Lấy 89+1=90

Vậy năm 8922 thuộc thế kỉ thứ 90 

\(a,892^2+216.892+108^2=892^2+2.108.892+108^2=\left(892+108\right)^2=1000^2=1000000\)

b, \(9x^2+42x+49=9.1^2+42.1+49=9+42+49=100\)

c,\(\left(x^2-3x\right)+x-3=x\left(x-3\right)+\left(x-3\right)=\left(x+1\right)\left(x-3\right)=\left(6+1\right)\left(6-3\right)=7. 3=21\)

3660 nha bạn

Mahoustakai Precure

\(1980+1680=3660\)

1980+1680

=2660

Đặt \(\frac{x}{5}=\frac{y}{6}=\frac{z}{7}=k\Rightarrow\hept{\begin{cases}x=5k\\y=6k\\z=7k\end{cases}}\)

Khi đó xyz = -1680

<=> 5k.6k.7k = -1680

=> 210k³ = -1680

=> k³ = -8

=> k = -2

=> \(\hept{\begin{cases}x=-10\\y=-12\\z=-14\end{cases}}\)

              Bài làm :

\(\text{Đặt : }\frac{x}{5}=\frac{y}{6}=\frac{z}{7}=k\Rightarrow\hept{\begin{cases}x=5k\\y=6k\\z=7k\end{cases}}\text{(*)}\)

Vì xyz=-1680 nên :

\(5k.6k.7k=-1680\Leftrightarrow k^3=-8\Leftrightarrow k=-2\)

Thay k=-2 vào (*) ; ta được :

\(\hept{\begin{cases}x=-10\\y=-12\\z=-14\end{cases}}\)

Đặt x/5=y/6=z/7=k

x=5.k

y=6.k

z=7.k

Mà x.y.z=(-1680)

=>5k.6y.7k=(-1680)

=>x=2.3=6

y=2.5=10

z=2.6=12

Vậy x=6

       y=10

       z=12

A\(=\left(x-4\right)\left(x-7\right)\left(x-5\right)\left(x-6\right)-1680\)

\(=\left(x^2-11x+28\right)\left(x^2-11x+30\right)-1680\)

Đặt \(\left(x^2-11x+28\right)=t\)

A\(=t\left(t+2\right)-1680=\left(t+1\right)^2-41^2=\left(t-40\right)\left(t+42\right)\)

Thay \(\left(x^2-11x+28\right)=t\)

A\(=\left(x^2-11x-12\right)\left(x^2-11x+70\right)=\left(x-12\right)\left(x+1\right)\left(x^2-11x+70\right)\)

1680 - 135 - 178 + 73

= 1680 - ( 135 + 178 - 73 )

= 1680 - ( 135 + 105 )

= 1680 - 240 = 1440

1618 nhé em