Ta có: B = (2 +1)(2² +1)(2⁴ +1)(2⁸ +1) 

              = (2² - 1)(2² +1)(2⁴ +1)(2⁸ +1) 

              = (2⁴ - 1)(2⁴ +1)(2⁸ +1) 

              = (2⁸ - 1)(2⁸ +1)

              = 2¹⁶ - 1

 \(\Rightarrow\)B = 2¹⁶ - 1 \(< \)2¹⁶ = A 

ai mà bt được :)

Lớp 10 hay xao vậy.?.

\(\left(2x-1\right)^2=3x^2-4x+5\)

\(\Rightarrow4x^2-4x+1-3x^2+4x-5=0\)

\(\Rightarrow x^2=4\)

\(\Rightarrow x=\pm2\)

nhớ cho mik nha =)

\(xyz\le\frac{\left(x+y\right)^2z}{4}=\frac{\left(6-z\right)^2z}{4}=\frac{\left(6-z\right)\left[\left(6-z\right)z\right]}{4}\)

\(\le\frac{\left(6-z\right)\left(6-z+z\right)^2}{16}=\frac{\left(6-z\right).36}{16}\le\frac{\left(6-3\right).36}{16}=\frac{27}{4}\)

Dấu \(=\)khi \(\hept{\begin{cases}x=y\\z=3\\x+y+z=6\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y=\frac{3}{2}\\z=3\end{cases}}\).

thanks

HF là đường trung bình tam giác ABC 

=> AB=2HF 

hay x=2.1=2(cm)

EH là đường trung bình tam giác ADC

=>EH=1/2DC

hay y=1/2.4=2(cm)

\(\Leftrightarrow4x^2+4x+1-\left(4x^2-9\right)=10\)

\(\Leftrightarrow4x=0\Rightarrow x=0\)

e) \(a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b-c\right)\left(a+b\right)\)

f) \(x^2-2x-4y^2-4y=x^2-2x+1-\left(4y^2+4y+1\right)=\left(x-1\right)^2-\left(2y+1\right)^2\)

\(=\left(x-2y-2\right)\left(x+2y\right)\)

g) \(x^2y-x^3-9y+9x=x^2\left(y-x\right)-9\left(y-x\right)=\left(x^2-9\right)\left(y-x\right)=\left(x-3\right)\left(x+3\right)\left(y-x\right)\)

h) \(x^2\left(x-1\right)+16\left(1-x\right)=\left(x^2-16\right)\left(x-1\right)=\left(x-4\right)\left(x+4\right)\left(x-1\right)\)