Đáp án:
$H7)\\
A>B\\
H8)\\
b)
P=-35\\
Q=100$
Giải thích các bước giải:
$H7)\\
B=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)\\
\Rightarrow 2B=2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)\\
\Rightarrow 2B=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)\\
\Rightarrow 2B=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)\\
\Rightarrow 2B=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)\\
\Rightarrow 2B=(3^8-1)(3^8+1)(3^{16}+1)\\
\Rightarrow 2B=(3^{16}-1)(3^{16}+1)\\
\Rightarrow 2B=(3^{32}-1)\\
\Rightarrow B=\dfrac{3^{32}-1}{2}\\
\Rightarrow A>B\\
H8)\\
a)
a+b+c=2p\Rightarrow p=\dfrac{a+b+c}{2}\\
\Rightarrow VT=(p-a)^2+(p-b)^2+(p-c)^2\\
=(\dfrac{a+b+c}{2}-a)^2+(\dfrac{a+b+c}{2}-b)^2+(\dfrac{a+b+c}{2}-c)^2\\
=\dfrac{1}{4}(a+b+c-2a)^2+\dfrac{1}{4}(a+b+c-2b)^2+\dfrac{1}{4}(a+b+c-2c)^2\\
=\dfrac{1}{4}(b+c-a)^2+\dfrac{1}{4}(a+c-b)^2+\dfrac{1}{4}(a+b-c)^2\\
=\dfrac{1}{4}.\left ( b^2+(c-a)^2+2b(c-a)+a^2+(c-b)^2+2a(c-b)+a^2+(b-c)^2+2a(b-c)\right )\\
=\dfrac{1}{4}.\left ( b^2+c^2-2ac+a^2+2bc-2ab+a^2+c^2-2bc+b^2+2ac-2ab+a^2+b^2-2bc+c^2+2ab-2ac\right )\\
=\dfrac{1}{4}.\left ( 3b^2+3c^2-2ac+3a^2-2ab-2bc\right )\\
VP=a^2+b^2+c^2-\dfrac{a+b+c}{2}^2\\
=\dfrac{1}{4}.\left ( 4a^2+4b^2+4c^2-(a+b+c)^2 \right )\\
=\dfrac{1}{4}.\left ( 4a^2+4b^2+4c^2-(a^2+b^2+c^2+2ab+2bc+2ac) \right )\\
=\dfrac{1}{4}.\left ( 4a^2+4b^2+4c^2-a^2-b^2-c^2-2ab-2bc-2ac \right )\\
=\dfrac{1}{4}.\left ( 3a^2+3b^2+3c^2-2ab-2bc-2ac \right )=VT\Rightarrow đpcm\\
b)
x+y=5\Rightarrow x=5-y\\
\Rightarrow P=3x^2-2x+3y^2-2y+6xy-100\\
=3(5-y)^2-2(5-y)+3y^2-2y+6(5-y)y-100\\
=3.(25-10y+y^2)-10+2y+3y^2-2y+30y-6y^2-100\\
=75-30y+3y^2-10+2y+3y^2-2y+30y-6y^2-100\\
=(-30y+2y-2y+30y)+(3y^2+3y^2-6y^2)+(-10-100+75)\\
=-35\\
Q=x^3+y^3-2x^2-2y^2+3xy(x+y)-4xy+3(x+y)+10\\
=(5-y)^3+y^3-2(5-y)^2-2y^2+3y(5-y).5-4.(5-y).y+3.5+10\\
=125-75y+15y^2-y^3+y^3-2(25-10y+y^2)-2y^2+15y(5-y)-4y(5-y)+15+10\\
=125-75y+15y^2-y^3+y^3-50+20y-2y^2-2y^2+75y-15y^2-20y+4y^2+15+10\\
=(125-50+15+10)+(-75y+75y+20y-20y)+(15y^2-2y^2-2y^2-15y^2+4y^2)+(-y^3+y^3)\\
=100$
