a: Ta có: x=31

nên x-1=30

Ta có: \(A=x^3-30x^2-31x+1\)

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

\(=x^3-x^3+x^2-x^2+1\)

=1

c: Ta có: x=16

nên x+1=17

Ta có: \(C=x^4-17x^3+17x^2-17x+20\)

\(=x^4-x^3\left(x+1\right)+x^2\left(x+1\right)-x\left(x+1\right)+20\)

\(=x^4-x^4-x^3+x^3+x^2-x^2-x+20\)

\(=20-x=4\)

d: Ta có: x=12

nên x+1=13

Ta có: \(D=x^{10}-13x^9+13x^8-13x^7+...+13x^2-13x+10\)

\(=x^{10}-x^9\left(x+1\right)+x^8\left(x+1\right)-x^7\left(x+1\right)+...+x^2\left(x+1\right)-x\left(x+1\right)+10\)

\(=10-x\)

=-2

d: Ta có: x=12

nên x+1=13

Ta có: \(D=x^{10}-13x^9+13x^8-13x^7+...+13x^2-13x+10\)

\(=x^{10}-x^9\left(x+1\right)+x^8\left(x+1\right)-x^7\left(x+1\right)+...+x^2\left(x+1\right)-x\left(x+1\right)+10\)

\(=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...+x^3+x^2-x^2-x+1+9\)

\(=-x+10=-2\)

Ta có: \(\left(0,125\right)^3\cdot8^3+\left(\frac34-\frac56\right)^2-\frac{50^3}{125}+20^5\cdot\frac{5^{10}}{100^5}\)

\(=\left(0,125\cdot8\right)^3+\left(\frac{9}{12}-\frac{10}{12}\right)^2-\frac{\left(5^2\cdot2\right)^3}{5^3}+\frac{5^5\cdot2^{10}\cdot5^{10}}{\left(2^2\cdot5^2\right)^5}\)

\(=1+\left(-\frac{1}{12}\right)^2-\frac{5^6\cdot2^3}{5^3}+5^5=1+\frac{1}{144}-5^3\cdot2^3+5^5\)

\(=\frac{145}{144}+5^3\left(5^2-2^3\right)=\frac{145}{144}+125\cdot17=\frac{306145}{144}\)

Ta có: \(\left(0,125\right)^3\cdot8^3+\left(\frac34-\frac56\right)^2-\frac{50^3}{125}+20^5\cdot\frac{5^{10}}{100^5}\)

\(=\left(0,125\cdot8\right)^3+\left(\frac{9}{12}-\frac{10}{12}\right)^2-\frac{\left(5^2\cdot2\right)^3}{5^3}+\frac{5^5\cdot2^{10}\cdot5^{10}}{\left(2^2\cdot5^2\right)^5}\)

\(=1+\left(-\frac{1}{12}\right)^2-\frac{5^6\cdot2^3}{5^3}+5^5=1+\frac{1}{144}-5^3\cdot2^3+5^5\)

\(=\frac{145}{144}+5^3\left(5^2-2^3\right)=\frac{145}{144}+125\cdot17=\frac{306145}{144}\)

Ta có: \(\left(0,125\right)^3\cdot8^3+\left(\frac34-\frac56\right)^2-\frac{50^3}{125}+20^5\cdot\frac{5^{10}}{100^5}\)

\(=\left(0,125\cdot8\right)^3+\left(\frac{9}{12}-\frac{10}{12}\right)^2-\frac{\left(5^2\cdot2\right)^3}{5^3}+\frac{5^5\cdot2^{10}\cdot5^{10}}{\left(2^2\cdot5^2\right)^5}\)

\(=1+\left(-\frac{1}{12}\right)^2-\frac{5^6\cdot2^3}{5^3}+5^5=1+\frac{1}{144}-5^3\cdot2^3+5^5\)

\(=\frac{145}{144}+5^3\left(5^2-2^3\right)=\frac{145}{144}+125\cdot17=\frac{306145}{144}\)

Bài 2: 

a: =>x-45=-20

hay x=25

b: =>35+30-5x=-12+112

=>65-5x=100

=>5x=-35

hay x=-7

c: =>x-124=1000

hay x=1124

d: \(\Leftrightarrow46-\left(3x-2\right)^3=-18\)

\(\Leftrightarrow\left(3x-2\right)^3=64\)

=>3x-2=4

hay x=2

Bài 2: 

a: =>x-45=-20

hay x=25

b: =>35+30-5x=-12+112=100

=>65-5x=100

=>5x=-35

hay x=-7

c: =>x-124=1000

hay x=1124

d: \(\Leftrightarrow46-\left(3x-2\right)^3=-18\)

\(\Leftrightarrow\left(3x-2\right)^3=64\)

=>3x-2=4

hay x=2

sao bạn ko làm bài 1 cho mình zậy

17                         

c thôy giùm ik, cảm ơn giề nh` dzữ zợ