Question

tìm giá trị nhỏ nhất của biểu thức

A=\(3x^2-5x+3\)

B=\(2x^2+2x+1\)

C=\(2x^2+y^2+10x-2xy+27\)

D=\(x^2+4x+y^2+10y+30\)

E=\(\left(x-3\right)^2+\left(x-5\right)^2\)

H=\(\left(x-7\right)\left(x-5\right)\left(x-4\right)\left(x-2\right)-70\)

Answer 1

\(A=3x^2-5x+3=3(x^2-\frac{5}{3}x)+3\)

\(=3(x^2-\frac{5}{3}x+\frac{5^2}{6^2})+\frac{11}{12}=3(x-\frac{5}{6})^2+\frac{11}{12}\)

Vì \((x-\frac{5}{6})^2\geq 0, \forall x\Rightarrow A\geq 3.0+\frac{11}{12}=\frac{11}{12}\)

Vậy A(min)$=\frac{11}{12}$ khi $x=\frac{5}{6}$

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

\(=2(x+\frac{1}{2})^2+\frac{1}{2}\geq 2.0+\frac{1}{2}=\frac{1}{2}\)

Vậy \(B_{\min}=\frac{1}{2}\) tại \((x+\frac{1}{2})^2=0\Leftrightarrow x=\frac{-1}{2}\)

Answer 2

C)

\(C=2x^2+y^2+10x-2xy+27\)

\(=(x^2+10x+25)+(x^2+y^2-2xy)+2\)

\(=(x+5)^2+(x-y)^2+2\)

Vì \((x+5)^2\ge 0, (x-y)^2\geq 0\Rightarrow C\geq 0+0+2=2\)

Vậy \(C_{\min}=2\) tại \(\left\{\begin{matrix} (x+5)^2=0\\ (x-y)^2=0\end{matrix}\right.\Leftrightarrow x=y=-5\)

Answer 3

E)

\(E=(x-3)^2+(x-5)^2\)

\(=x^2-6x+9+x^2-10x+25\)

\(=2x^2-16x+34=2(x^2-8x+16)+2\)

\(=2(x-4)^2+2\geq 2.0+2=2\)

Vậy $E_{\min}=2$ tại $(x-4)^2=0$ hay $x=4$

H)

\(H=(x-7)(x-5)(x-4)(x-2)-70\)

\(=[(x-7)(x-2)][(x-5)(x-4)]-70\)

\(=(x^2-9x+14)(x^2-9x+20)-70\)

\(=a(a+6)-70\) (đặt \(a=x^2-9x+14)\)

\(=(a+3)^2-79\geq 0-79=-79\)

Vậy $H_{\min}=-79$ tại \(a=x^2-9x+14=-3\Leftrightarrow x=\frac{9\pm \sqrt{13}}{2}\)