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a) \(\left(2x-1\right)\left(4x^2+2x+1\right)=8x^3-1\)
b) \(\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2\)
* Đặt tên các biểu thức theo thứ tự là A,B,C,D,E.
Câu a)
Theo hằng đẳng thức đáng nhớ ta có:
\(a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(c+a)\)
\(=(a+b+c)^3-3[ab(a+b)+bc(b+c)+ca(c+a)+2abc]\)
\(=(a+b+c)^3-3[ab(a+b+c)+bc(b+c+a)+ca(c+a+b)-abc]\)
\(=(a+b+c)^3-3[(a+b+c)(ab+bc+ac)]+3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=(a+b+c)^3-3(ab+bc+ac)(a+b+c)\)
\(=(a+b+c)[(a+b+c)^2-3(ab+bc+ac)]\)
\(=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)\) (*)
Do đó:
\(A=\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ac)}{a^2+b^2+c^2-ab-bc-ac}=a+b+c\)
Câu b)
\(x^3-y^3+z^3+3xyz=x^3+(-y)^3+z^3-3x(-y)z\)
Sử dụng kết quả (*) của câu a. Với \(a=x, b=-y, c=z\)
\(\Rightarrow x^3+(-y)^3+z^3-3x(-y)z=(x-y+z)(x^2+y^2+z^2+xy+yz-xz)\)
Mặt khác xét mẫu số:
\((x+y)^2+(y+z)^2+(x-z)^2=x^2+2xy+y^2+y^2+2yz+z^2+x^2-2xz+z^2\)
\(=2(x^2+y^2+z^2+xy+yz-xz)\)
Do đó: \(B=\frac{(x-y+z)(x^2+y^2+z^2+xy+yz-xz)}{2(x^2+y^2+z^2+xy+yz-xz)}=\frac{x-y+z}{2}\)
Câu c) Sử dụng kết quả (*) của phần a:
\(x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)\)
Và mẫu số:
\((x-y)^2+(y-z)^2+(z-x)^2=2(x^2+y^2+z^2-xy-yz-xz)\)
Do đó: \(C=\frac{(x+y+z)(x^2+y^2+z^2-xy-yz-xz)}{2(x^2+y^2+z^2-xy-yz-xz)}=\frac{x+y+z}{2}\)
Câu d)
Xét tử số:
\(a^2(b-c)+b^2(c-a)+c^2(a-b)\)
\(=a^2(b-c)-b^2[(b-c)+(a-b)]+c^2(a-b)\)
\(=(b-c)(a^2-b^2)-(b^2-c^2)(a-b)\)
\(=(b-c)(a-b)(a+b)-(b-c)(b+c)(a-b)\)
\(=(a-b)(b-c)[a+b-(b+c)]=(a-b)(b-c)(a-c)\) (1)
Xét mẫu số:
\(a^4(b^2-c^2)+b^4(c^2-a^2)+c^4(a^2-b^2)\)
\(=a^4(b^2-c^2)-b^4[(b^2-c^2)+(a^2-b^2)]+c^4(a^2-b^2)\)
\(=(a^4-b^4)(b^2-c^2)-(b^4-c^4)(a^2-b^2)\)
\(=(a^2-b^2)(a^2+b^2)(b^2-c^2)-(b^2-c^2)(b^2+c^2)(a^2-b^2)\)
\(=(a^2-b^2)(b^2-c^2)[a^2+b^2-(b^2+c^2)]\)
\(=(a^2-b^2)(b^2-c^2)(a^2-c^2)\)
\(=(a-b)(b-c)(a-c)(a+b)(b+c)(c+a)\)(2)
Từ (1)(2) suy ra \(D=\frac{1}{(a+b)(b+c)(c+a)}\)
Câu e)
Theo phần d ta có:
\(TS=(a-b)(b-c)(a-c)\)
\(MS=ab^2-ac^2-b^3+bc^2\)
\(=b^2(a-b)-c^2(a-b)=(a-b)(b^2-c^2)=(a-b)(b-c)(b+c)\)
Do đó: \(E=\frac{(a-b)(b-c)(a-c)}{(a-b)(b-c)(b+c)}=\frac{a-c}{b+c}\)
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) \(\left(\frac{2}{3}x-\frac{3}{2}y\right)\left(\frac{2}{3}x+\frac{3}{2}y\right)=\left(\frac{2}{3}x\right)^2-\left(\frac{3}{2}y\right)^2=\frac{4}{9}x^2-\frac{9}{4}y^2\)
\(\left(a+b-c\right)^2=\left(\left(a+b\right)-c\right)^2\)
\(=\left(a+b\right)^2+c^2-2\left(a+b\right)c\)
\(=a^2+b^2+2ab+c^2-2ac-2bc\)
\(=a^2+b^2+c^2+2ab-2bc-2ca\)
\(\left(a-b+c\right)^2=\left(\left(a-b\right)+c\right)^2\)
\(=\left(a-b\right)^2+c^2+2\left(a-b\right)c\)
\(=a^2+b^2-2ab+c^2+2ac-2bc\)
\(=a^2+b^2+c^2-2ab-2bc+2ca\)
\(\left(x-y+z\right)\left(x-y-z\right)=\left(\left(x-y\right)+z\right)\left(\left(x-y\right)-z\right)\)
\(=\left(x-y\right)^2-z^2\)
\(=x^2+y^2-2xy-z^2\)
( 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