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: \(413^2+213^2-326\cdot213\)

\(=413^2-2\cdot413\cdot213+213^2+106500\)

\(=200^2+106500\)

\(=40000+106500\)

\(=146500\)

Bài 4:

\(N=3x^2+x\left(x-4y\right)-\left(x+y\right)\left(x-y\right)+x^2+1\)

\(=3x^2+x^2-4xy-x^2+y^2+x^2+1=4x^2-4xy+y^2+1\)

\(=\left(2x-y\right)^2+1\ge1>0\forall x,y\)

=>N luôn dương với mọi x,y

Bài 3:

1: A+B

\(=x^2-4xy+4y^2+4x^2+4xy+y^2=5x^2+5y^2\)

2: Thay x=1;y=-2 vào M, ta được:

\(M=2\cdot1^2+4\cdot1\cdot\left(-2\right)-4\cdot\left(-2\right)^2\)

=2-8-16

=-6-16

=-22

Bài 1:

a; \(\frac12xy\).( - 2\(x^2y\) + \(\frac12y\))

= \(\frac12xy\) .(-2\(x^2y\)) + \(\frac12xy\).\(\frac12y\)

= [\(\frac12.\left(-2\right)\)] (\(x.x^2\)).(y.y) + (\(\frac12.\frac12\)).\(x\).(y.y)

= -\(x^3y^2\) + \(\frac14xy^2\)

b; (\(\frac{x}{2}-2y\))\(^2\)

= \(\left(\frac{x}{2}\right)^2\) - 2.\(\frac{x}{2}\).2y+ (2y)\(^2\)

= \(\frac14x^2\) - (2.\(\frac12.2\)).\(x.y\) + 4y\(^2\)

= \(\frac14x^2\) - 2\(xy\) + 4y\(^2\)

c; (12\(x^6\).y\(^4+9x^5y^3-15x^2y^3):\left(3x^2y^3\right)\)

Câu c đề bài phải như này mới hợp lý em ơi

d; (\(x+2)^2\) - (\(x-3)\left(x+1\right)\)

= (\(x^2\) + 4\(x\) + 4) - (\(x^2\) + \(x\) - 3\(x-3\))

= \(x^2\) + 4\(x+4\) - \(x^2\) - \(x\) + 3\(x\) + 3

= (\(x^2\) - \(x^2\)) + (4\(x\) - \(x+3x\)) + (4 + 3)

= 0 + (3\(x+3x\)) + 7

= 6\(x+7\)