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

\(=2\left(x^2-\frac{x}{2}+\frac{5}{2}\right)\)

\(2\left(x^2-2.x.\frac{1}{4}+\frac{1}{16}+\frac{39}{16}\right)\)

\(=2\left[\left(x-\frac{1}{4}\right)^2+\frac{39}{16}\right]\)

\(=2\left(x-\frac{1}{4}\right)^2+\frac{39}{8}\ge\frac{39}{8}\)

Dấu '' ='' xảy ra 

\(\Leftrightarrow2\left(x-\frac{1}{4}\right)^2\Leftrightarrow x=\frac{1}{4}\)

Vậy........

1²+5

=1+5

=6
 

\(B=2x^2+40x-1\)

\(=2\left(x^2+20x-\frac{1}{2}\right)\)

\(=2\left(x^2+20x+100-\frac{201}{2}\right)\)

\(=2\left[\left(x+10\right)^2-\frac{201}{2}\right]\)

\(=2\left[\left(x+10\right)^2\right]-201\ge-201\)

Vậy \(B_{min}=-201\Leftrightarrow x+10=0\Leftrightarrow x=-10\)

\(B=2x^2+40x-1\)\(=2\left(x^2+2.10x+100\right)-201\)

\(=2\left(x+10\right)^2-201\)

Vì \(2\left(x+10\right)^2\ge0\forall x\)=>\(2\left(x+10\right)^2-201\ge-201\forall x\)

hay \(B\ge-201\forall x\)

\(MinB=-201\Leftrightarrow x+10=0\Leftrightarrow x=-10\)

câu a) mẫu chung: \(x^3x^2\left(x-1\right)^2\)  hơi dài nên mk ko làm ^_^

b)

\(=\frac{xy\left(a-b\right)+\left(x-a\right)\left(y-a\right)b-\left(x-b\right)\left(y-b\right)a}{ab\left(a-b\right)}\)

\(=\frac{xya-xyb+bxy-abx-aby+a^2b-axy+axb+aby-ab^2}{ab\left(a-b\right)}\)

\(=\frac{ab\left(a-b\right)}{ab\left(a-b\right)}=1\)

\(\left(2x-3\right)^2+\left(2x+1\right)^2-2\left(4x^2-9\right)\)

\(=\left(4x^2-12x+9\right)+\left(4x^2+4x+1\right)-\left(8x^2-18\right)\)

\(=\left(4x^2+4x^2-8x^2\right)+\left(4x-12x\right)+\left(9+1+18\right)\)

\(=-8x+38\)

Thay x = 3 vào bt trên

\(-8.3+38=38-24=14\)

Best_Suarez

Hàng 4 : 9 + 1 + 18 = 28. Bị nhầm rồi :)

Ta có: \(\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow\sqrt{\left(a+b\right)^2}\ge\sqrt{4ab}\)

\(\Leftrightarrow a+b\ge2\sqrt{ab}\)(Vì BĐT Cauchy chỉ áp dụng cho 2 số dương)

\(\Leftrightarrow\frac{a+b}{2}\ge\sqrt{ab}\)

Chứng minh áp dụng với n số không âm đi

\(x^2\left(x-2\right)-\left(x-1\right)\left(x^2+x+1\right)\)

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

\(=x^3-2x^2-x^3+1\)

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