Tính theo hằng đẳng thức : a) $(a+b+c)^{2}$ b) $(a-b+c)^{2}$ c) (x+y+z) (x+y-z) d) $(x+y+z)^{2}$ e) $(a-b-c)^{2}$ 17/08/2021 Bởi Remi Tính theo hằng đẳng thức : a) $(a+b+c)^{2}$ b) $(a-b+c)^{2}$ c) (x+y+z) (x+y-z) d) $(x+y+z)^{2}$ e) $(a-b-c)^{2}$
Giải thích các bước giải:`a)(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc` `b)(a−b+c)^2=a^2+b^2+c^2+2ac−2ab−2bc` `c)(x+y+z)(x+y−z)=(x+y)^2-z^2=x^2+2xy+y^2−z^2` `d)(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz` `e)(a-b-c)^2=a^2+b^2+c^2-2ab-2ac+2bc` e)(a−b−c)2=(a−b−c)(a−b−c)=a2−ab−ac−ab+b2+bc−ac+bc+c2=a2+b2+c2−2ac+2bc−2ab Bình luận
Đáp án: $a) (a+b+c)^2 = (a+b+c)(a+b+c) = a^2 +ab+ac+ab+b^2+bc+ac+bc+c^2 = a^2+b^2+c^2 +2ab+2ac+2bc$ $b) (a-b+c)^2 = (a-b+c)(a-b+c) = a^2 -ab+ac-ab+b^2-bc+ac-bc+c^2 = a^2+b^2+c^2 +2ac-2ab-2bc$ $c)( x+y+z)(x+y-z) = x^2 +xy -xz +xy +y^2 -yz +xz+yz-z^2 = x^2 +2xy+y^2 -z^2$ $d) (x+y+z)^2 = (x+y+z)(x+y+z) = x^2 +xy +xz +xy +y^2+yz +xz +yz +z^2 = x^2+y^2+z^2 +2xy+2xz+2yz$ $e) (a-b-c)^2= (a-b-c)(a-b-c) = a^2 -ab-ac-ab+b^2+bc-ac+bc+c^2 = a^2+b^2+c^2-2ac+2bc-2ab$ Bình luận
Giải thích các bước giải:`a)(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc`
`b)(a−b+c)^2=a^2+b^2+c^2+2ac−2ab−2bc`
`c)(x+y+z)(x+y−z)=(x+y)^2-z^2=x^2+2xy+y^2−z^2`
`d)(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz`
`e)(a-b-c)^2=a^2+b^2+c^2-2ab-2ac+2bc`
e)(a−b−c)2=(a−b−c)(a−b−c)=a2−ab−ac−ab+b2+bc−ac+bc+c2=a2+b2+c2−2ac+2bc−2ab
Đáp án:
$a) (a+b+c)^2 = (a+b+c)(a+b+c) = a^2 +ab+ac+ab+b^2+bc+ac+bc+c^2 = a^2+b^2+c^2 +2ab+2ac+2bc$
$b) (a-b+c)^2 = (a-b+c)(a-b+c) = a^2 -ab+ac-ab+b^2-bc+ac-bc+c^2 = a^2+b^2+c^2 +2ac-2ab-2bc$
$c)( x+y+z)(x+y-z) = x^2 +xy -xz +xy +y^2 -yz +xz+yz-z^2 = x^2 +2xy+y^2 -z^2$
$d) (x+y+z)^2 = (x+y+z)(x+y+z) = x^2 +xy +xz +xy +y^2+yz +xz +yz +z^2 = x^2+y^2+z^2 +2xy+2xz+2yz$
$e) (a-b-c)^2= (a-b-c)(a-b-c) = a^2 -ab-ac-ab+b^2+bc-ac+bc+c^2 = a^2+b^2+c^2-2ac+2bc-2ab$