Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(2x-72x^3=2x\left(1-36x^2\right)=2x\left(1-6x\right)\left(1+6x\right)\)
f) \(4x^4+1=4x^4+4x^2+1-4x^2=\left(2x^2+1\right)^2-\left(2x\right)^2=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)
1) \(xy\left(a^2+2b^2\right)-ab\left(2x^2+y^2\right)\)
\(=a^2xy+2b^2xy-2abx^2-aby^2\)
\(=\left(a^2xy-aby^2\right)+\left(2b^2xy-2abx^2\right)\)
\(=ay\left(ax-by\right)+2bx\left(by-ax\right)\)
\(=ay\left(ax-by\right)-2bx\left(ax-by\right)\)
\(=\left(ax-by\right)\left(ay-2bx\right)\)
2) Sửa đề \(\left(xy+ab\right)^2+\left(bx-ay\right)^2\)
\(=\left(xy\right)^2+2xyab+\left(ab\right)^2+\left(bx\right)^2-2xyab+\left(ay\right)^2\)
\(=x^2y^2+a^2b^2+b^2x^2+a^2y^2\)
\(=\left(x^2y^2+b^2x^2\right)+\left(a^2b^2+a^2y^2\right)\)
\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)
\(=\left(b^2+y^2\right)\left(x^2+a^2\right)\)
3) \(\left(2xy+ab\right)^2+\left(2ay-bx\right)^2\)
\(=\left(2xy\right)^2+2.2xyab+\left(ab\right)^2+\left(2ay\right)^2-2.2xyab+\left(bx\right)^2\)
\(=4x^2y^2+4xyab+a^2b^2+4a^2y^2-4xyab+b^2x^2\)
\(=4x^2y^2+4a^2y^2+a^2b^2+b^2x^2\)
\(=4y^2\left(x^2+a^2\right)+b^2\left(a^2+x^2\right)\)
\(=\left(a^2+x^2\right)\left(4y^2+b^2\right)\)
1) \(xy\left(a^2+2b^2\right)-ab\left(2x^2+y^2\right)\)
\(=a^2xy+2b^2xy-2x^2ab-y^2ab\)
\(=\left(a^2xy-y^2ab\right)+\left(2b^2xy-2x^2ab\right)\)
\(=ay\left(ax-by\right)+2bx\left(by-ax\right)\)
\(=ay\left(ax-by\right)-2bx\left(ax-by\right)\)
\(=\left(ax-by\right)\left(ay-2bx\right)\)
2) Sửa đề \(\left(xy+ab\right)^2+\left(bx-ay\right)^2\)
\(=\left(xy\right)^2+2xyab+\left(ab\right)^2+\left(bx\right)^2-2xyab+\left(ay\right)^2\)
\(=x^2y^2+a^2b^2+b^2x^2+a^2y^2\)
\(=\left(x^2y^2+b^2x^2\right)+\left(a^2b^2+a^2y^2\right)\)
\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)
\(=\left(b^2+y^2\right)\left(a^2+x^2\right)\)
3) \(\left(2xy+ab\right)^2+\left(2ay-bx\right)^2\)
\(=\left(2xy\right)^2+2.2xyab+\left(ab\right)^2+\left(2ay\right)^2-2.2xyab+\left(bx\right)^2\)
\(=4x^2y^2+a^2b^2+4a^2y^2+b^2x^2\)
\(=\left(4x^2y^2+b^2x^2\right)+\left(4a^2y^2+a^2b^2\right)\)
\(=x^2\left(4y^2+b^2\right)+a^2\left(4y^2+b^2\right)\)
\(=\left(4y^2+b^2\right)\left(a^2+x^2\right)\)
\(1,\\ a,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ b,=a^2\left(a-x\right)-y\left(a-x\right)=\left(a^2-y\right)\left(a-x\right)\\ c,=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\\ d,=x\left(x-2y\right)+t\left(x-2y\right)=\left(x+t\right)\left(x-2y\right)\\ 2,\\ \Rightarrow x^2-4x+4-x^2+9=6\\ \Rightarrow-4x=-7\Rightarrow x=\dfrac{7}{4}\\ 3,\\ a,x^2+2x+2=\left(x+1\right)^2+1\ge1>0\\ b,-x^2+4x-5=-\left(x-2\right)^2-1\le-1< 0\)
a: \(x^2-4xy+4y^2-2x+4y-35\)
\(=\left(x^2-4xy+4y^2\right)-\left(2x-4y\right)-35\)
