Câu hỏi của nguyễn bảo anh

Question

Bài toán 1: Phân tích thành nhân tử.

a) 20x – 5y                                                    e) 4x2y – 8xy2 + 10x2y2

b) 5x(x – 1) – 3x(x – 1)                                  g) 20x2y – 12x3...

@Nk>↑@ · Accepted Answer

1.a)$20x-5y=5\left(4x-y\right)$

b)$5x\left(x-1\right)-3x\left(x-1\right)=\left(5x-3x\right)\left(x-1\right)=2x\left(x-1\right)$

c)$x\left(x+y\right)-6x-6y=x\left(x+y\right)-6\left(x+y\right)=\left(x-6\right)\left(x+y\right)$

d)$6x^3-9x^2=3x^2\left(2x-3\right)$

e)$4x^2y-8xy^2+10x^2y^2=2xy\left(2x-8y+10xy\right)$

g)$20x^2y-12x^3=4x^2\left(5y-3x\right)$

h)$8x^4+12x^2y-16x^3y^4=4x^2\left(2x^2+12y-16xy^4\right)$Câu trả lời: 1.a)$20x-5y=5\left(4x-y\right)$

b)$5x\left(x-1\right)-3x\left(x-1\right)=\left(5x-3x\right)\left(x-1\right)=2x\left(x-1\right)$

c)$x\left(x+y\right)-6x-6y=x\left(x+y\right)-6\left(x+y\right)=\left(x-6\right)\left(x+y\right)$

d)$6x^3-9x^2=3x^2\left(2x-3\right)$

e)$4x^2y-8xy^2+10x^2y^2=2xy\left(2x-8y+10xy\right)$

g)$20x^2y-12x^3=4x^2\left(5y-3x\right)$

h)$8x^4+12x^2y-16x^3y^4=4x^2\left(2x^2+12y-16xy^4\right)$

@Nk>↑@ · Answer

2.a)$3x\left(x+1\right)-5y\left(x+1\right)=\left(3x-5y\right)\left(x+1\right)$

b)$3x\left(x-6\right)-2\left(x-6\right)=\left(3x-2\right)\left(x-6\right)$

c)$4y\left(x-1\right)-\left(1-x\right)=4y\left(x-1\right)+\left(x-1\right)=\left(4y+1\right)\left(x-1\right)$

d)$\left(x-3\right)^3+3-x=\left(x-3\right)^3-\left(x-3\right)=\left(x-3\right)\left[\left(x-3\right)^2-1\right]=\left(x-3\right)\left(x-2\right)\left(x-4\right)$

e)$7x\left(x-y\right)-\left(y-x\right)=7x\left(x-y\right)+\left(x-y\right)=\left(7x+1\right)\left(x-y\right)$

h)$3x^3\left(2y-3z\right)-15x\left(2y-3z\right)^2=3x\left(2y-3z\right)\left[x^2-5\left(2y-3z\right)\right]$

k)Sai đề: $3x\left(z+2\right)+5\left(-z-2\right)=3x\left(z+2\right)-5\left(z+2\right)=\left(3x-5\right)\left(z+2\right)$

l)$18x^2\left(3+x\right)+3\left(x+3\right)=3\left(x+3\right)\left(6x^2+1\right)$

m)$14x^2y-21xy^2+28x^2y^2=7xy\left(2x-3y+4xy\right)$

n)$10x\left(x-y\right)-8y\left(y-x\right)=10x\left(x-y\right)+8y\left(x-y\right)=2\left(5x+4y\right)\left(x-y\right)$Câu trả lời: 2.a)$3x\left(x+1\right)-5y\left(x+1\right)=\left(3x-5y\right)\left(x+1\right)$

b)$3x\left(x-6\right)-2\left(x-6\right)=\left(3x-2\right)\left(x-6\right)$

c)$4y\left(x-1\right)-\left(1-x\right)=4y\left(x-1\right)+\left(x-1\right)=\left(4y+1\right)\left(x-1\right)$

d)$\left(x-3\right)^3+3-x=\left(x-3\right)^3-\left(x-3\right)=\left(x-3\right)\left[\left(x-3\right)^2-1\right]=\left(x-3\right)\left(x-2\right)\left(x-4\right)$

