Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\left(x-2\right)^2-\left(2x+1\right)^2=x^2-4x+4-4x^2-4x-1=-3x^2+3=-3\left(x^2-1\right)\)
\(=-3\left(x-1\right)\left(x+1\right)\)
\(B=\left(x-2y\right)^2-\left(x-2y\right)\left(x+2y\right)=\left(x-2y\right)\left(x-2y-x-2y\right)=-4y\left(x-2y\right)\)
\(C=\left(x+1\right)^3-\left(x-2\right)^3=\left(x^3+3x^2+3x+1\right)-\left(x^3-6x^2+12x-8\right)\)
\(=x^3+3x^2+3x+1-x^3+6x^2-12x+8=9x^2-9x+9=9\left(x^2-x+1\right)\)
\(D=\left(x-1\right)^2-2\left(x-1\right)\left(x+1\right)+\left(x+1\right)^2=\left(x-1-x-1\right)^2=-2^2=4\)
\(E=\left(x+2y\right)^2+2\left(x+2y\right)\left(x-2y\right)+2y-x=x^2+4xy+4y^2+2\left(x^2-4y^2\right)+2y-x\)
\(=x^2+4xy+4y^2+2x^2-8y^2+2y-x=3x^2-4y^2+4xy+2y-x\)
\(G=\left(2x+1\right)^3-\left(2x-1\right)=8x^3+12x^2+6x+1-2x+1=8x^3+12x^2+4x+2\)
\(=2\left(4x^3+6x^2+2x+1\right)=2\left(4x\left(x+1\right)^2+1\right)\)
A = ( x - 2 )2 - ( 2x + 1 )2
A = x2 - 4x + 4 - 4x2 + 4x + 1
A = - 3x2 + 5
B = ( x - 2y )2 - ( x - 2y ) . ( 2y + x )
B = x2 - 4xy + 4y2 - ( 2xy + x2 - 4y2 - 2xy )
B = x2 - 4xy + 4y2 - 2xy - x2 + 4y2 + 2xy
B = 8y2 - 4xy
\(x^2+3x+2\) =\(x^2+2.\frac{3}{2}x+\left(\frac{3}{2}\right)^2-\frac{5}{4}\)=\(\left(x+\frac{3}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)
Dấu "=" xảy ra <=>\(x+\frac{3}{2}=0\)<=>\(x=-\frac{3}{2}\)
Bài 2:
a) \(x^2-4x+y^2+2y+5=0\)
=> \(\left(x^2-4x+4\right)+\left(y^2+2y+1\right)=0\)
=>\(\left(x-2\right)^2+\left(y+1\right)^2=0\)
Vì \(\left(x-2\right)^2+\left(y+1\right)^2\ge0\)nên:
=>\(\hept{\begin{cases}x-2=0\\y+1=0\end{cases}}\)<=>\(\hept{\begin{cases}x=2\\y=-1\end{cases}}\)
b)\(2x^2+y^2-2xy+10x+25=0\)
=>\(\left(x^2-2xy+y^2\right)+\left(x^2+10x+25\right)=0\)
=>\(\left(x-y\right)^2+\left(x+5\right)^2=0\)
Tới đây thì dễ nhá !
aVT=.\(\left(a+b+c\right)^2+a^2+b^2+c^2\)
=\(a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)
=\(2a^2+2b^2+2c^2+2ab+2ac+2bc\)
VP=\(\left(a+b\right)^2+\left(b+c\right)^2+\left(a+c\right)^2\)=\(a^2+2ab+b^2+b^2+2bc+b^2+a^2+2ac+c^2\)
=\(2a^2+2b^2+2c^2+2ab+2bc+2ac\)
Vậy VT=VP
a)\(\text{(a+b+c)^2 +a^2+b^2+c^2=(a+b)^2+(b+c)^2+(c+a)^2}\)
Ta có:
\(\left(a+b+c\right)^2+a^2+b^2+c^2=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2\)
\(=\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)+\left(c^2+2ca+a^2\right)\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
Vậy \(\left(a+b+c\right)^2+a^2+b^2+c^2=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
b) Câu b sao chỉ có một vế vậy , hằng đẳng thức thì phải có hai vế chứ
a) \(\left(x-2y\right)^2+\left(x+2y\right)^2=x^2-4xy+4y^2+x^2+4xy+4y^2=2x^2+8y^2\)
b) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2=2\left(x^2-y^2\right)+x^2+2xy+y^2+x^2-2xy^2+y^2\)
\(=2x^2-2y^2+2x^2+2y^2=4x^2\)
\(a,\left(x-2y\right)^2+\left(x+2y\right)^2\)
\(=\left(x^2-4xy+4y^2\right)
+\left(x^2+4xy+4y^2\right)\)
\(=2x^2+8y^2\)
\(b,2\left(x-y\right).\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=2\left(x^2-y^2\right)+\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)\)
\(=2x^2-2y^2+2x^2+2y^2\)
\(=4x^2\)
\(\frac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}\)
\(=\frac{\left(y-x\right)\left(y+x\right)}{\left(x-y\right)^3}\)
\(=-\frac{\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^3}\)
\(=-\frac{x+y}{\left(x-y\right)^2}\)
Rút gọn biểu thức
\(=\left(1-y^2\right)z+2y^2+\left(-x^2\right)y+2x^2-2\)