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\(=\frac{41^2-39}{41^2+39^2+2.41.39}\)
\(=\frac{41^2-39}{41^2+2.41.39+39^2}\)
\(=\frac{41^2-39}{\left(41+39\right)^2}\)
\(\)
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,\(=\left(\frac{3}{5}x+\frac{2}{7}y\right)^2=\left(\frac{3}{5}.5+\frac{2}{7}.\left(-7\right)\right)^2=0\)
\(b,=\left(\frac{5}{4}u^2v+\frac{2}{25}v^2\right)^2=\left(\frac{5}{4}.\left(\frac{2}{5}\right)^2.5+\frac{2}{25}.5^2\right)^2=3^2=9\)
Pythagorean theorem: \(AD=\sqrt{BD^2-AB^2}=4\) (cm)
\(\Rightarrow BC=AD=4\left(cm\right)\)
\(CC'=\sqrt{BC'^2-BC^2}=4\sqrt{2}\)
The lateral surface area: \(2CC'.\left(BC+AB\right)=56\sqrt{2}\left(cm^2\right)\)
Bài 1: \(\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\) (1)
Từ \(\frac{a}{c}=\frac{c}{b}\Rightarrow ab=c^2\)
Thay vào (1) ta có:
\(\frac{a^2+ab}{b^2+ab}=\frac{a}{b}\Rightarrow\frac{a\left(a+b\right)}{b\left(a+b\right)}=\frac{a}{b}\) (luôn đúng)
Vậy ta có điều phải chứng minh
A=\(413^2-413.26+13^2=413^2-2.413.13+13^2\)
=> A=\(\left(413-13\right)^2=400^2=160000\)
chuc ban hoc tot
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) \(\frac{41^2+39^2+8^2.39}{41^2-39^2}=\frac{\left(41+39\right)^2}{\left(41-39\right)\left(41+39\right)}=\frac{41+39}{41-39}=\frac{80}{2}=40\)