x+(x-1)+(x-5)+...+(x-101)=-516

Số số hạng là (101-1):4+1=26(số)

Tổng là (101+1)*26/2=1326

Theo đề, ta có: x+26x-1326=-516

=>x+26x=810

=>27x=810

=>x=30

\(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\left(5^{128}-1\right)=2.5^{128}-2\)

c: Ta có: \(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^2-1\right)\left(5^2+1\right)\cdot\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^{16}-1\right)\cdot\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^{32}-1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^{64}-1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^{128}-1\right)\)

\(=2\cdot5^{128}-2\)

d: Ta có: \(E=\left(x-2\right)^3-\left(x+1\right)\left(x^2-x+1\right)+6\left(x-1\right)^2\)

\(=x^3-6x^2+12x-8-x^3-1+6x^2-12x+6\)

\(=-3\)

Câu 2: 

\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)

Có \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge0\)

do đó phương trình ban đầu tương đương với: 

\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)

\(\Leftrightarrow100x+\left(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\right)=101x\)

\(\Leftrightarrow x=\frac{100.101}{2.101}=50\)

a: S=1(1+1)+2(1+2)+...+100(1+100)

=1+2+...+100+1^2+2^2+...+100^2

\(=\dfrac{100\cdot102}{2}+\dfrac{100\cdot\left(100+1\right)\cdot\left(2\cdot100+1\right)}{6}\)

\(=100\cdot51+\dfrac{100\cdot101\cdot201}{6}\)

=343450

b: \(A=1\cdot2\cdot3+2\cdot3\cdot4+...+100\cdot101\cdot102\)

=>\(4\cdot A=1\cdot2\cdot3\cdot\left(4-0\right)+2\cdot3\cdot4\left(5-1\right)+...+100\cdot101\cdot102\left(103-99\right)\)

=>4*A=100*101*102*103

=>A=25*101*102*103

Với x > 0  

ta có 

x + 1/101 + x  + 2/101 + ... + x + 100/ 101  = 101x 

=> 100x  + ( 1 + 2 + 3 + ... + 100)/101  = 101x 

=>  5050/101 = 101 x - 100x 

=> x = 50 

x < 0 ta có :

   -x - 1/101 - x - 2/101 - ... - x - 100/101 = 101x 

=> - 100x - ( 1 + 2 + .. + 100)/101  = 101x 

=> 5050/101  = -100x - 101x

=> 50          = -201x 

=> x = 

thang Tran trả lời sai, x chỉ có thể lớn hơn 0 thôi, ta có : VT= |x+1/101|+|x+2/101|+|x+3/101|+...+|x+100/101| >= 0

Mà VT=VP =)) VP= 101x >= (lớn hơn hoặc bằng) 0 mà 101 >= 0 =)) x >= 0

<sau đó mới làm giống TH x>0 của bn í>

 SAi vậy mà bn vẫn ak???

Do |x + 1/101| + |x + 2/101| + |x + 3/101| + ... + |x + 100/101| > 0 với mọi x 
mà |x + 1/101| + |x + 2/101| + |x + 3/101| + ... + |x + 100/101| = 101x 
=> x > 0 
Với x > 0 
=> x + 1/101 + x + 2/101 +....+ x + 100/101 = 101x 
<=> x = (1 + 2 + 3 + ... + 100)/101 = 50

Tham Khảo:https://olm.vn/hoi-dap/detail/10864465705.html

=>4y=516

hay y=129