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1: a) Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=7^2+2.7+37\) (Vì \(x-y=7\))
\(=100\)
Vậy \(A=100\)
b) Ta có: \(B=x^2+4y^2-2x+10+4xy-4y\)
\(=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10\)
\(=25\)
Vậy \(B=25\)
c) Ta có : \(C=\left(x-y\right)^2\)
\(=x^2-2xy+y^2\)
\(=\left(x^2+y^2\right)-2xy\)
\(=26-2.5\) (Vì \(x^2+y^2=26\) ; \(xy=5\))
\(=16\)
Vậy \(C=16\)
2: a) \(\left(x+y\right)^2-y^2=x^2+2xy+y^2-y^2\)
\(=x^2+2xy\)
\(=x\left(x+2y\right)\) \(\left(dpcm\right)\)
b) \(\left(x^2+y^2\right)^2-2xy^2=\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x-y\right)^2\left(x+y\right)^2\) \(\left(dpcm\right)\)
c) \(\left(x+y\right)^2=x^2+2xy+y^2\)
\(=\left(x^2-2xy+y^2\right)+4xy\)
\(=\left(x-y\right)^2+4xy\) \(\left(dpcm\right)\)
Chúc bn học tốt ✔✔✔
a. -(b-a)3= -b3+a3 (phá ngoặc trước có dấu trừ nên đổi dấu)
= a3 - b3 = (a-b)3
b)
\(\left(-a-b\right)^2=\left(-a\right)^2-2.\left(-a\right)b+b^2\\ =a^2+2ab+b^2=\left(a+b\right)^2\)
a ) \(\left(x+y\right)^3+\left(x-y\right)^3-2x^3\)
\(=x^3+3x^2y+3y^2x+y^3+x^3-3x^2y+3y^2x-y^3-2x^3\)
\(=\left(x^3+x^3-2x^3\right)+\left(y^3-y^3\right)+\left(3x^2y-3x^2y\right)+\left(3y^2x+3y^2x\right)\)
\(=6y^2x\)
b ) \(\left(x+y\right)^2-\left(x-y\right)^2+\left(x+y\right)\left(x-y\right)\)
\(=\left(x+y-x+y\right)\left(x+y+x-y\right)+x^2-y^2\)
\(=2y.2x+x^2-y^2\)
\(=x^2-y^2+4xy\)
c ) \(\left(3x+1\right)^2+2\left(9x^2-1\right)+\left(3x-1\right)^2\)
\(=\left(3x+1\right)^2+2\left(3x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(3x+1+3x-1\right)^2\)
\(=\left(6x\right)^2=36x^2\)
d ) \(\left(a+b+c\right)^2-2\left(a+b+c\right)\left(b+c\right)+\left(b+c\right)^2\)
\(=\left(a+b+c-b-c\right)^2\)
\(=a^2\)
a) \(\left(a-b\right)^3=-\left(b-a\right)^3\)
Ta có: \(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3\)
\(=-\left(b^3-3ab^2+3a^2b-a^3\right)\)
\(=-\left(b-a\right)^3\)
Vậy..
c) \(\left(x+y\right)^3=x\left(x-3y\right)^2+y\left(y-3x\right)^2\)
Ta có: \(x\left(x-3y\right)^2+y\left(y-3x\right)^2\)
\(=x^3-6x^2y+9xy^2+y^3+y^3-6xy^2+9x^2y\)
\(=x^3-3x^2y\left(2-3\right)+3xy^2\left(3-2\right)+y^3\)
\(=x^3+3x^2y+3xy^2+y^3\)
\(=\left(x+y\right)^3\)
Vậy..
d)\(\left(x+y\right)^3-\left(x-y\right)^3=2y\left(y^2+3x^2\right)\)
Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=\left(x+y-x+y\right)\left(x^2+2xy+y^2+x^2-y^2+x^2+x^2+y^2\right)\)
\(=2y\left(y^2+3x^2\right)\)
Vậy...
Bài 1 :
a ) \(2x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(2x+2\right)=2\left(x+1\right)^2\)
b ) \(y^2\left(x^2+y\right)-zx^2-zy=y^2\left(x^2+y\right)-z\left(x^2+y\right)=\left(x^2+y\right)\left(y^2-z\right)\)
c ) \(4x\left(x-2y\right)+8y\left(2y-x\right)=4x\left(x-2y\right)-8y\left(x-2y\right)=4\left(x-2y\right)^2\)
d ) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)=\left(x+1\right)\left(3x-2x^2+7\right)\)
e ) \(x^2-6xy+9y^2=\left(x-3x\right)^2\)
Bài 1 :
f ) \(x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)
g ) \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
h ) \(125x^3+y^6=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
* 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) \(\left(x+y\right)^2-y^2=x\left(x+y^2\right)\)
\(\Leftrightarrow\left(x+y+y\right)\left(x+y-y\right)=x^2+xy^2\)
\(\Leftrightarrow\left(x+2y\right)x=x^2+xy^2\)
\(\Leftrightarrow x^2+2xy-x^2-xy=0\)
\(\Leftrightarrow xy=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\y=0\\x=y=0\end{matrix}\right.\)
Chứng minh đẳng thức mà, làm kì quá ông ơi