4=0\Leftrightarrow\left[{}\begin{matrix}y=4\\y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3-4=-1\\x=3-\left(-1\right)=4\end{matrix}\right.\)
\(TH1:x^3-y^3=\left(4^3\right)-\left(-1\right)^3=65\)
\(TH2:x^3-y^3=\left(-1\right)^3-4^3=-65\)
Bài 2:
\(A=x^2-3x=\left(x^2-3x+\frac{9}{4}\right)-\frac{9}{4}\)\(=\left(x+\frac{3}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\)
Dấu = xảy ra \(\Leftrightarrow x=-\frac{3}{2}\)
Vậy \(Min_A=-\frac{9}{4}\Leftrightarrow x=-\frac{3}{2}\)
\(B=2x^2+x=2\left(x^2+\frac{1}{2}x+\frac{1}{16}\right)-\frac{1}{8}\)
\(=2\left(x+\frac{1}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)
Dấu = xảy ra \(\Leftrightarrow x=-\frac{1}{4}\)
\(Min_B=-\frac{1}{8}\Leftrightarrow x=-\frac{1}{4}\)
Bài 2:
a, Ta có:
\(A=x^2-3x=x^2-2x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}=\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\)
Vì \(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow A\ge-\frac{9}{4}\)
\(''=''\Leftrightarrow x=\frac{3}{2}\)
b, Ta có:
\(B=2x^2+x=2\left(x^2+\frac{1}{2}x\right)=2\left(x^2+2x.\frac{1}{4}+\frac{1}{16}\right)-\frac{1}{8}\)\(=2\left(x+\frac{1}{4}\right)^2-\frac{1}{8}\)
Vì \(2\left(x+\frac{1}{4}\right)^2\ge0\Rightarrow B\ge-\frac{1}{8}\)
\(''=''\Leftrightarrow x=-\frac{1}{4}\)
