tính nhanh ( -2002 ) - ( 15 -2002 )
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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
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P=\(\frac{\left(2002+1\right)\times14+1988+2001\times2002}{2002\times\left(1+503+504\right)}\)
\(=\frac{2002\times14+2002+2001\times2002}{2002\times1008}\)
\(=\frac{2002\times\left(14+1+2001\right)}{2002\times1008}=\frac{2016}{1008}=2\)
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(–2002) – (57 – 2002)
= –2002 – 57 + 2002
= 2002 – 2002 – 57
= 0 – 57 = –57.
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\(f_{\left(x\right)}=x^6-2002x^5+2002x^4-2002x^3+2002x^2-2002x+2006\)
\(=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+x+5\)
\(=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+x+5\)
\(=5\)
Vậy \(f_{\left(x\right)}=5\)Tại x = 2001
Lạ OLM ghê làm sai mà vẫn được k ???
Ta có : x=2001 \(\Rightarrow\)x+1=2002
\(F\left(x\right)=x^6-\left(x-1\right).x^5+\left(x-1\right).x^4-\left(x-1\right).x^3+\left(x-1\right).x^2-\left(x-1\right).x+2006\)
\(F\left(x\right)=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+2006\)
\(F\left(2001\right)=-2001+2006=5\)
( - 2002 ) - ( 15 - 2002)
= - 2002 - 15 + 2002
= -15
\(\left(-2002\right)-\left(15-2002\right)\)
\(=-2002-15+2002\)
\(=\left(-2002+2002\right)-15\)
\(=0-15\)
\(=-15\)