4: Đặt 315=a; 651= b
\(A=2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)
\(=\left(2+\frac{1}{a}\right)\cdot\frac{1}{b}-\frac{3}{a}\cdot\left(3+\frac{b-1}{b}\right)-\frac{4}{ab}+\frac{12}{a}\)
\(=\frac{2a+1}{a}\cdot\frac{1}{b}-\frac{3}{a}\cdot\frac{3b+b-1}{b}-\frac{4}{ab}+\frac{12}{a}\)
\(=\frac{2a+1}{ab}-\frac{3\left(4b-1\right)}{ab}-\frac{4}{ab}+\frac{12b}{ab}=\frac{2a+1-12b+3-4+12b}{ab}=\frac{2a}{ab}=\frac{2}{b}\)
=2/651
3:
a: \(A=x^3-30x^2-31x+1\)
\(=x^3-31x^2+x^2-31x+1\)
\(=\left(x-31\right)\left(x^2+x\right)+1=\left(31-31\right)\left(31^2+31\right)+1=1\)
b: \(B=x^5-15x^4+16x^3-29x^2+13x\)
\(=x^5-14x^4-x^4+14x^3+2x^3-28x^2-x^2+14x-x\)
\(=\left(x-14\right)\left(x^4-x^3+2x^2-x\right)-x\)
\(=\left(14-14\right)\left(x^4-x^3+2x^2-x\right)-14\)
=-14
c: x=9
=>x+1=10
\(C=x^{14}-10x^{13}+10x^{12}-10x^{11}+\cdots+10x^2-10x+10\)
\(=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-x^{11}\left(x+1\right)+\cdots+x^2\left(x+1\right)-x\left(x+1\right)+\left(x+1\right)\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+\cdots+x^3+x^2-x^2-x+x+1\)
=1
