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Giờ mình phải ôn
Tại hông có thời gian để lm. Mà mình hứa mai sẽ làm cho bn <3
\(x^3+y^3+z^3-3xyz\)
\(=x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^2-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)\cdot z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2+z^2-zx-yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
x3 + y3 + z3 - 3xyz
= ( x3 + y3 ) + z3 - 3xyz
= ( x + y )3 - 3xy( x + y ) + z3 - 3xyz
= [ ( x + y )3 + z3 ] - [ 3xy( x + y ) + 3xyz ]
= ( x + y + z )[ ( x + y )2 - ( x + y )z + z2 ] - 3xy( x + y + z )
= ( x + y + z )( x2 + 2xy + y2 - xz - yz + z2 - 3xy )
= ( x + y + z )( x2 + y2 + z2 - xy - yz - xz )
mk chỉnh đề
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
Ta có :
\(x^3+y^3+z^3-3xyz\)
\(\Rightarrow\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(\Rightarrow\left(x+y+z\right)\left[\left(x+y^2\right)-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
P/s tham khảo nha, Tớ sửa đề dấu - thành dấu +nha
hok tốt
x^3+y^3+z^3-3xyz
= (x^3+3x^2y+3xy^2+y^3)+z^3-(3x^2y+3xy^2+3xyz)
= (x+y)^3+z^3 -3xy(x+y+z)
= (x+y+z)(x+y)^2-(x+y)z+z^2)-3xy(x+y+z)
=(x+y+z)(x^2+y^2+2xy-xz-yz+z^2-3xy)
=(x+y+z)(x^2+y^2+z^2-xz-yx-xy)
x^3+y^3+z^3-3xyz
= (x^3+3x^2y+3xy^2+y^3)+z^3-(3x^2y+3xy^2+3xyz)
= (x+y)^3+z^3 -3xy(x+y+z)
= (x+y+z)(x+y)^2-(x+y)z+z^2)-3xy(x+y+z)
=(x+y+z)(x^2+y^2+2xy-xz-yz+z^2-3xy)
=(x+y+z)(x^2+y^2+z^2-xz-yx-xy)
\(2,25x^2-12x-13\)
\(=25x^2-25x+13x-13\)
\(=25x\left(x-1\right)+13\left(x-1\right)\)
\(=\left(x-1\right)\left(25x+13\right)\)
\(3,2y^2-3y-5\)
\(=2y^2+2y-5y-5\)
\(=2y\left(y+1\right)-5\left(y+1\right)\)
\(=\left(y+1\right)\left(2y-5\right)\)
Còn bài 1 mik đang nghĩ, khi nào biết mik trả lời nha!!!
Chúc bn học giỏi!!!
\(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\)
\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3\)
\(=6a^2b+2b^3\)
\(=2b\left(3a^2+b^2\right)\)
a/\(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=\left(a^3+3a^2b+3ab^2+b^3\right)-\left(a^3-3a^2b+3ab^2-b^3\right)\)\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^2\)
\(=6ab^2+2b^3\)(rút gọn hết)
b/\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x-y\right)+z^3-3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x-y-z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-2xz+2xz+2xy-3xz-3yz-3xy\right).\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
Hok tốt
a. x3+y3+z3-3xyz
=(x3+3x2y+3xy2+y3)+z3+(-3xyz-3x2y-3xy2)
=((x+y)3+z3)-3xy(x+y+z)
=(x+y+z)((x+y)2-z(x+y)+z2)-3xy(x+y+z)
=(x+y+z)(x2+2xy+y2-zx-zy+z2-3xy)
=(x+y+z)(x2-xy+y2+z2-zx-zy)
b. (x2-8)2+36
=x4-16x2+64+36
=x4-16x2+100
=(x4+20x2+100)-36x2
=(x2+10)2-36x2
=(x2-6x+10)(x2+6x+10)
Chúc bạn học giỏi, k cho mình nhé!!!
a)(a+b+c)3 - a3 - b3 - c3
= (a+b+c-a)( a2+b2+c2+2ab+2bc+2ac-a2-ab-ac+a2) - (b+c)(b2-bc+c2)
=(b+c)(a2+ab+ac+bc)
b) x3+y3+z3-3xyz
= (x+y)3-3xy(x+y) +z3-3xyz
= (x+y+z)(x2+y2+2xy-xz-yz+z2) - 3xy(x+y+z)
=(x+y+z)( x2+y2+z2-xy-yz-xz)
x^3+y^3+z^3-3xyz= (x^3+3x^2y+3xy^2+y^3)+z^3-(3x^2y+3xy^2+3xyz)
= (x+y)^3+z^3 -3xy(x+y+z)
= (x+y+z)(x+y)^2-(x+y)z+z^2)-3xy(x+y+z)
=(x+y+z)(x^2+y^2+2xy-xz-yz+z^2-3xy)
=(x+y+z)(x^2+y^2+z^2-xz-yx-xy)
x3+y3+z3-xyz