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\(a,=\left(93-7\right)\left(93+7\right)=100\cdot86=8600\\ b,=\left(45+55\right)^2=100^2=10000\)
\(Q=x^2-y^2-2y-1\)
\(\Rightarrow Q=x^2-\left(y^2+2y+1\right)\)
\(\Rightarrow Q=x^2-\left(y+1\right)^2\)
\(\Rightarrow Q=\left(x-y-1\right)\left(x+y+1\right)\)
Thay \(x=93;y=6\)vào \(Q\)ta được :
\(Q=\left(93-6-1\right)\left(93+6+1\right)\)
\(\Rightarrow Q=86.100\)
\(\Rightarrow Q=8600\)
Vậy \(Q=8600\)
=2004^2-2003^2+2002^2-2001^2+....+1
=(2004+2003)(2004-2003)+(2002+2001)(2002-2001)+.....+1
=2004+2003+...+1
=2009010
a) ( 100 – 1 ) 3 = 970299. b) ( 91 + 9 ) 3 = 100 3 .
c) ( 1000 + 1 ) 3 = 1003003001. d) ( 102 – 2 ) 3 = 100 3 .
a) Ta có 15.64 + 25.100 + 36.15 + 60.100
= (15.64 + 36.15) + (25.100 + 60.100)
= 100.(15 + 85) = 10000.
b) Ta có 47 2 + 48 2 - 25 + 94.48
= ( 47 2 +2.47.48+ 48 2 ) - 5 2 = ( 47 + 48 ) 2 - 5 2 =9000.
c) Ta có 93 -92.(-l)-9.11 + (-l).ll
= (93 +92)-(9.11 + 1.11)
= 92(9 +1) -ll.(9 + l) = 700.
\(201^2=\left(200+1\right)^2=200^2+2.200.1+1^2=40000+400+1=40401\)
\(498^2=\left(500-2\right)^2=500^2-2.500.2+2^2=250000-2000+4=248004\)
\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)
\(\Rightarrow\frac{x+2}{2002}+1+\frac{x+5}{1999}+1+\frac{x+201}{1803}+1=0\)
\(\Rightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)
Dễ thấy \(\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)>0\)nên x + 2004 = 0
Vậy x = -2004
Ta có:
\(1999^2=\left(2000-1\right)^2\)
\(=2000^2-2\cdot2000+1\)
\(=4000000-4000+1\)
\(=3996001\)
\(104^2-16=104^2-4^2\)
\(=\left(104-4\right)\left(104+4\right)\)
\(=100\cdot108\)
\(=10800\)
\(87\cdot93=\left(90-3\right)\left(90+3\right)\)
\(=90^2-3^2\)
\(=8100-9\)
\(=8091\)
Học tốt nhé!