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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. 4-32x3
= 4.(1-8x3)
= 4.[13-(2x)3 ]
= 4.(1-2x).(1+2x+4x2)
2. b. \(\left(\frac{x}{xy-y^2}-\frac{2x-y}{xy-x^2}\right):\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left[\frac{x}{y\left(x-y\right)}+\frac{2x-y}{x\left(x-y\right)}\right]:\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left[\frac{x.x}{y\left(x-y\right).x}+\frac{\left(2x-y\right).y}{x\left(x-y\right).y}\right]:\left(\frac{x+y}{xy}\right)\)
\(=\left[\frac{x^2+2xy-y^2}{xy\left(x-y\right)}\right]:\left(\frac{x+y}{xy}\right)\)
\(=\left[\frac{-\left(x-y\right)^2}{xy\left(x-y\right)}\right].\frac{xy}{x+y}\)
\(=\frac{-\left(x-y\right)}{xy}.\frac{xy}{x+y}\)
\(=\frac{y-x}{x+y}\)
1: Sửa đề: 3x-5
\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)
2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
=5x^2+14x^2+12x+8
3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)
5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)
1: \(=\left(x-1\right)^2\)
2: \(x\in\left\{0;20\right\}\)
Câu 13:
\(1,=\left(x-1\right)^2\\ 2,\Leftrightarrow x\left(x-20\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=20\end{matrix}\right.\\ 3,\text{Đề lỗi}\)
Câu 14:
\(1,ĐK:x\ne-2\\ 2,=\dfrac{\left(x+2\right)^2}{x+2}=x+2\\ 3,\Leftrightarrow x+2=0\Leftrightarrow x=-2\left(ktm\right)\Leftrightarrow x\in\varnothing\)
Câu 16:
\(A=x^2-4x+4+20=\left(x-2\right)^2+20\ge20\)
Dấu \("="\Leftrightarrow x=2\)
\(1.\) Phân tích đa thức thành nhân tử
\(4-32x^3=-\left(32x^3-4\right)=\left(-4\right)\left(8x^3-1\right)=\left(-4\right)\left(2x-1\right)\left(4x^2+2x+1\right)\)
\(2.\) Thực hiện phép tính
Ta có: \(-x^2+6x^3-26x+21=6x^3-x^2-26x+21=\left(x-1\right)\left(2x-3\right)\left(3x+7\right)\)
Do đó:
\(\frac{-x^2+6x^3-26x+21}{2x-3}=\frac{\left(x-1\right)\left(2x-3\right)\left(3x+7\right)}{2x-3}=\left(x-1\right)\left(3x+7\right)=3x^2+4x-7\)