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2:
a: \(9x^2-1=\left(3x\right)^2-1=\left(3x-1\right)\left(3x+1\right)\)
b: \(2\left(x-1\right)+x^2-x\)
\(=2\left(x-1\right)+x\left(x-1\right)\)
\(=\left(x-1\right)\left(x+2\right)\)
c: \(3x^2+14x-5\)
\(=3x^2+15x-x-5\)
\(=3x\left(x+5\right)-\left(x+5\right)=\left(x+5\right)\left(3x-1\right)\)
3:
a: \(2x\left(x-1\right)-2x^2=4\)
=>\(2x^2-2x-2x^2=4\)
=>-2x=4
=>x=-2
b: \(x\left(x-3\right)-\left(x+2\right)\left(x-1\right)=5\)
=>\(x^2-3x-\left(x^2+x-2\right)=5\)
=>\(x^2-3x-x^2-x+2=5\)
=>-4x=3
=>x=-3/4
c: \(4x^2-25+\left(2x+5\right)^2=0\)
=>\(\left(2x-5\right)\left(2x+5\right)+\left(2x+5\right)^2=0\)
=>\(\left(2x+5\right)\left(2x-5+2x+5\right)=0\)
=>4x(2x+5)=0
=>\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
\(a,6x^2-15x^3y\\ b,=-\dfrac{2}{3}x^2y^3+\dfrac{2}{3}x^4y-\dfrac{8}{3}xy\)
Bài 2:
\(a,=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\\ b,=3x^2-6x-5x+5x^2-8x^2+24=24-11x\\ c,=x^5+x^3-2x^3-2x=x^5-x^3-2x\)
Bài 2:
\(\Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
Câu 1:
a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x
b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)
\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)
\(=2x^2+6x+17\)
c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)
Bài 1:
a: \(=6x^3-10x^2+6x\)
b: \(=-2x^3-10x^2-6x\)
Bài 4:
a: =>3x+10-2x=0
=>x=-10
c: =>3x2-3x2+6x=36
=>6x=36
hay x=6
Bài 1:
\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)
Bài 4:
\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)
Bài 1:
\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)
\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Câu 1
a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)
b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)
a) \(\left(x^5+4x^3-6x^2\right):4x^2\)
\(=\left(x^5:4x^2\right)+\left(4x^3:4x^2\right)+\left(-6x^2:4x^2\right)\)
\(=\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)
b)
Vậy \(\left(x^3+x^2-12\right):\left(x-2\right)=x^2+3x+6\)
c) (-2x5 : 2x2) + (3x2 : 2x2) + (-4x^3 : 2x^2)
= \(-x^3+\dfrac{3}{2}-2x\)
d) \(\left(x^3-64\right):\left(x^2+4x+16\right)\)
\(=\left(x-4\right)\left(x^2+4x+16\right):\left(x^2+4x+16\right)\)
\(=x-4\)
(dùng hẳng đẳng thức thứ 7)
Bài 2 :
a) 3x(x - 2) - 5x(1 - x) - 8(x2 - 3)
= 3x2 - 6x - 5x + 5x2 - 8x2 + 24
= (3x2 + 5x2 - 8x2) + (-6x - 5x) + 24
= -11x + 24
b) (x - y)(x2 + xy + y2) + 2y3
= x3 - y3 + 2y3
= x3 + y3
c) (x - y)2 + (x + y)2 - 2(x - y)(x + y)
= (x - y)2 - 2(x - y)(x + y) + (x + y)2
= [(x - y) + x + y)2 = [x - y + x + y] = (2x)2 = 4x2
Bài 1 :
a]= \(\frac{1}{4}\)x3 + x - \(\frac{3}{2}\).
b] => [x3 + x2 -12 ] = [ x2 +3 ][x-2] + [-6]
c]= -x3 -2x +\(\frac{3}{2}\).
d] = [ x3 - 64 ] = [ x2 + 4x + 16][ x- 4].
Bài 2:
a) \(x^2+y^2-9-2xy\)
\(=\left(x^2-2xy+y^2\right)-3^2\)
\(=\left(x-y\right)^2-3^2\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
b) \(4x^2-5x-9\)
\(=4x^2+4x-9x-9\)
\(=4x\left(x+1\right)-9\left(x+1\right)\)
\(=\left(x+1\right)\left(4x-9\right)\)
\(\left(2x-3\right)^2-\left(4x-1\right)\left(x+2\right)=4x^2-12x+9-4x^2-7x+2=-19x+11\)
\(\left(3x+2\right)\left(3x-2\right)-\left(3x-1\right)^2=9x^2-4-9x^2+6x-1=6x-5\)
\(x^2+y^2-9-2xy=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\)
\(4x^2-5x-9=\left(4x-9\right)\left(x+1\right)\)
\(\left(x-3\right)^2-\left(x-1\right)\left(x-2\right)=5\Leftrightarrow x^2-6x+9-x^2+3x-2=5\)
\(\Leftrightarrow-3x=-2\Leftrightarrow x=x=\frac{2}{3}\)
\(3x^2+5x-8=0\Leftrightarrow\left(x-1\right)\left(3x+8\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{8}{3}\end{cases}}\)