2001 x 767 + 2002 x 233 TÍNH NHANH
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a) 2001 x 757 + 2002 x 233
= 2001 x 757 + (2001 + 1) x 233
= 2001 x 757 + 2001 x 233 + 1 x 233
= 2001 x (757+233) + 233
= 2001 x 1000 + 233
= 2001000 + 233
= 2001233
a) 2001x757+2002x243
=2001 x 757+ (2001+1) x233
=2001 x757+2001x233+1 x233
=2001x(757+233)+233
=2001x1000 +233
=201000+233
=2001233
b)(m:1-m x1):(mx2001+m+1)
=(m-m):(mx2001+m+1)
=0:(mx2001+m+1)
=0
Học tốt
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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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\(f_{\left(x\right)}=x^6-2002x^5+2002x^4-2002x^3+2002x^2-2002x+2006\)
\(=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+x+5\)
\(=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+x+5\)
\(=5\)
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
\(F\left(x\right)=x^6-\left(x-1\right).x^5+\left(x-1\right).x^4-\left(x-1\right).x^3+\left(x-1\right).x^2-\left(x-1\right).x+2006\)
\(F\left(x\right)=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+2006\)
\(F\left(2001\right)=-2001+2006=5\)
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A = 2003 × 2002 - 2/2001 × 2003 + 2001
A = 2003 × (2001 + 1) - 2/2001 × 2003 + 2001
A = 2003 × 2001 + (2003 - 2)/2001 × 2003 + 2001
A = 2003 × 2001 + 2001/2001 × 2003 + 2001
A = 1
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=2001*(767+233)*2002
=2001*1000*2002
=4006002000
nha! sao bạn ko thay số 2001=2002 hoặc thay 2002=2001 để cho dễ tính hơn?
chúc bạn học giỏi
2001*( 767 + 233) + 2002
=2001 * 1000 + 2002
= 2001 * 3002
= 6007002
chúc bạn học tốt