Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(Q=x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)
\(Q=x^3+y^3+3xy\left(x+y\right)-2\left(x^2+y^2+2xy\right)+3\left(x+y\right)+10\)
\(Q=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10\)
Thay x + y = 5 vào ta có :
\(Q=5^3-2.5^2+3.5+10\)
\(Q=100\)
x11+x4+1
= x11+x10+x9-x10-x9-x8+x8+x7+x6-x7-x6-x5+x5+x4+x3-x3-x2-x+x2+x+1
= x9(x2+x+1)-x8(x2+x+1)+x6(x2+x+1)-x5(x2+x+1)+x3(x2+x+1)-x(x2+x+1)+(x2+x+1)
= (x2+x+1)(x9-x8+x6-x5+x3-x+1)
\(x^2-3x+1\)
\(=x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{5}{4}\)
\(=\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\)
\(=\left(x-\frac{3}{2}-\frac{\sqrt{5}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{5}}{2}\right)\)
a) \(7x^2-28=0\Leftrightarrow7\left(x^2-4\right)=0\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) vậy \(x=2;x=-2\)
b) \(\left(2x+1\right)+x\left(2x+1\right)=0\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\2x=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=\dfrac{-1}{2}\end{matrix}\right.\) vậy \(x=-1;x=\dfrac{-1}{2}\)
c) \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-5=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\) vậy \(x=0;x=5;x=-5\)
d) \(9\left(3x-2\right)=x\left(2-3x\right)\Leftrightarrow9\left(3x-2\right)=-x\left(3x-2\right)\)
\(\Leftrightarrow9\left(3x-2\right)+x\left(3x-2\right)=0\Leftrightarrow\left(9+x\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}9+x=0\\3x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\3x=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x=-9;x=\dfrac{2}{3}\)
e) \(5x\left(x-3\right)-2x+6=0\Leftrightarrow5x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(5x-2\right)\left(x-3\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}5x-2=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=2\\x=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\x=3\end{matrix}\right.\) vậy \(x=\dfrac{2}{5};x=3\)
Đặt
\(A=x^4-4x^3+8x+3\)
Giả sử
\(A=\left(x^2+ax+b\right)\left(x^2+cx+d\right)\)
\(=x^4+cx^3+dx^2+ax^3+acx^2+adx+bx^2+bcx+bd\)
\(=x^4+\left(a+c\right)x^3+\left(b+ac+d\right)x^2+\left(ad+bc\right)x+bd\)
\(\left[\begin{array}{nghiempt}a+c=-4\\b+ac+d=0\\ad+bc=8\\bd=3\end{array}\right.\)
\(\left[\begin{array}{nghiempt}a=-2\\b=-3\\c=-2\\d=-1\end{array}\right.\)
\(A=\left(x^2-2x-3\right)\left(x^2-2x-1\right)\)
dài dòng
\(x^4-4x^3+8x+3=x^4-2x^3-2x^3-x^2+4x^2-3x^2+2x+6x+3\)
\(=\left(x^4-2x^3-x^2\right)-\left(2x^3-4x^2-2x\right)-\left(3x^2-6x-3\right)\)
\(=x^2\left(x^2-2x-1\right)-2x\left(x^2-2x-1\right)-3\left(x^2-2x-1\right)\)
\(=\left(x^2-2x-1\right)\left(x^2-2x-3\right)\)
1
a . X.(X-2)-Y.(X-2)=(X-Y).(X-2)
b .(X2 +1+2X).(X2 +1-2X)
2
3X2 +2X+X2 +2X+1-4X2 -10X+10X+5=(-12)
4X+6= -12
X=9/2
1. a, x2-2x+2y-xy = x(x-2)+y(2-y) = x(x-2)-y(x-2) = (x-y)(x-2)
b, (x2+1)2-4x2 = (x2+1-2x)(x2+1+2x) = (x-1)2(x+1)2
2. x(3x+2)+(x+1)2-(2x-5)(2x+5) = -12
=> (3x2+2x+x2+2x+1)-(2x)2-52 = -12
=> 3x2+2x+x2+2x+1-4x2-25 = -12
=> 4x-24 = -12 => 4x = 12 => x = 3
\(x^4+2014x^2+2013x+2014\)
\(=x^4+2014x^2+2014x-x+2014\)
\(=\left(x^4-x\right)+\left(2014x^2+2014x+2014\right)\)
\(=x\left(x^3-1\right)+2014\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2014\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2014\right)\)
b)\(x^8+7x^4+6\)
\(=x^8+x^4+6x^4+6\)
\(=x^4\left(x^4+1\right)+6\left(x^4+1\right)\)
\(=\left(x^4+1\right)\left(x^4+6\right)\)
b) \(x^8+7x^4+16\)
\(=\left(x^8+8x^4+16\right)-x^4\)
\(=\left[\left(x^4\right)^2+2.x^4.4+4^2\right]-x^4\)
\(=\left(x^4+4\right)^2-\left(x^2\right)^2\)
\(=\left(x^4+4-x^2\right)\left(x^4+4+x^2\right)\)
a)\(x^4-7x^2+1\)
\(=x^4+2x^2+1-9x^2\)
\(=\left(x^2+1\right)^2-\left(3x\right)^2\)
\(=\left(x^2+1-3x\right)\left(x^2+1+3x\right)\)
b)\(4x^4-12x^2+1\)
\(=4x^4+4x^2+1-16x^2\)
\(=\left(2x^2+1\right)^2-\left(4x\right)^2\)
\(=\left(2x^2+1+4x\right)\left(2x^2+1-4x\right)\)