c= -(a+b) ; a+b =-c

a³+b³+c³=a³+b³-(a+b)³=a³+b³- [ a³+ 3ab(a+b)+b³ ] = - 3ab(-c) = 3 abc = 3.3 =9

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

Vì  \(\left(x-1\right)\ge0\Rightarrow\left(x-1\right)+1\ge1\)

vậy GTNN của  \(x^2-2x+2\) là 1 tại \(x-1=0\Rightarrow x=1\)

A = sin⁶ + cos⁶ + sin⁴ + cos⁴ + 5sin²cos²

= (sin² + cos²)(s​in⁴ - sin² cos² + cos⁴) + s​in⁴ + 5sin² cos² + cos⁴​

= 2(sin⁴ + 2sin² cos² + cos⁴) = 2

g: \(x\left(x-5\right)-3\left(x-5\right)=\left(x-5\right)\left(x-3\right)\)

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

i: \(x\left(x+3\right)+5\left(x+3\right)=\left(x+3\right)\left(x+5\right)\)

k: \(m\left(x-3\right)-n\left(x-3\right)=\left(x-3\right)\left(m-n\right)\)

l: \(5x-10=5\left(x-2\right)\)

\(a)5m-5n=5(m-n)\\b) -2x-2y=-2(x+y)\\c)-7+7y=-7(1-y)\\d)10x^3-15x^2=5x^2(2x-3)\\e) x^2-xy=x(x-y)\\f)9x^4-6x^2=3x^2(3x^2-2)\\g)x(x-5)-3(x-5)=(x-3)(x-5)\\h)x(x-y)-2(y-x)=x(x-y)+2(x-y)=(x+2)(x-y)\\i)x(x+3)+5(3+x)=(x+5)(x+3)\\k)m(x-3)+n(3-x)=m(x-3)-n(x-3)=(m-n)(x-3)\\l)5x-10=5(x-2) \)

giúp tớ với, tớ đang cần gấp lắm ạ.

x^2 -x - 2x + 2

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

=( x - 2 ) . (x -1)

tick nha

\(a,x^2-2x=0< =>x\left(x-2\right)=0< =>\orbr{\begin{cases}x=0\\x-2=0\end{cases}}< =>\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy nghiệm của phương trình là.....

\(b,x^2-7x-10=0< =>x^2-2x-5x-10=0< =>x\left(x-2\right)-5\left(x+2\right)=0\)

bn xem lại đề câu b, chút

a) <=> x*(x-2)=0

x=0 hoa8c5  x=2

b) luo7i2