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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: \(A=\left(2x-5\right)^2-4x\left(x-5\right)\)
\(=4x^2-20x+25-4x^2+20x\)
=25
b: \(B=\left(4-3x\right)\left(4+3x\right)+\left(3x+1\right)^2\)
\(=16-9x^2+9x^2+6x+1\)
=6x+17
c: \(C=\left(x+1\right)^3-x\left(x^2+3x+3\right)\)
\(=x^3+3x^2+3x+1-x^3-3x^2-3x\)
=1
d: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)
\(=\left(2021x-2020-2020x+2021\right)^2\)
\(=\left(x+1\right)^2\)
\(=x^2+2x+1\)
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.\(x^3-x=0 \)
\(x(x^2-1)=0\)
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
\(1,\left(x+2022\right)\left(x-1\right)=x^2+2021x-2022\left(B\right)\\ 2,\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\left(A\right)\)
Sao đề bài... nó khó hiểu quá!