A= 75. (4²⁰⁰⁴+.......+4+1) + 25

   = 25 . (4-1) . (4²⁰⁰⁴+.....+4+1) +  25

   = 25.[4.(4²⁰⁰⁴+......+4+1) - (4²⁰⁰⁴+......+4+1)] + 25

   = 25.[ (4+ 4²+........+ 4²⁰⁰⁵ ) - ( 1+ 4 +........+4²⁰⁰⁴)] + 25

   = 25.(4²⁰⁰⁵ - 1) + 25 

   = 25. 4²⁰⁰⁵- 25 +25

   = 25. 4²⁰⁰⁵

   = (25. 4). 4²⁰⁰⁴

    = 100. 2²⁰⁰⁴

Mà 100 chia hết cho 100 => 100. 2²⁰⁰⁴ chia hết cho 100 

                                         => A chia hết cho 100 ( đccm)

ĐÂY LÀ TOÁN LỚP 6 !

ĐÂY LÀ TOÁN 6:

a) Ta có : \(\orbr{\begin{cases}\left(x+20\right)^{100}\ge0\\\left|y+4\right|\ge0\end{cases}}\)

=> \(\left(x+20\right)^{100}+\left|y+4\right|\ge0\)

Do đó \(\left(x+20\right)^{100}=0\)=> \(x=-20\)

\(y+4=0\Rightarrow y=-4\)

Vậy x = -20 và y = -4

b) \(\left(x-\frac{2}{5}\right)\left(x+\frac{3}{7}\right)=0\)

=> \(\orbr{\begin{cases}x-\frac{2}{5}=0\\x+\frac{3}{7}=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{3}{7}\end{cases}}\)

1)A) vì \(\Delta ABC\)CÓ \(\widehat{B}=\widehat{C}\)

\(\Rightarrow\Delta ABC\)CÂN TẠI A

\(\Rightarrow AB=AC\)

XÉT \(\Delta ADB\)VÀ\(\Delta ADC\)CÓ

\(AB=AC\left(CMT\right)\)

\(\widehat{ADB}=\widehat{ADC}\left(GT\right)\)

\(AD\)LÀ CẠNH CHUNG

\(\Rightarrow\Delta ADB=\Delta ADC\left(C-G-C\right)\)

B)VÌ\(\Delta ABC\)CÓ \(\widehat{B}=\widehat{C}\)

\(\Rightarrow\Delta ABC\)CÂN TẠI A

=> AB=AC

​​

1, There did not use to be many vehicles on the road now.

2.We used to cycle to school two years ago.

3. Now, there are more traffic accidents.

4.My brother used to go swimming when he was a child.

5. I used to walk to school everyday.

Câu cuối mk ko chắc lắm!

1.There did not use to be (as) many vehicles on the roads.

2.We used to cycle to school two years ago.

3.Now there are more traffic accidents than there used to be.

4.My brother used to go swimming when he was a child

5.I used to ...... ( câu này mk k lm đc used to nha)

=a-b+c-d-b+c-2d+3d-3a

=(a-3a)-(b+b)+(c+c)+(-d-2d+3d)

=-2a-2b+2c

=-2(a+b-c)

( a - b + c - d ) - ( b - c + 2d ) + 3( d - a )

= a - b + c - d - b + c - 2d + 3d - 3a 

= ( a - 3a ) - ( b + b ) + ( c + c ) - ( d + 2d - 3d )

= -2a - 2b + 2c - 0

= -2a + ( -2b ) - ( -2c )

= -2 ( a + b - c )