\(\left(A+B\right)^2=A^2+2AB+B^2\)

\(\left(A+B\right)^3=A^3+3A^2B+3AB^2+B^3\)

\(A^2-B^2=\left(A-B\right)\left(A+B\right)\)

2

a

\(15x^2y^3z^2-20x^2yz^2+10xy^3z\)

\(=5xyz\left(3xy^2z-4xz+2y^2\right)⋮5xyz\)

b

\(13ab^2+abc+32a=a\left(13b^2+bc+32\right)\) 

TH1:\(13b^2+bc+32=7b\cdot P\left(x\right)\) thì A chia hết cho B

TH2:\(13b^2+bc+32=7b\cdot Q\left(x\right)+r\left(r>0\right)\) thì A không chia hết cho B

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

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

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

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

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

Dấu '' ='' xảy ra 

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

Vậy........

1²+5

=1+5

=6
 

https://olm.vn/tin-tuc /Bat-dang-thuc-Cauchy-(-Co-si)

#Trang

Bất đẳng thức Cauchy ( Cô-si) - Học toán với OnlineMath

#Trang

a)x^5+x+1

=x⁵-x²+x²+x+1

=x²(x³-1)+x²+x+1

=x²(x+1)(x²+x+1)+x²+x+1

=(x²+x+1)(x³+x²+1)

b)(x+1)(x+3)(x+5)(x+7)+15

=(x²+8x+7)(x²+8x+15)+15

Đặt x²+8x+7=t

=> t(t+8)+15=t²+8t+15

=(t+3)(t+5)

=(x²+8x+10)(x²+8x+12)

1; 1; 2; 4; 6; 10; 16; 26;..42..; 68.

2; 8; 14; 20; .26....; .32..... 

Học tốt!!!!!!!!!

- 42; 68
- 26; 32