Bài 1: Nhân đơn thức với đa thức

a.

\(M=\left(4a^2-4a+1\right)+\left(a^2+6ab+9b^2\right)-3\)

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

\(M_{min}=-3\) khi \(\left(a;b\right)=\left(\dfrac{1}{2};-\dfrac{1}{6}\right)\)

b.

Đặt \(P=k^3+3k^2-k-3=\left(k-1\right)\left(k+1\right)\left(k+3\right)\)

Do k lẻ \(\Rightarrow k=2n+1\)

\(\Rightarrow P=2n.\left(2n+2\right)\left(2n+4\right)=8n\left(n+1\right)\left(n+2\right)\)

Do \(n\left(n+1\right)\left(n+2\right)\) là tích 3 số tự nhiên liên tiếp nên luôn chia hết cho 6

\(\Rightarrow P=8n\left(n+1\right)\left(n+2\right)⋮\left(8.6\right)\Rightarrow P⋮48\)

5:

\(\dfrac{3x^4-8x^3-10x^2+ax+b}{3x^2-2x+1}\)

\(=\dfrac{3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5+\left(a-8\right)x+b+5}{3x^2-2x+1}\)

\(=x^2-2x-5+\dfrac{\left(a-8\right)x+b+5}{3x^2-2x+1}\)

Để đây là phép chia hết hết thì a-8=0 và b+5=0

=>a=8 và b=-5

=>(x^2+4x+3)(x^2+4x-3)=16

=>(x^2+4x)^2-9=16

=>(x^2+4x)^2=25

=>x^2+4x=-5 hoặc x^2+4x=5

=>x^2+4x-5=0

=>(x+5)(x-1)=0

=>x=1 hoặc x=-5

\(=-15x^4y^7+10x^5y^6+5x^3y^5\)

\(=-15x^4y^7+10x^5y^6+5x^3y^5\)

\(3x^4.\left(-2x^3+5x^2-\dfrac{2}{3}x+\dfrac{1}{3}\right)=-6x^7+15x^6-2x^5+x^4\)

\(=-6x^7+15x^6-2x^5+x^4\)

a: =>2x^2-2x-2x^2=7

=>-2x=7

=>x=-7/2

b: =>6x^2-3x+10x-5-6x^2-12x=x

=>-5x-5=x

=>-6x=5

=>x=-5/6

c: =>(x+3)(2-x)=0

=>x=2 hoặc x=-3

tách ra em nhé!

a) x³+y³+z³-3xyz

= (x+y)³ - 3x²y-3xy²+z²-3xyz

=(x+y+z)[(x+y)²-(x+y)z+z²]-3x(x+y+z)

=(x+y+z)(x²+y²+z²-xy-yz-zx)

b) 

=>(a+b+c)3=0

=>a3+b3+c3+3a2b+3ab2+3b2c+3bc2+3a2c+3ac2+6abc=0

=>a3+b3+c3+(3a2b+3ab2+3abc)+(3b2c+3bc2+3abc)+(3a2c+3ac2+3abc)-3abc=0

=>a3+b3+c3+3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)=3abc

=>a3+b3+c3=3abc \(\left(đpcm\right)\)

a) x³+y³+z³-3xyz

= (x+y)³ - 3x²y-3xy²+z²-3xyz

=(x+y+z)[(x+y)²-(x+y)z+z²]-3x(x+y+z)

=(x+y+z)(x²+y²+z²-xy-yz-zx)

b) 

=>(a+b+c)³=0

=>a³+b³+c³+3a²b+3ab²+3b²c+3bc²+3a²c+3ac³+6abc=0

=>a³+b³+c³+(3a²b+3ab²+3abc)+(3b²c+3bc²+3abc)+(3a²c+3ac²+3abc)-3abc=0

=>a³+b³+c³+3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)=3abc

=>a³+b³+c³=3abc (đpcm)

a: =3x(x-2)

b: =(x-1)^2-y^2

=(x-1-y)(x-1+y)

c: \(=9x^2\left(x-y\right)-4\left(x-y\right)=\left(x-y\right)\left(3x-2\right)\left(3x+2\right)\)

d: \(=x\left(x^2-2x-8\right)=x\left(x-4\right)\left(x+2\right)\)

a: \(=2\left(x^2+\dfrac{5}{2}x+\dfrac{1}{2}\right)\)

\(=2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}-\dfrac{17}{16}\right)\)

\(=2\left(x+\dfrac{5}{4}\right)^2-\dfrac{17}{8}>=-\dfrac{17}{8}\)

Dấu = xảy ra khi x=-5/4

b: \(=2\left(x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}\right)=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}>=-\dfrac{9}{2}\)

Dấu = xảy ra khi x=3/2