x=99

=>x+1=100

thay x+1=100 và 99=x vào B ta được:

x⁹⁹-(x+1).x⁹⁸+(x+1).x⁹⁷-(x+1).x⁹⁶+...+(x+1).x-1

=x⁹⁹-x⁹⁹-x⁹⁸+x⁹⁸+x⁹⁷-x⁹⁷-x⁹⁶+...+x²+x-1

=x+1

=100

Vậy B=100

SỬA

x=99

=>x+1=100

thay x+1=100 và 99=x vào B ta được:

x⁹⁹-(x+1).x⁹⁸+(x+1).x⁹⁷-(x+1).x⁹⁶+...+(x+1).x-1

=x⁹⁹-x⁹⁹-x⁹⁸+x⁹⁸+x⁹⁷-x⁹⁷-x⁹⁶+...+x²+x-1

=x-1

=99-1

=98

Vậy B=98

A=x¹⁰-100x⁹+100x⁸-100x⁷+...+100x²-100x+1 tại x=99

=>A=x¹⁰-(x+1)x⁹+(x+1)x⁸-(x+1)x⁷+...+(x+1)x²-(x+1)x+1

A=x¹⁰-x¹⁰-x⁹+x⁹+x⁸-x⁸-x⁷+...+x³+x²-x²-x+1

= 1-x

=1-99

=-98

Bài 1:

a, \(A=3^{100}+3^{99}+...+3+1\)

\(\Rightarrow3A=3^{101}+3^{100}+...+3^2+3\)

\(\Rightarrow3A-A=\left(3^{101}+3^{100}+...+3^2+3\right)-\left(3^{100}+3^{99}+...+3+1\right)\)

\(\Rightarrow2A=3^{101}+1\Rightarrow A=\dfrac{3^{101}+1}{2}\)

b, \(B=\dfrac{15^9.2^{18}.9^8}{3^{15}.4^8.25^4}=\dfrac{3^9.5^9.2^{18}.3^{16}}{3^{15}.2^{16}.5^8}\)

\(=3^{10}.5.2^2=472392\)

c, \(C=\dfrac{2^{10}.10^{17}.7^9}{5^{15}.14^9.64^9}=\dfrac{2^{10}.2^{17}.5^{17}.7^9}{5^{15}.2^9.7^9.2^{54}}\)

\(=\dfrac{5^2}{2^{36}}\)

Chúc bạn học tốt!!!

1.

\(A=3^{100}+3^{99}+3^{98}+...+3^2+3+1\\ A=\dfrac{3-1}{2}\cdot\left(3^{100}+3^{99}+3^{98}+...+3^2+3+1\right)\\ =\dfrac{\left(3-1\right)\cdot\left(3^{100}+3^{99}+3^{98}+...+3^2+3+1\right)}{2}\\ =\dfrac{3^{101}-3^{100}+3^{100}-3^{99}+...+3^2-3+3-1}{2}\\ =\dfrac{3^{101}-1}{2}\)

\(B=\dfrac{15^9\cdot2^{18}\cdot9^8}{3^{15}\cdot4^8\cdot25^4}\\ =\dfrac{\left(3\cdot5\right)^9\cdot2^{18}\cdot\left(3^2\right)^8}{3^{15}\cdot\left(2^2\right)^8\cdot\left(5^2\right)^4}\\ =\dfrac{3^9\cdot5^9\cdot2^{18}\cdot3^{16}}{3^{15}\cdot2^{16}\cdot5^8}\\ =\dfrac{3^9\cdot5\cdot2^2\cdot3}{1\cdot1\cdot1}\\ =3^{10}\cdot5\cdot2^2\\ =59049\cdot5\cdot4\\ =59049\cdot\left(5\cdot4\right)\\ =59049\cdot20\\ =1180980\)

\(C=\dfrac{2^{10}\cdot10^{17}\cdot7^9}{5^{15}\cdot14^9\cdot64^9}\\ =\dfrac{2^{10}\cdot\left(2\cdot5\right)^{17}\cdot7^9}{5^{15}\cdot\left(2\cdot7\right)^9\cdot\left(2^6\right)^9}\\ =\dfrac{2^{10}\cdot2^{17}\cdot5^{17}\cdot7^9}{5^{15}\cdot2^9\cdot7^9\cdot2^{54}}\\ =\dfrac{2\cdot1\cdot5^2\cdot1}{1\cdot1\cdot1\cdot2^{37}}\\ =\dfrac{5^2}{2^{36}}\\ =\dfrac{25}{2^{36}}\)

\(x+\frac{1}{100}+x+\frac{2}{100}+...+x+\frac{99}{100}=100x\)

\(\Rightarrow99x+\frac{1+2+...+99}{100}=100x\)