e)$7x\left(x-y\right)-\left(y-x\right)=7x\left(x-y\right)+\left(x-y\right)=\left(7x+1\right)\left(x-y\right)$

h)$3x^3\left(2y-3z\right)-15x\left(2y-3z\right)^2=3x\left(2y-3z\right)\left[x^2-5\left(2y-3z\right)\right]$

k)Sai đề: $3x\left(z+2\right)+5\left(-z-2\right)=3x\left(z+2\right)-5\left(z+2\right)=\left(3x-5\right)\left(z+2\right)$

l)$18x^2\left(3+x\right)+3\left(x+3\right)=3\left(x+3\right)\left(6x^2+1\right)$

m)$14x^2y-21xy^2+28x^2y^2=7xy\left(2x-3y+4xy\right)$

n)$10x\left(x-y\right)-8y\left(y-x\right)=10x\left(x-y\right)+8y\left(x-y\right)=2\left(5x+4y\right)\left(x-y\right)$

@Nk>↑@ · Answer

3.a)$4x\left(x+1\right)=8\left(x+1\right)$

$\Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0$

$\Leftrightarrow4\left(x+1\right)\left(x-2\right)=0$

$\Rightarrow\left\{{}\begin{matrix}x+1=0\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\x=2\end{matrix}\right.$

b)$x\left(x-1\right)-2\left(1-x\right)=0$

$\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0$

$\Leftrightarrow\left(x+2\right)\left(x-1\right)=0$

$\Rightarrow\left\{{}\begin{matrix}x+2=0\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\x=1\end{matrix}\right.$

c)$2x\left(x-2\right)-\left(2-x\right)^2=0$

$\Leftrightarrow2x\left(x-2\right)-\left(x-2\right)^2=0$

$\Leftrightarrow\left(2x-x+2\right)\left(x-2\right)=0$

$\Rightarrow\left\{{}\begin{matrix}x+2=0\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\x=2\end{matrix}\right.$

d)$\left(x-3\right)^3+3-x=0$

$\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)=0$

$\Leftrightarrow\left(x-3\right)\left[\left(x-3\right)^2-1\right]=0$

$\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x-4\right)=0$

$\Rightarrow\left\{{}\begin{matrix}x-3=0\x-2=0\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\x=2\x=4\end{matrix}\right.$

e)$5x\left(x-2\right)-\left(2-x\right)=0$

$\Leftrightarrow5x\left(x-2\right)+\left(x-2\right)=0$

$\Leftrightarrow\left(5x+1\right)\left(x-2\right)=0$

$\Rightarrow\left\{{}\begin{matrix}5x+1=0\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{5}\x=2\end{matrix}\right.$

g)$5x\left(x-2000\right)-x+2000=0$

$\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0$

$\Leftrightarrow\left(5x-1\right)\left(x-2000\right)=0$

$\Rightarrow\left\{{}\begin{matrix}5x-1=0\x-2000=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{5}\x=2000\end{matrix}\right.$

h)$x^2-4x=0$

$\Leftrightarrow x\left(x-4\right)=0$

$\Rightarrow\left\{{}\begin{matrix}x=0\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\x=4\end{matrix}\right.$

k)$\left(1-x\right)^2-1+x=0$

$\Leftrightarrow\left(1-x\right)^2-\left(1-x\right)=0$

$\Leftrightarrow\left(1-x\right)\left(1-x-1\right)=0$

$\Rightarrow\left\{{}\begin{matrix}1-x=0\-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\x=0\end{matrix}\right.$

m)$x+6x^2=0$

$\Leftrightarrow x\left(1+6x\right)=0$

$\Rightarrow\left\{{}\begin{matrix}x=0\1+6x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\x=-\dfrac{1}{6}\end{matrix}\right.$

n)$\left(x+1\right)=\left(x+1\right)^2$

$\Leftrightarrow\left(x+1\right)-\left(x+1\right)^2=0$

$\Leftrightarrow\left(x+1-1\right)\left(x+1\right)=0$

$\Rightarrow\left\{{}\begin{matrix}x=0\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\x=-1\end{matrix}\right.$

Bạn nhớ ghi vậy x cần tìm là ... nhaCâu trả lời: 3.a)$4x\left(x+1\right)=8\left(x+1\right)$