Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,=\dfrac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\\ =\dfrac{1}{2}\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\\ b,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ c,=\dfrac{1}{2}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)^2\)
2x2 + 2y2 + b2 + 3xy - bx - by = 0
<=> 4x2 + 4y2 + 2b2 + 6xy - 2bx - 2by = 0
<=> (x2 - 2bx + b2) + (y2 - 2by + y2) + (3x2 + 6xy + 3y2) = 0
<=> (x - b)2 + (y - b)2 + 3(x + y)2 = 0
Ta thấy VT > 0 nên không có nghiệm.
PS: Không phải phân tích nhân tử mà là giải phương trình nhé.
a.
\(2x^3-x^2y+x^2+y^2-2xy-y=0\)
\(\Leftrightarrow x^2\left(2x-y+1\right)-y\left(2x-y+1\right)=0\)
\(\Leftrightarrow\left(x^2-y\right)\left(2x-y+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-y=0\\2x-y+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=x^2\\y=2x+1\end{matrix}\right.\)
Thế vào pt đầu:
\(\left[{}\begin{matrix}x^3+x-2=0\\x\left(2x+1\right)+x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x^2+x+2\right)=0\\x^2+x-1=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
\(x^2-2xy+x=-y\)
Thế vào \(y^2\) ở pt dưới:
\(x^2\left(x^2-4y+3\right)+\left(x^2-2xy+x\right)^2=0\)
\(\Leftrightarrow x^2\left(x^2-4y+3\right)+x^2\left(x-2y+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\Rightarrow y=0\\x^2-4y+3+\left(x-2y+1\right)^2=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x^2-4xy+2x+4y^2-8y+4=0\)
\(\Leftrightarrow2\left(x^2-2xy+x\right)+4y^2-8y+4=0\)
\(\Leftrightarrow-2y+4y^2-8y+4=0\)
\(\Leftrightarrow...\)
\(a,=2\left(\dfrac{1}{4}x^2-y^2\right)=2\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)\\ b,=\dfrac{1}{3}x\left(y+3xz+3z\right)\\ c,=2x\left(9x^2-\dfrac{4}{25}\right)=2x\left(3x-\dfrac{2}{5}\right)\left(3x+\dfrac{2}{5}\right)\)
\(d,=x^2\left(\dfrac{2}{5}+5x+y\right)\\ e,=\dfrac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\\ =\dfrac{1}{2}\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\\ f,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ g,=\dfrac{1}{2}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)^2\)
a) x4 + 2x3 + x2
= x2 ( x2 + 2x + 1 )
= x2 ( x + 1 )2
b) 5x2 - 10xy + 5y2 - 20z2
= 5 [(x2 - 2xy + y2 ) - 4z2 ]
= 5 [( x - y )2 - ( 2z )2 ]
= 5 ( x - y - 2z ) ( x - y + 2z )
c) x3 - x + 3x2y + 3xy2+ y3- y
= ( x3 + 3x2y + 3xy2 + y3 ) - ( x + y )
= (x + y )3 - ( x + y)
= ( x + y ) [( x + y )2 - 1 ]
= ( x + y ) ( x + y + 1 ) ( x + y - 1 )