(2x-2y)\(^3\)-xy(x-5y)-(x-y)\(^3\)

a.100-[60:(15625:625-3.5)]
=100-[60:(25-15)]
=100-(60:10)
=100-6
=94
b.50-[(50-8.5):2+3]
=50-[(50-40):2+3]
=50-(10:2+3)
=50-(5+3)
=50-8
=42

a.47-[(45.16-25.12):14]
=47-[(720-300):14]
=47-(420:14)
=47-30
=17
b.50-[(20-8):2+34]
=50-(12:2+34)
=50-(6+34)
=50-40
=10

a.[(6x-39):7].4=12
(6x-39):7=12:4
(6x-39):7=3
(6x-39)=3.7
6x-39=21
6x=21+39
6x=60
x=10
vậy x=10
b.(x:3-4).5=15
x:3-4=15:5
x:3-4=3
x:3=3+4
x:3=7
x=7.3
x=21
vậy x=21
c.\(\left|x-5\right|=\left|7\right|\)
\(\left|x-5\right|=7\)
\(\Rightarrow\left[{}\begin{matrix}x-5=7\\x-5=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=7+5\\x=-7+5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=12\\x=-2\end{matrix}\right.\)
vậy x\(\in\left\{12;-2\right\}\)

1,Dãy A có số số hạng là:
(100-1):1+1=100(số hạng)
Vì 100:4=25 nên ta nhóm như sau:
A=(1+2-3-4)+(5+6-7-8)+...+(97+98-99-100)
A=-4+(-4)+...+(-4)
A=-4.25
A=-100

=>\(-1+2^{100}=2^{100}-1\)
=>\(2^{100}-1=2^{100}-1\left(đpcm\right)\)

Đặt A=\(2+2^2+2^3+...+2^{99}\)
2A=\(2^2+2^3+2^4+...+2^{100}\)
=>2A-A=(\(2^2+2^3+2^4+...+2^{100}\))-(\(2+2^2+2^3+...+2^{99}\))
=>A=\(2^{100}-2\)
Thay vào đề bài :
\(1+2^{100}-2=2^{100}-1\)
=>\(\left(1-2\right)+2^{100}=2^{100}-1\)
=>