1)
\(A=x^2-3x+5\)
\(=x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2+\frac{11}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)
Có: \(\left(x-\frac{3}{2}\right)^2\ge0\)
\(\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)
Dấu = xảy ra khi:
\(x-\frac{3}{2}=0\)
\(\Rightarrow x=\frac{3}{2}\)
Vậy: \(MIN_A=\frac{11}{4}\) tại \(x=\frac{3}{2}\)
b) \(B=6x-4x^2-2\)
\(=-\left(4x^2-6x+2\right)\)
\(=-\left[\left(2x\right)^2-2.2x.2+4-2\right]\)
\(=-\left(2x-2\right)^2+2\)
Có: \(-\left(2x-2\right)^2\le0\)
\(\Rightarrow-\left(2x-2\right)^2+2\le2\)
Dấu = xảy ra khi:
\(2x-2=0\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
Vậy: ..................
1)
a) \(A=x^2-3x+5=x^2-2\cdot x\cdot\frac{3}{2}+\frac{9}{4}+\frac{11}{4}\)\(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)
Dấu bằng xảy ra \(\Leftrightarrow x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)
Vậy \(Min_A=\frac{11}{4}\) khi \(x=\frac{3}{2}\)
b) \(B=6x-4x^2-2=-4\left(x^2-\frac{3}{2}x+\frac{1}{2}\right)\) \(=-4\left(x^2-2\cdot x\cdot\frac{3}{4}+\frac{9}{16}-\frac{1}{16}\right)\) \(=\frac{1}{4}-4\left(x-\frac{3}{4}\right)^2\le\frac{1}{4}\)
Dấu bằng xảy ra \(\Leftrightarrow x-\frac{3}{4}=0\Leftrightarrow x=\frac{3}{4}\)
Vậy \(Max_B=\frac{1}{4}\) khi \(x=\frac{3}{4}\)
Bài 2:
a) Sửa đề: \(x^3+y^3+z^3-3xyz\)
Ta có: \(x^3+y^3+z^3-3xyz\)
\(=\left(x^3+y^3\right)+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy^2-3x^2y+z^3-3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-\left(3xy^2+3x^2y+3xyz\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)\cdot z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
2)
a) Sửa đề: \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3x^2y-3xy^2-3xyz\)
\(=\left(x+y+z\right)^3\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)\left(x^2+y^2+2xy-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
b) \(x^4+64=\left(x^2+8\right)^2-16x^2\) \(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
Cảm ơn các bạn nhiều!
