Aufgaben zu den binomischen Formeln

$(a+b)^2$

$\left(a+bc\right)^2$

$\left(ac+bd\right)^2$

$\left(ad+cb\right)^2$

$5(6x+4y)^2$

$(3-8n)^2$

$(u+5v)^2$

$(z-1)^2$

$(2s+r)^2$

$(4a-5b)^2$

$(a+b)(u+v)^2$

$(3b-5c)(2x+5y)^2$

$(a+7)(a-7)$

$\left(a+4b\right)^2$

$\left(2r-12\right)^2$

$\left(r^2+5\right)^2$

$\left(r-\frac13\right)^2$

$\left(r+8\right)\left(r-8\right)$

$\left(2r+9\right)\left(2r-9\right)$

$\left(-z+9\right)^2$

$\left(-a-2{,}5\right)^2$

$\left(r+4\right)^3$

$\left(2r-\frac12\right)^3$

$\left(2+r\right)^2-\left(2-r\right)^2$

$16r^2-\left(3a-4r\right)^2$

$\left(5r-19\right)^2-\left(r-3\right)\left(3+r\right)-\left(3r+4\right)\left(4r-5\right)+\left(2r+3\right)^2+179r+1$

$(4x+5y)\;(5y+4x)$

$(8x^2-3x)(8x^2-3x)$

$(4p+5q)(4q+5p)$

$(8x^2-3x)(3x+8x^2)$

$(-x-3y)(-3y-x)$

$(\frac34m-2n)(\frac43m-2n)$

$16a^2-16ab+4b^2$

$\frac14a^2-4ab+4b^2$

$25a^2+50ab-4b^2$

$\left(3x-5y\right)^2-\left(3x+5y\right)^2$

$\frac23\left(6a-1{,}5b\right)^2$

$\left(0{,}5x-y\right)^2-\left(0{,}5x-y\right)\cdot\left(0{,}5x+y\right)$

$\left(\frac13x-2\right)^2+\frac13\cdot\left(x+2\right)^2$

$\left(2x-3\right)^2-2\cdot\left(x+3\right)^2-\frac12\cdot\left(6-2x\right)^2$

$\frac13\cdot\left(1{,}5a-b\right)^2-\frac34\cdot\left(\frac13b+a\right)^2$

$\left(a-0{,}4b\right)^2-2\cdot\left(0{,}3b-0{,}5a\right)^2+0{,}2\cdot\left(a+0{,}1b\right)^2$

$…\;+\;14r+49=\left(…….\right)^2$

$r^2-\frac13r\;…..=\left(……\right)^2$

$225+30a+a^2$

$4m^2+28m+49$

$9a^2-16b^2$

$81u^2-36u+4$

$36u^2-289w^2$

$324+36x+x^2$

$49k^2-70ku+25u^2$

$64y^2-160yz+100z^2$

$361m^2-256n^2$

$121x^2+44xy+4y^2$

$100r^2-225$

$4r^2+4r+1$

$r^2-7r+12\frac14$

$48r^3-147ry^2$

$49p^2-112pq+64q^2$

$24a^2r^2+120ar+150$

$\frac14a^2-3\mathrm{ab}+9b^2$

$4b^2-c^2$

$4x^2-12\mathrm{xy}+9y^2$

$\frac14a^2-\mathrm{ab}+b^2$

$a^2-b^2$

$49p^2-81q^2$

$a^2+16-8a$

$a^2+10a+25$

$5x^2+3xy+y^2+xy-x^2$

$\frac{36}{49}m^2-\frac3{14}mn+\frac1{64}n^2$

$\frac19u^2-\frac4{15}uv+\frac4{25}v^2$

$ca+a$

$4a+2ab$

$4a+6b$

$-3ac+9ab$

$(a+b)^2$

$(c-d)^2$

$(2+a)\cdot(2-a)$