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\(ĐK:x\ne y;x\ne-y;x^2+xy+y^2\ne0;x^2-xy+y^2\ne0\)
\(A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\left[1:\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)}\right]\\ A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+y^2\right)}\\ A=x-y=B\)
\(x=0;y=0\Leftrightarrow B=0\)
Giá trị của A không xác định vì \(x=y\) trái với ĐK:\(x\ne y\)
Vậy \(A\ne B\)
a) \(x^2+y^2=x^2+y^2+2xy-2xy=\left(x+y\right)^2-2xy\)
b) \(\left(a+b\right)^2-\left(a-b\right)\left(a+b\right)=\left(a+b\right)^2-\left(a^2-b^2\right)=a^2+2ab+b^2-a^2+b^2\)
\(=2ab+2b^2=2b\left(a+b\right)\)
c)\(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b-a+b\right)\left(a+b+a-b\right)\)
\(=2b.2a=4ab\)
a: \(\left(x+y\right)^2-2xy\)
\(=x^2+2xy+y^2-2xy\)
\(=x^2+y^2\)
b: \(\left(a+b\right)^2-\left(a-b\right)\left(a+b\right)\)
\(=\left(a+b\right)\left(a+b-a+b\right)\)
\(=2b\left(a+b\right)\)
c: \(\left(a+b\right)^2-\left(a-b\right)^2\)
\(=\left(a+b-a+b\right)\left(a+b+a-b\right)\)
\(=4ab\)
a)
\(A=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)
\(=x^3-3x^2+9x+3x^2-9x+27-54-x^3\)
\(=-27\)
or
\(A=x^3+27-54-x^3=-27\)
b)
\(B=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3=2y^3\)
c)
\(C=\left(2x+1\right)^2+\left(1-3x\right)^2+2\left(2x+1\right)\left(3x-1\right)\)
\(=\left(2x+1+3x-1\right)^2=\left(5x\right)^2=25x^2\)
d)
\(D=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=x^3-8-\left(x-1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=6x^2-3x-10\)
a) \(\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x^3+1\right)-\left(x^3-1\right)\)
\(=x^3+1-x^3+1\)
\(=2\)
Biểu thức trên có giá trị bằng 2 với mọi x nên không phụ thuộc vào biến.
b) \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)-\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)-27\left(2y^3-1\right)\)
\(=\left(8x^3+27y^3\right)-\left(8x^3-27y^3\right)-27\left(2y^3-1\right)\)
\(=8x^3+27y^3-8x^3+27y^3-54y^3+27\)
\(=27\)
Biểu thức trên có giá trị bằng 27 với mọi x nên không phụ thuộc vào biến.
c) \(\left(x-1\right)^3-\left(x+4\right)\left(x^2-4x+16\right)+3x\left(x-1\right)\)
\(=x^3-3x^2+3x-1-x^3-64+3x^2-3x\)
\(=-65\)
Biểu thức trên có giá trị bằng -65 với mọi x nên không phụ thuộc vào biến.
d) \(\left(x+y+z\right)^2+\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2-3\left(x^2+y^2+z^2\right)\)
\(=x^2+y^2+z^2+2\left(xy+yz+xz\right)+\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2-3\left(x^2+y^2+z^2\right)\)
\(=2\left(xy+yz+xz\right)-2\left(x^2+y^2+z^2\right)+x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2\)
\(=2\left(xy+yz+xz\right)-2\left(x^2+y^2+z^2\right)+2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\)
\(=0\)
Biểu thức trên có giá trị bằng 0 với mọi x nên không phụ thuộc vào biến.
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.
\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)
b.
\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
TH1:
\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
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...
1.a, VT= \(\left(x^2+y^2\right)^2-\left(2xy\right)^2=\)\(\left(x^2+y^2-2xy\right)\left(x^2+y^2+2xy\right)=\left(x-y\right)^2\left(x+y\right)^2=VP.\left(đpcm\right)\)
b, VP=\(x\left(x-3y\right)^2+y\left(y-3x\right)^2\)\(=x\left(x^2-6xy+9y^2\right)+y\left(y^2-6xy+9x^2\right)\)\(=x^3-6x^2y+9xy^2+y^3-6xy^2+9x^2y\)
\(=x^3+3x^2y+3xy^2+y^3\)\(=\left(x+y\right)^3=VT\left(đpcm\right)\)
2. VT=\(\left(a+b\right)^3-\left(a-b\right)^3\)\(=\left(a+b-a+b\right)\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
\(2b\left(b^2+3a^2\right)\)\(=VP\left(đpcm\right)\).
a) (x2 + y2)2 - (2xy)2
= [(x2 + y2) - 2xy].[(x2 + y2) + 2xy]
= [x2 + y2 - 2xy].[(x2 + y2 + 2xy]
= (x - y)2 . (x + y)2