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a, (x+y+z)2
=\(x^2+y^2+z^2+2xy+2xz+2yz\)
b, (x+y−z)2
=\(x^2+y^2+z^2+2xy-2xz-2yz\)
c, (x−y−z)2
=\(x^2+y^2+z^2-2xy-2xz+2yz\)
chúc bạn học tốt ạ
a) Ta có: \(\left(x+y+z\right)^2=\left[\left(x+y\right)+z\right]^2\)
\(=\left(x+y\right)^2+2\left(x+y\right)z+z^2\)
\(=x^2+2xy+y^2+2xz+2yz+z^2\)
\(=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)
b) Ta có: \(\left(x+y-z\right)^2=\left[\left(x+y\right)-z\right]^2\)
\(=\left(x+y\right)^2-2\left(x+y\right)z+z^2\)
\(=x^2+2xy+y^2-2xz-2yz+z^2\)
\(=x^2+y^2+z^2+2\left(xy-yz-zx\right)\)
c) Ta có: \(\left(x-y-z\right)^2=\left[\left(x-y\right)-z\right]^2\)
\(=\left(x-y\right)^2-2\left(x-y\right)z+z^2\)
\(=x^2-2xy+y^2-2xz-2yz+z^2\)
\(=x^2+y^2+z^2-2\left(xy+yz+zx\right)\)
a) \(\left(\frac{1}{3}u+3v\right)^2=\frac{1}{9}u^2+2uv+9v^2\)
b) \(\left(\frac{1}{2}x^2-6x\right)^2=\frac{1}{4}x^4-6x^3+36x^2\)
c) \(\left(-\frac{1}{2}a+b\right)^2=\frac{1}{4}a^2-ab+b^2\)
d) \(\left(-\frac{4}{3}a-\frac{1}{3}b\right)^2=\frac{16}{9}a^2+\frac{8}{9}ab+\frac{1}{9}b^2\)
e) \(\left(\frac{2}{3}x-\frac{3}{2}y\right)\left(\frac{2}{3}x+\frac{3}{2}y\right)=\frac{4}{9}x^2-\frac{9}{4}y^2\)
a) \(\left(\frac{1}{3}u+3v\right)^2=\frac{1}{9}u^2+2uv+9v^2\)
b) \(\left(\frac{1}{2}x^2-6x\right)^2=\frac{1}{4}x^4-6x^3+36x^2\)
c) \(\left(-\frac{1}{2}a+b\right)^2=\frac{1}{4}a^2-ab+b^2\)
d) \(\left(-\frac{4}{3}a-\frac{1}{3}b\right)^2=\frac{16}{9}a^2+\frac{8}{9}ab+\frac{1}{9}b^2\)
e) \(\left(\frac{2}{3}x-\frac{3}{2}y\right)\left(\frac{2}{3}x+\frac{3}{2}y\right)=\left(\frac{2}{3}x\right)^2-\left(\frac{3}{2}y\right)^2=\frac{4}{9}x^2-\frac{9}{4}y^2\)
a) \(\left(2x-1\right)\left(4x^2+2x+1\right)=8x^3-1\)
b) \(\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2\)
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
a) Ta có: \(\left(x-3\right)^3\)
\(=x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3\)
\(=x^3-9x^2+27x^2-27\)
b) Ta có: \(\left(2x-3\right)^3\)
\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2-3^3\)
\(=8x^3-36x^2+54x-27\)
c) Ta có: \(\left(x-\frac{1}{2}\right)^3\)
\(=x^3-3\cdot x^2\cdot\frac{1}{2}+3\cdot x\cdot\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3\)
\(=x^3-\frac{3}{2}x^2+\frac{3}{4}x-\frac{1}{8}\)
d) Ta có: \(\left(x^2-2\right)^3\)
\(=\left(x^2\right)^3-3\cdot\left(x^2\right)^2\cdot2+3\cdot x^2\cdot2^2-2^3\)
\(=x^6-6x^4+12x^2-8\)
e) Ta có: \(\left(2x-3y\right)^3\)
\(=\left(2x\right)^3-2\cdot\left(2x\right)^2\cdot3y+2\cdot2x\cdot\left(3y\right)^2-\left(3y\right)^3\)
\(=8x^3-24x^2y+36xy^2-27y^3\)
f) Ta có: \(\left(\frac{1}{2}x-y^2\right)^3\)
\(=\left(\frac{1}{2}x\right)^3-3\cdot\left(\frac{1}{2}x\right)^2\cdot y^2+3\cdot\frac{1}{2}x\cdot\left(y^2\right)^2-\left(y^2\right)^3\)
\(=\frac{1}{8}x^3-\frac{3}{4}x^2y^2+\frac{3}{2}xy^4-y^6\)
a,\(\left(2x-1\right)\left(4x^2+2x+1\right)=\left(2x-1\right)\left[\left(2x\right)^2+2x.1+1^2\right]\)
\(=\left(2x\right)^3-1=8x^3-1\)
b,\(\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2\)
\(=x^2+2.x.2y+\left(2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)
`a)(2x-1)(4x^2+2x+1)`
`=(2x-1)[(2x)^2+2x.1+1^2]`
`=(2x)^3-1^3`
`=8x^3-1`
Áp dụng HĐT:`A^3-B^3=(A-B)(A^2+AB+B^2)`
`b)(x+2y+z)(x+2y-z)`
`=[(x+2y)+z][(x+2y)-z]`
`=(x+2y)^2-z^2`
`=x^2+2.x.2y+(2y)^2-z^2`
`=x^2+4xy+4y^2-z^2`
Áp dụng HĐT:`A^2-B^2=(A+B)(A-B)`
`(A+B)^2=A^2+2AB+B^2`
Câu hỏi của Yến Trần - Toán lớp 8 - Học toán với OnlineMath
Trả lời :
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\(\left(a+b-c\right)^2=\left(\left(a+b\right)-c\right)^2\)
\(=\left(a+b\right)^2+c^2-2\left(a+b\right)c\)
\(=a^2+b^2+2ab+c^2-2ac-2bc\)
\(=a^2+b^2+c^2+2ab-2bc-2ca\)
\(\left(a-b+c\right)^2=\left(\left(a-b\right)+c\right)^2\)
\(=\left(a-b\right)^2+c^2+2\left(a-b\right)c\)
\(=a^2+b^2-2ab+c^2+2ac-2bc\)
\(=a^2+b^2+c^2-2ab-2bc+2ca\)
\(\left(x-y+z\right)\left(x-y-z\right)=\left(\left(x-y\right)+z\right)\left(\left(x-y\right)-z\right)\)
\(=\left(x-y\right)^2-z^2\)
\(=x^2+y^2-2xy-z^2\)
( a + b - c )2 = [ ( a + b ) - c ]2
= ( a + b )2 - 2( a + b )c + c2
= a2 + b2 + c2 + 2ab - 2bc - 2ac
( a - b + c )2 = [ ( a- b ) + c ]2
= ( a - b )2 + 2( a - b )c + c2
= a2 + b2 + c2 - 2ab - 2bc + 2ca
( x - y + z )( x - y - z ) = [ ( x - y ) + z ][ ( x - y ) - z ]
= ( x - y )2 - z2
= x2 + y2 - z2 - 2xy