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a) \(413.\left(413-26\right)+169=413^2-2.13.413+13^2=\left(413-13\right)^2=160000\)
b) \(\left(625^2+3\right).\left(25^4-3\right)-5^{16}+10\)
\(=\left(5^8+3\right)\left(5^8-3\right)-5^{16}+10\)
\(=5^{16}-9-5^{16}+10=1\)
c) \(\frac{41^2+39^2+8^2.39}{41^2-39^2}=\frac{\left(41+39\right)^2}{\left(41-39\right)\left(41+39\right)}=\frac{41+39}{41-39}=\frac{80}{2}=40\)
Bài 1:
\(a,413\left(413-26\right)+169\)
\(=413^2-26.413+13^2\)
\(=413^2-2.413.13+13^2\)
\(=\left(413-13\right)^2\)
\(=400^2\)
\(=160000\)
\(b,x^2+2x+1\)
\(=x^2+2.x.1+1^2\)
\(=\left(x+1\right)^2\)
\(=\left(99+1\right)^2\)
\(=100^2=10000\)
a) 9x2 - 36
= ( 3x)2 - 62
= ( 3x - 6)( 3x + 6)
b) ax - ay - bx + by
= a( x - y) - b( x - y)
= ( x - y)( a - b)
c) y3 - 4y2 + 3y
= y3 - y2 - 3y2 + 3y
= y2( y - 1) - 3y( y - 1)
= ( y - 1)( y2 - 3y)
= y( y - 3)( y - 1)
a) Ta có: \(413^2+213^2-326\cdot213\)
\(=413^2-2\cdot413\cdot213+213^2+106500\)
\(=200^2+106500\)
\(=40000+106500\)
\(=146500\)
a) Ta có: \(A=\dfrac{37^3+12^3}{49}-37\cdot12\)
\(=\dfrac{\left(37+12\right)\left(37^2-37\cdot12+12^2\right)}{49}-37\cdot12\)
\(=37^2-2\cdot37\cdot12+12^2\)
\(=\left(37-12\right)^2\)
\(=25^2=625\)
\(a.75^2-50.75+25^2\\ =75^2-2.25.75+25^2\\ =\left(75-25\right)^2=50^2=2500\)
\(b.103.97\\ =\left(100+3\right)\left(100-3\right)\\ =100^2-3^2\\ =9991\)
\(a,75^2-50.75+25^2\\ =75^2-2.75.25+25^2\\ =\left(75+25\right)^2=100^2=10000\\ b,103.97\\ =\left(100+3\right).\left(100-3\right)\\ =100^2-3^2=10000-9=9991\)
A=\(413^2-413.26+13^2=413^2-2.413.13+13^2\)
=> A=\(\left(413-13\right)^2=400^2=160000\)
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