\(\Rightarrow100x-99x=\frac{\frac{\left(1+99\right).99}{2}}{100}\)

\(\Rightarrow x=\frac{99}{2}\)

Vậy \(x=\frac{99}{2}\)

\(x+\frac{1}{100}+x+\frac{2}{100}+x+\frac{3}{100}+...+x+\frac{99}{100}=100x\)

\(\Leftrightarrow99x+\frac{1+2+3+...+99}{100}=100x\)

\(\Leftrightarrow x=\frac{1+2+3+...+99}{100}\)

\(\Leftrightarrow x=\frac{\frac{99\left(99+1\right)}{2}}{100}\)

\(\Leftrightarrow x=\frac{4950}{100}\)

\(\Leftrightarrow x=\frac{99}{2}\)

Ta có : \(x=99\Rightarrow x+1=100\)

\(\Leftrightarrow P\left(99\right)=x^{99}-\left(x+1\right)x^{98}+\left(x+1\right)x^{97}-...+\left(x+1\right)x-1\)

\(\Leftrightarrow x^{99}+x^{98}+x^{97}+...+x^2+x-1\)

\(\Leftrightarrow x-1\) Thay x = 99 vào x - 1 ta có 

\(\Leftrightarrow P\left(99\right)=99-1=98\)

a)A(-1)=-1+(-1)²+(-1)³+...+(-1)¹⁰⁰

          =50(-1)+50.1

          =-50+50

          =0

\(A=x+x^2+x^3+...+x^{100}\)

\(A=x\left(1+x+x^2+...+x^{99}\right)\)

\(A=x\left(1+A-x^{100}\right)\)

\(\left(1-x\right)A=x-x^{101}\)

\(A=\frac{x-x^{101}}{1-x}\)

a) Với x = -1, ta có \(A=\frac{\left(-1\right)-\left(-1\right)101}{2}=0\)

Vậy nên x = -1 là một nghiệm của A(x)

b) Với x = 1/2 thì \(A=\frac{\frac{1}{2}-\left(\frac{1}{2}\right)^{101}}{1-\frac{1}{2}}=\frac{\frac{1}{2}-\frac{1}{2^{101}}}{\frac{1}{2}}=\frac{2^{100}-1}{2^{100}}\)

a ) \(A\left(-1\right)=-1+\left(-1\right)^2+\left(-1\right)^3+\left(-1\right)^4+....+\left(-1\right)^{99}+\left(-1\right)^{100}\)

\(=-1+1-1+1-1+1-....-1+1\)

\(=\left(-1+1\right)+\left(-1+1\right)+.....+\left(-1+1\right)\)

\(=0\)

Hay \(x=-1\) là nguyện của A(x) (đpcm )

b ) \(A\left(\frac{1}{2}\right)=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+....+\left(\frac{1}{2}\right)^{100}\)

\(=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{100}}\)

\(2A\left(\frac{1}{2}\right)=1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{99}}\)

\(\Rightarrow2A\left(\frac{1}{2}\right)-A\left(\frac{1}{2}\right)=1-\frac{1}{2^{100}}\)

\(\Rightarrow A\left(\frac{1}{2}\right)=\frac{2^{100}-1}{2^{100}}\)

Tại \(x=\frac{1}{2}\) thì A(x) = \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+.......+\left(\frac{1}{2}\right)^{100}\)

=> 2A(x) = \(1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+.......+\left(\frac{1}{2}\right)^{99}\)

=> 2A(x) - A(x) =\(1-\left(\frac{1}{2}\right)^{100}\) 

=> A(x) = \(1-\left(\frac{1}{2}\right)^{100}\)

x=99

=>x+1=100

thay x+1=100 và 99=x vào B ta được:

x⁹⁹-(x+1).x⁹⁸+(x+1).x⁹⁷-(x+1).x⁹⁶+...+(x+1).x-1

=x⁹⁹-x⁹⁹-x⁹⁸+x⁹⁸+x⁹⁷-x⁹⁷-x⁹⁶+...+x²+x-1

=x-1

=99-1

=98

Vậy B=98

trời                

\(f\left(x\right)=x^{99}-100x^{98}+100x^{97}-...+100x-1\)

\(f\left(99\right)=99^{99}-100\cdot99^{98}+100\cdot99^{97}-...+100\cdot99-1\)

\(f\left(99\right)=99^{99}-\left(99+1\right)\cdot99^{98}+\left(99+1\right)\cdot99^{97}-...+\left(99+1\right)\cdot99-1\)

\(f(99)= 99^{99}-99^{99}-99^{98}+99^{98}+99^{97}-99^{97}-99^{96}+...+99^2+99-1\)

\(f\left(99\right)=99-1=98\)

Chắc là -100 nhé bn!!!