dua vao t/c dg trung bình

a chia 5 dư 4 thì a có dạng: 5k+4

\(=>a^2=\left(5k+4\right)^2=\left(5k\right)^2+2.5k.4+4^2=25k^2+40k+16\)

\(=5\left(5k^2+8k+3\right)+1\) chia 5 dư 1 (vì 5(5k²+8k+3) chia hết cho 5)

Vậy................

 \(-x^2+4x-5< 6=>-x^2+4x< 11=>-x^2+4x-11< 0\)

Ta có \(-x^2+4x-11=-\left(x^2-4x+11\right)=-\left(x^2-4x+4+7\right)=-\left(x^2-2.x.2+2^2+7\right)\)

\(=-\left[\left(x-2\right)^2+7\right]=-7-\left(x-2\right)^2< 0\) (với mọi x)

\(=>-x^2+4x-11< 0=>-x^2+4x< 11=>-x^2+4x-5< 6\) (đpcm)

Cách thông dụng nhất:
a3+b3+c3−3abca3+b3+c3−3abc
=a3+3ab(a+b)+b3+c3−3abc−3ab(a+b)=a3+3ab(a+b)+b3+c3−3abc−3ab(a+b)
=(a+b)3+c3−3ab(a+b+c)=(a+b)3+c3−3ab(a+b+c)
=(a+b+c)(a2+2ab+b2−ab−ac+c2)−3ab(a+b+c)=(a+b+c)(a2+2ab+b2−ab−ac+c2)−3ab(a+b+c)
=(a+b+c)(a2+b2+c2−ab−bc−ca)=(a+b+c)(a2+b2+c2−ab−bc−ca)

=(x+y)^3-3xy(x+y)-3xyz+z^3

=((x+y)^3-3xy(x+y+z)+(x+y+z)[(x+y)^2-(x+y)z+z^2]ư

=(x+y+z)[(x+y)^2-(x+y)z+z^2-3xy]

=(x+y+z)(x^2+y^2+z^2+xy+yz+zx)

5x² - 10xy + 5y² - 20z²

= 5.(x² - 2xy + y² - 4z²)

= 5.[(x² - 2xy + y²) - (2z)²]

= 5.[(x - y)² - (2z)²]

= 5.(x - y - 2z).(x - y + 2z)

x².(1 - x²) - 4 + 4x²

= x².(1 - x²) - 4.(1 - x²)

= (1 - x²).(x² - 4)

= (1 - x)(1 + x)(x - 2)(x + 2)

a) phân tích đc \(\left(x+y+2z\right)\left(x+y-2z\right)\)

b) phân tích đc \(\left(1-x^2\right)\left(x-2\right)\left(x+2\right)\)

k chị nha

mk làm cho 1) các phần sau cũng z

1) = x² - 2² + (x-2)²

= (x+2)(x-2) +(x-2)(x-2)

= (x-2)(x+2+x-2)

2x(x-2)

Áp dụng hằng đẳng thức : \(\left(x+y\right)^3=x^3+3x^2y+3xy^2+y^3=x^3+y^3+3xy\left(x+y\right)\)

Ta có: \(\left(x+y+z\right)^3-x^3-y^3-z^3=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)

\(=\left(x+y\right)^3+z^3+3\left(x+y\right).z.\left(x+y+z\right)-x^3-y^3-z^3\)

\(=x^3+y^3+3xy\left(x+y\right)+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)

\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)

\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)