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\(41^2+82.59+59^2\)
= \(41^2+2.41.59+59^2\)
= \(\left(41+59\right)^2\)
= \(100^2=10000\)
\(29^2+41^2-30^2+41.58\)
\(=\left(41^2+41.58+29^2\right)-30^2\)
\(=\left(41^2+2.41.29+29^2\right)-30^2\)
\(=\left(41+29\right)^2-30^2\)
\(=70^2-30^2\)
\(=\left(70-30\right)\left(70+30\right)\)
\(=40.100=4000\)
2. Đặt \(x-1996=t\)
\(\Rightarrow\left(x-1996\right)^3+\left(x-1997\right)^3-1=t^3+\left(t-1\right)^2-1\)
\(=t^3+t^2-2t+1-1=t^3+t^2-2t=t\left(t^2+t-2\right)\)
\(=t.\left[\left(t^2-t\right)+\left(2t-2\right)\right]=t\left[t\left(t-1\right)+2\left(t-1\right)\right]\)
\(=t\left(t-1\right)\left(t+2\right)=\left(x-1996\right)\left(x-1996-1\right)\left(x-1996+2\right)\)
\(=\left(x-1996\right)\left(x-1997\right)\left(x-1994\right)\)
1. Đặt x2 + 4x + 8 = y
bthuc ⇔ y2 + 3xy + 2x2
= y2 + xy + 2xy + 2x2
= ( xy + y2 ) + ( 2x2 + 2xy )
= y( x + y ) + 2x( x + y )
= ( x + y )( y + 2x )
= ( x + x2 + 4x + 8 )( x2 + 4x + 8 + 2x )
= ( x2 + 5x + 8 )( x2 + 6x + 8 )
= ( x2 + 5x + 8 )( x2 + 2x + 4x + 8 )
= ( x2 + 5x + 8 )[ x( x + 2 ) + 4( x + 2 ) ]
= ( x2 + 5x + 8 )( x + 2 )( x + 4 )
2. Đặt t = x - 1996
bthuc ⇔ t3 + ( t - 1 )2 - 1
= t3 + t2 - 2t + 1 - 1
= t3 + t2 - 2t
= t( t2 + t - 2 )
= t( t2 - t + 2t - 2 )
= t( t - 1 )( t + 2 )
= ( x - 1996 )( x - 1996 - 1 )( x - 1996 + 2 )
= ( x - 1996 )( x - 1997 )( x - 1994 )
3. 4( x2 + 15x + 59 )( x2 + 18x + 72 ) - 3x2 < bó tay :)) >
\(A=5^{n+2}+26.5^n+8^{2n+1}\)
\(A=5^n\left(5^2+26\right)+\left(8^2\right)^n.8\)
\(A=5^n.51+64^n.8\)
\(A=5^n.59-5^n.8+64^n.8\)
\(A=5^n.59+8.\left(-5^n+64^n\right)\)
Ta có: \(\left(5^n.59\right)⋮59\left(1\right)\)
mà \(\left(-5^n+64^n\right)\) luôn chia hết cho \(\left(-5+64\right)=59\Leftrightarrow8.\left(-5^n+64^n\right)⋮59\left(2\right)\)
Từ (1)(2)⇒ A\(⋮\)59
a) 412 = (40+1)2
= 402+2.40.1+12
= 1600+80+1
= 1681
b) 2012 = (200+1)2
= 2002+2.200.1+12
= 40000+400+1
= 40401
c)992 = ( 100-1)2
= 1002-2.100.1+12
= 10000-200+1
= 9800+1
= 9801
a , \(5x^2+9y^2-12xy-6x+9=0\)
\(\Leftrightarrow25x^2+45y^2-60xy-30x+45=0\)
\(\Leftrightarrow\left(5x\right)^2-2.5.\left(6y+3\right)+\left(6y+3\right)^2+9y^2-36y+36=0\)
\(\Leftrightarrow\left(5x-6y-3\right)^2+9\left(y^2-4y+4\right)=0\)
\(\Leftrightarrow\left(5x-6y-3\right)^2+9\left(y-2\right)^2=0\)
Vì \(\left\{{}\begin{matrix}\left(5x-6y-3\right)^2\ge0\\9\left(y-2\right)^2\ge0\end{matrix}\right.\Rightarrow\left(5x-6y-3\right)^2+9\left(y-2\right)^2\ge0\)
Dấu ''='' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}5x-6y-3=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
Vậy ...
\(41^2+82.59+59^2=41^2+2.41.59+59^2=\left(41+59\right)^2\)