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a: Ta có: \(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
b: Ta có: \(\left(x-1\right)\cdot x-2\left(1-x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
c: Ta có: \(x^3+2x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
d: Ta có: \(x^3-3x^2=0\)
\(\Leftrightarrow x^2\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
a.\(x^3-x=0 \)
\(x(x^2-1)=0\)
x=0 hay x2-1=0
x=0 hay x2=1
x=0 hay x=1
Vậy x=0 hay x=1
b.\(x^3+1=0\)
\(x(x^2+1)=0\)
\(x=0 hay x^2+1=0\)
\(x=0 hay x^2=-1\)(vô lí vì x2≥0)
Vậy x=0
c.\(x^2-4x=0\)
\(x(x-4)=0\)
x=0 hay x-4=0
x=0 hay x=4
Vậy x=0 hay x=4
d.\(x(x-1)-2(1-x)=0\)
\(x(x-1)+2(x-1)=0 \)
\((x-1)(x+2)=0\)
x-1=0 hay x+2=0
x=1 hay x=-2
Vậy x=1 hay x=-2
e.\(2x(x-2)-(2-x)^2=0\)
\(2x(x-2)+(x-2)^2=0\)
\((x-2)(2x+x-2)=0\)
\((x-2)(3x-2)=0\)
x-2=0 hay 3x-2=0
x=2 hay 3x=2
x=2 hay x=2/3
Vậy x=2 hay x=2/3
f.\(4x(x+1)=8(x+1)\)
\(4x(x+1)-8(x+1)=0\)
\(4(x+1)(x-2)=0\)
4(x+1)=0 hay x-2=0
x+1=0 hay x=2
x=-1 hay x=2
Vậy x=-1 hay x=2
g.\(5x(x-2)-x+2=0\)
\(5x(x-2)-(x-2)=0\)
\((x-2)(5x-1)=0\)
x-2=0 hay 5x-1=0
x=2 hay 5x=1
x=2 hay x=1/5
Vậy x=2 hay x=1/5
h.\((x+1)=(x+1)^2\)
\((x+1)-(x+1)^2=0\)
\((x+1)(1-x-1)=0\)
\((x+1)(-x)=0\)
x+1= 0 hay -x=0
x=-1 hay x=0
Vậy x=-1 hay x=0
a.16x-5x2-3 = - ( 5x2-16x+3) = -( 5x2-15x-x+3)= -[ 5x(x-3)-(x-3)] = -(5x-1)(x-3)
b.x^3-x+3x^2y+3xy^2+y^3-y = \(\left(x^3+3x^2y+3xy^2+y^3\right)-\)\(\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)=\)\(\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
c.x^4+8x = \(x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\)
d.x^2+x-6 = \(x^2+3x-2x-6=x\left(x+3\right)-2\left(x+3\right)\)
\(=\left(x+3\right)\left(x-2\right)\)
e.5x^2-10xy+5y^2-20z^2\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y+2z\right)\left(x-y-2z\right)\)
f.2(x^5)-x^2-5x ( mik ko bik làm)
g.x^3-3x^2-4x+12 = \(x^2\left(x-3\right)-4\left(x-3\right)=\left(x^2-2^2\right)\left(x-3\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
h.x^4-5x^2+4 \(=\left(x^2\right)^2-4x^2+4-x^2\)
\(=\left(x^2-2\right)-x^2=\left(x^2-2+x\right)\left(x^2-2-x\right)\)
để pt được xác định thì :
\(x-2\ne0;x^2-1\ne0\)
=>\(\left\{{}\begin{matrix}x\ne2\\x\ne-1\\x\ne1\end{matrix}\right.\)
Vậy chọn B
a) \(\left(x-9\right)\left(x-7\right)+1\)
\(=x^2-16x+63+1\)
\(=x^2-16x+64\)
\(=\left(x-8\right)^2\)
b) \(x^3+2x^2-3x-6\)
\(=x^2\left(x+2\right)-3x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-3x\right)\)
\(=x\left(x+2\right)\left(x-3\right)\)
c) \(x^2-y^2+xz-yz\)
\(=x\left(x+z\right)-y\left(y+z\right)\)
\(=\left(x-y\right)\left(y+z\right)\)
d) \(x^3-x+3x^2y+y^3-y\)
botay:(
Sao đề bài... nó khó hiểu quá!