bài này giải nhanh giúp mình nha
2001 +2001 + 2001 + 2001 + 2001 x 7 - 2001
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Dễ thấy 2001=2000+1=x+1,thay vào C ta có:
\(C=x^{20}-\left(x+1\right)x^{19}+\left(x+1\right)x^{18}-\left(x+1\right)x^{17}+...-\left(x+1\right)x^3+\left(x+1\right)x^2\)
\(=x^{20}-x^{20}-x^{19}+x^{19}+x^{18}-x^{18}-x^{17}+...-x^4-x^3+x^3+x^2=x^2=2001^2=4004001\)
Vậy C=4004001

–2001 + (1999 + 2001)
= –2001 + 1999 + 2001
= 2001 – 2001 + 1999 = 1999.



\(\frac{2001}{2000}-\frac{2002}{2001}=\frac{2001.2001}{2000.2001}-\frac{2002.2000}{2000.2001}\)
\(=\frac{\left(2002-1\right).\left(2000+1\right)-2002.2000}{2000.2001}\)
\(=\frac{2002.\left(2000+1\right)-\left(2000+1\right)-2002.2000}{2000.2001}\)
\(=\frac{2002.2000+2002-2000-1-2002.2000}{2000.2001}\)
\(=\frac{2002.2000+1-2002.2000}{2000.2001}\)
\(=\frac{1}{2000.2001}\)


2001 . 2022 + 1981+2003 . 21/ 2002 . 2003 - 2001. 2002
= ( 2001. 2002 - 2001 . 2022 ) + ( 1981 + 2003 . 21/ 2002 . 2003)
= 0+( 1981 + ( 2003 . 21 / 2002 + 1)
= 0 + 1981+( 2002 . 21/2002+1+1)
= 1981 + ( 21+2)
= 1981+ 23
= 2004

\(A=2001+2001^2+...+2001^9\)
\(\Rightarrow2001A=2001^2+2001^3+...+2001^{10}\)
\(\Rightarrow2001A-A=\left(2001^2+2001^3+...+2001^{10}\right)-\left(2001+2001^2+...+2001^9\right)\)\(\Rightarrow2000A=2001^{10}-1\)
\(\Rightarrow A=\frac{2001^{10}-1}{2000}\)
\(\Rightarrow K=2000.\frac{2001^{10}-1}{2000}+1=2001^{10}-1+1=2001^{10}\)
Vậy K=200110
\(2001+2001+2001+2001+2001\times7-2001\)
\(=2001\times\left(1+1+1+1+7-1\right)\)
\(=2001\times10\)
\(=20010\)
2001 +2001+2001+2001+2001x7 -2001
=2001 x 4 +2001 x7 -2001
=2001 x(4+7) -2001
=2001 x 11 -2001
=22011 - 2001
=2010