A= (2x-1)(3x²-5x+6) = 6x³- 10x² +12x-3x² +5x -6

A = 6x³ -15x² +17x +6

vậy hệ số của x² là -15

\(\left(x-3\right)^2\) luôn lớn hơn hoặc bằhg 0

=> -17-(x-3)^2 luôn nhỏ hơn hoặc bằng -17

dấu "=" xảy ra khi (x-3)^2=0 

=>x=3

vớii x=3, ta có: -17-(3-3)^2=-17-0=-17

vậy GTLN của -17-(x-3)^2 là -17 tại x=3

a⁵+a+1=a⁵+a⁴+a³+a²+a+1-a⁴-a³-a²

=a³.(a²+a+1)+(a²+a+1)-a².(a²+a+1)

=(a²+a+1)(a³-a²+1)

Ta có : \(a^5+a+1=\left(a^5-a^2\right)+\left(a^2+a+1\right)\)

\(=a^2\left(a^3-1\right)+\left(a^2+a+1\right)\)

\(=a^2\left(a-1\right)\left(a^2+a+1\right)+\left(a^2+a+1\right)\)

\(=\left(a^2+a+1\right)\left(a^3-a^2+1\right)\)

a) \(x^2+4y^2+4xy\)

\(=x^2+2.x.2y+\left(2y\right)^2\)

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

b) \(\left(x+y\right)^2-\left(x-y\right)^2\)

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

\(=2y.2x\)

\(=4xy\)

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

\(=\left(3x+1-x-1\right)\left(3x+1+x-1\right)\)

a) \(x^6-y^6=\left(x^2\right)^3-\left(y^2\right)^3\)

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

a) \(2x\left(3x+5\right)=3x\left(10+2x\right)+15\)

\(\Leftrightarrow6x^2+10x=30x+6x^2+15\)

\(\Leftrightarrow6x^2+10x-30x-6x^2=15\)

\(\Leftrightarrow-20x=15\)

\(\Leftrightarrow x=-0.75\)

b) \(3x\left(x+5\right)-x\left(3x-10\right)+7=0\)

\(\Leftrightarrow3x^2+15x-3x^2+10x+7=0\)

\(\Leftrightarrow25x+7=0\)

\(\Leftrightarrow25x=-7\)

\(\Leftrightarrow x=-0.28\)

Ta có : \(\left(x-3\right)^2\ge0\Leftrightarrow-\left(x-3\right)^2\le0\Leftrightarrow-\left(x-3\right)^2-17\le-17\)

Vậy Max = -17 <=> x = 3