Có số số hạng ở dãy là:

(2017 - 1) : 3 + 1 = 673

Vậy tổng là:
(2017+1) x 673 : 2 = 679 057

vaqy aii ncha biet nnhung phai viet ro ra co nhu the nay nay :(1+2017)+(4+2014+...+bao nhieu so o giua ay minh tinh ko ra

\(\dfrac{\left(3\cdot2^{20}+7\cdot2^{19}\right)\cdot52}{\left(13\cdot8^4\right)^2}=\dfrac{52\cdot2^{19}\cdot\left(3\cdot2+7\right)}{13^2\cdot2^{24}}\)

\(=\dfrac{2^{21}\cdot13\cdot13}{13^2\cdot2^{24}}=\dfrac{1}{2^3}=\dfrac{1}{8}\)

căn bậc hai không có số âm

\(\sqrt{-1}\) đó

a) ĐK : x ≥ 0 ; x ≠ 1

A=\(\frac{x-\sqrt{x}}{\sqrt{x}-1}:\frac{x+\sqrt{x}}{\sqrt{x}+1}\)

=\(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}:\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

=\(\sqrt{x}:\sqrt{x}\)

=1

Vậy A=1 với x ≥ 0 ; x ≠ 1

b) Vì A=1 nên không thể thay x

\(4+\frac{11}{8}-\frac{5}{6}=\frac{96+33-20}{24}=\frac{109}{24}\)

\(\left(\frac{1}{2}+\frac{4}{7}\right)-\left(\frac{3}{7}-\frac{3}{10}\right)=\frac{1}{2}+\frac{4}{7}-\frac{3}{7}+\frac{3}{10}\)

\(\left(\frac{1}{2}+\frac{3}{10}\right)+\left(\frac{4}{7}-\frac{3}{7}\right)=\frac{4}{5}+\frac{1}{7}=\frac{28+5}{35}=\frac{33}{35}\)

ban Chitanda Eru oi to bao nhung ma nhan voi bao nhieu ma ra duoc 96

Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)

\(=\left(x+y\right)^3=1^3=1\)

Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)

\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)

Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)

\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)

\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)

\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)

\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)