ĐỀ : \(A=\frac{10^{11}-1}{10^{12}-1}\)và \(B=\frac{10^{10}+1}{10^{11}+1}\).

Ta có : \(10A=\frac{10\left(10^{11}-1\right)}{10^{12}-1}=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}\)\(=1-\frac{9}{10^{12}-1}\).

            \(10B=\frac{10\left(10^{10}+1\right)}{10^{11}+1}=\frac{10^{11}+10}{10^{11}+1}=\frac{10^{11}+1+9}{10^{11}+1}\)\(=1+\frac{9}{10^{11}+1}\).

Ta so sánh : \(1-\frac{9}{10^{12}-1}< 1+\frac{9}{10^{11}+1}\)hay \(10A< 10B.\)

\(\Rightarrow A< B\left(đpcm\right)\)

\(\frac{-8}{9}+\frac{1}{9}\times\frac{2}{9}+\frac{1}{9}\times\frac{7}{9}\)

\(=\frac{-8}{9}+\frac{2}{81}+\frac{7}{81}\)

\(=\frac{-72}{81}+\frac{2}{81}+\frac{7}{81}\)

\(=\frac{-70}{81}+\frac{7}{81}\)

\(=\frac{-63}{81}\)

\(=\frac{-7}{9}\)

\(-\frac{7}{9}\)

\(C\text{ó}:x+4⋮x+1\)

\(x+1⋮x+1\)

\(\Rightarrow\left(x+4\right)-\left(x+1\right)⋮x+1\)

\(x+4-x-1⋮x+1\)

\(3⋮x+1\)

\(\Rightarrow x+1\in\text{Ư}_{\left(3\right)}=\left\{\pm1;\pm3\right\}\)

Ta có bảng:

Mà x là số tự nhiên

=> x = 2

\(x+4⋮x+1\)

\(\Leftrightarrow\left(x+1\right)+3⋮x+1\)

\(\Leftrightarrow3⋮x+1\)

\(\Rightarrow\left(x+1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

Ta có bảng sau:

Mà \(x\inℕ^∗\)

\(\Rightarrow x\in\left\{0;2\right\}\) thì \(x+4⋮x+1\)

\(-47.69+31.\left(-47\right)-\left(-115\right)+5.\left(-3\right)^2\)

\(=\left(-47\right).69+31+115+5.9\)

\(=\left(-47\right).100+115+45\)

\(=\left(-4700\right)+115+45\)

\(=\left(-4700\right)+150\)

\(=\left(-4550\right)\)

=-47x(69+31)-(-115)+5x(-3)²

^{=-47x100-(-115)+5x9}

=-4700-(-115)+45

=-4700-[(-115)+45]

=-4700-(-70)

=-4770

    5/x = x/5

=> x.x = 5.5

=> x² = 25

=> x² = (+-5)²

=> x = +-5

cho mình nha

x mũ 2=5.5

x.x=5.5

x=5