Tính nhanh:
b. 2011 x 89 + 10 x 2011 + 2011
Tính nhanh:
a, 2010 x 3+ 2010 x 6 + 2010
b, 2011 x 89 + 10 x 2011 + 2011
a, 2010 x 3+ 2010 x 6 + 2010
= 2010 x ( 3 + 6 + 1)
= 2010 x 10
= 20100
b, 2011 x 89 + 10 x 2011 + 2011
= 2011 x (89 + 10 + 1)
= 2011 x 100
= 201100
Tính nhanh:
a, 2011 x 3+ 2011 x 6 + 2011
b, 2010 x 89 + 10 x 2010 + 2010
a, 2011 x 3+ 2011 x 6 + 2011
= 2011 x ( 3+6+1)
= 2011 x 10
= 20110
b, 2010 x 89 + 10 x 2010 + 2010
= 2010 x (89+10+1)
= 2010 x 100
= 201000
Tính bằng cách thuận tiện:
a/ 2010 x 3 + 2010 x 6 + 2010
b/ 2011 x 111 - 10 x 2011 - 2011
a,2010x3+2010x6+2010
=2010x(3+6+1)
=2010x10
=20100
b,2011x111-10x2011-2011
=2011x(111-10-1)
=2011x100
=201100
a)2010 x 3 + 2010 x 6 +2010
=2010 x 3 + 2010 x 6 + 2010 x 1
=2010 x (3 + 6 + 1)
=2010 x 10
=20100
b)2011 x 111 - 10 x 2011 - 2011
=2011 x 111 - 10 x 2011 - 2011 x 1
=2011 x (111 - 10 -1)
=2011 x 100
=201100
a.2010x(3+6+1)=20100
b.2011x(111-10-1)=201100
2009 x 89 +2011 x 11
\(2009x89+2011x11\)
\(=178801+22121\)
\(=200922\)
Cho hàm số f(x) thỏa mãn f(f(x)) = x + 10. Biết f(2001) = 2011. Tính f(2011)
\(f\left(2011\right)=f\left(f\left(2001\right)\right)=2001+10=2011\)
Vậy \(f\left(2011\right)=2011\)
Tính nhanh A= 2012 x 14 + 1997 + 2010 x 2011 / 2011 x5 + 2011 x 1008 + 1012 X 2011
Ta có :
\(A=\frac{2012.14+1997+2010.2011}{2011.5+2011.1008+1012.2011}\)
\(\Rightarrow A=\frac{\left(2011+1\right).14+1997+2010.2011}{2011.\left(5+1008+1012\right)}\)
\(\Rightarrow A=\frac{2011.14+14+1997+2010.2011}{2011.2025}\)
\(\Rightarrow A=\frac{2011.14+2011+2010.2011}{2011.2025}\)
\(\Rightarrow A=\frac{2011.\left(14+1+2011\right)}{2011.2025}\)
\(\Rightarrow A=\frac{2011.25}{2011.25}\)
\(\Rightarrow A=1\) ( tử số = mẫu số )
Vậy \(A=1\)
~ Ủng hộ nhé
_Vào phần câu hỏi tương tự là có nhé bạn
\(A=\frac{2012\times14+1997+2010\times2011}{2011\times5+2011\times1008+1012\times2011}\)
\(A=\frac{\left(2011+1\right)\times14+1997+2010\times2011}{2011\times\left(5+1008+1012\right)}\)
\(A=\frac{2011\times14+\left(14+1997\right)+2010\times2011}{2011\times2025}\)
\(A=\frac{2011\times14+2011+2010\times2011}{2011\times2025}\)
\(A=\frac{2011\times\left(14+1+2010\right)}{2011\times2025}\)
\(A=\frac{2011\times2025}{2011\times2025}\)
\(A=1\)
cho a,b,c,d # 0 và : (x^2011+y^2011+z^2011+t^2011)/a^2+b^2+c^2+d^2 = (x^2010)/a^2 + ( y^2010)/b^2 + (z^2010)/c^2 + (t^2010)/d^2. Tính T= x^2011 + y^2011 + z^2011 + t^2011
Tính: C= 2011/5 + 2011/10 + 2011/30 + 2011/60 + 2011/100 + 2011/150 +.... + 2011/46560 + 2011/47530 + 2011/48510
(x-3+(x-2)+(x-1)+...+9+10+2011=2011