E=13/12.14/13....200/199=200/12=50/3

\(A=\left(1-\frac{1}{15}\right)\left(1-\frac{1}{21}\right)\left(1-\frac{1}{28}\right)...\left(1-\frac{1}{79800}\right)\)

\(A=\frac{14}{15}.\frac{20}{21}.\frac{27}{28}...\frac{209}{210}\)

\(A=\frac{28}{30}.\frac{40}{42}.\frac{54}{56}...\frac{418}{240}\)

\(A=\frac{4.7}{5.6}.\frac{5.8}{6.7}.\frac{6.9}{7.8}...\frac{19.22}{20.21}\)

\(A=\frac{4.5.6...19}{5.6.7...20}.\frac{7.8.9...22}{6.7.8...21}\)

\(A=\frac{4}{20}.\frac{22}{6}\)

\(A=\frac{11}{15}\)

\(a,x+1=\left(x+1\right)^2\)

\(\Leftrightarrow x+1=x^2+2x+1\)

\(\Leftrightarrow x^2+2x+1-x-1\)

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\left(+\right)x=0\)

\(\left(+\right)x+1=0\Leftrightarrow x=-1\)

Vậy phương trình có tập nghiệm \(S=\left\{-1;0\right\}\)

\(b,x^3+x=0\Leftrightarrow x\left(x^2+1\right)=0\)

\(\left(+\right)x=0\)

\(\left(+\right)x^2+1=0\)

Vì \(x^2\ge0;1>0\Rightarrow x^2+1>0\)

\(\Rightarrow\) Phương trình \(x^2+1=0\) vô nghiệm 

Vậy Phương trình có tập nghiệm \(S=\left\{0\right\}\)

cảm ơn bn j đó nha :))))

a)   nguyên

⇔ 2n + 3 ∈ Ư(2) = {-2; -1; 1; 2}

⇔ 2n ∈ {-5; -4; -2; -1}

Vì n nguyên nên n ∈ {-2; -1}

Bài 2: 

a) Để B nguyên thì \(6n+7⋮2n+3\)

\(\Leftrightarrow-2⋮2n+3\)

\(\Leftrightarrow2n+3\in\left\{1;-1;2;-2\right\}\)

\(\Leftrightarrow2n\in\left\{-2;-4\right\}\)

hay \(n\in\left\{-1;-2\right\}\)