\(=\left(x-2y\right)^2-2\left(x-2y\right)-35\)
\(=\left(x-2y\right)^2-7\left(x-2y\right)+5\left(x-2y\right)-35\)
\(=\left(x-2y\right)\left(x-2y-7\right)+5\left(x-2y-7\right)\)
\(=\left(x-2y-7\right)\left(x-2y+5\right)\)
c: \(\left(xy+ab\right)^2+\left(ay-bx\right)^2\)
\(=x^2y^2+a^2b^2+2xyab+a^2y^2-2aybx+b^2x^2\)
\(=x^2y^2+a^2y^2+a^2b^2+b^2x^2\)
\(=y^2\left(x^2+a^2\right)+b^2\left(a^2+x^2\right)\)
\(=\left(x^2+a^2\right)\left(y^2+b^2\right)\)
Bài 4:
Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) x2-xy+5y-25
= x(2-y)+ 5(y-2)
= x(2-y)-5(2-y)
= (x-5)(2-y)
\(a,\dfrac{2x+2y}{a^2+2ab+b^2}.\dfrac{ax-ay+bx-by}{2x^2-2y^2}\)
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{a\left(x-y\right)+b\left(x-y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{\left(x-y\right)\left(a+b\right)}{2\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{1}{a+b}\)
\(b,\dfrac{a+b-c}{a^2+2ab+b^2-c^2}.\dfrac{a^2+2ab+b^2+ac+bc}{a^2-b^2}\)
\(=\dfrac{a+b-c}{\left(a+b\right)^2-c^2}.\dfrac{\left(a+b\right)^2+c\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{a+b-c}{\left(a+b-c\right)\left(a+b+c\right)}.\dfrac{\left(a+b\right)\left(a+b+c\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{1}{a-b}\)
\(c,\dfrac{x^3+1}{x^2+2x+1}.\dfrac{x^2-1}{2x^2-2x+2}\)
\(=\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)^2}.\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x^2-x+1\right)}\) \(=\dfrac{x-1}{2}\) \(d,\dfrac{x^8-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4\right)^2-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4-1\right)\left(x^4+1\right)}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{x+1}.\dfrac{1}{x^2+1}\) \(=\dfrac{\left(x-1\right)\left(x+1\right)}{x+1}\) \(=x-1\) \(e,\dfrac{x-y}{xy+y^2}-\dfrac{3x+y}{x^2-xy}.\dfrac{y-x}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x\left(x-y\right)}.\dfrac{-\left(x-y\right)}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x}.\dfrac{-1}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{-3x-y}{x\left(x+y\right)}\) \(=\dfrac{x\left(x-y\right)+y\left(3x+y\right)}{xy\left(x+y\right)}\) \(=\dfrac{x^2-xy+3xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{x^2+2xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{\left(x+y\right)^2}{xy\left(x+y\right)}=\dfrac{x+y}{xy}\)tìm giá trị của m để pt 2x-m=1-x nhận giá trị x=-2 là nghiệm
giải hộ e với :)
Phân tích đa thức thành nhân tử ( phối hợp các phương pháp )
1) x2 - ( a + b )xy + aby2
\(=x^2-axy-bxy+aby^2\)
\(=(x^2-axy)-(bxy+aby^2)\)
\(=x(x-ay)-by(x+ay)\)
\(=(x-ay)(x-by)\)
2) x2 + ( 2a + b )xy + 2aby2
=x2 + 2axy + bxy + 2aby2
=(x2+ bxy) +(2axy+ 2aby2 )
=x(x+ by) +2ay(x+ by)
=(x+ by)(x+2ay)