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Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\left(x-1\right)^2-3x\left(x-5\right)=1\)
\(\Leftrightarrow\)\(3x^2-6x+3-3x^2+15x=1\)
\(\Leftrightarrow\)\(9x=-2\)
\(\Leftrightarrow\)\(x=-\frac{2}{9}\)
Vậy...
\(x^2-2x+1=25\)
\(\Leftrightarrow\)\(x^2-2x-24=0\)
\(\Leftrightarrow\)\(\left(x+4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+4=0\\x-6=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-4\\x=6\end{cases}}\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-2x+1=25\)
\(\Leftrightarrow\)\(\left(x+1\right)^2=25\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=5\\x+1=-5\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=4\\x=-6\end{cases}}\)
Vậy...
\(3\left(x-1\right)^2-3x\left(x-5\right)=1\)
\(\Leftrightarrow\)\(3\left(x^2-2x+1\right)-\left(3x^2-15x\right)=1\)
\(\Leftrightarrow\)\(3x^2-6x+3-3x^2+15x=1\)
\(\Leftrightarrow\)\(9x=-2\)
\(\Leftrightarrow\)\(x=-\frac{2}{9}\)
Vậy....
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
mk làm lun nha
(4x^2-4x+1)-(2x-3x^2+2-3x)=(9+6x+x^2)-3x+15
4x^2-4x+1-2x+3x^2-2+3x=9+6x+x^2-3x+15
CHUYỂN VẾ,TA ĐƯỢC:
6x^2=25
x^2=25/6
X=5/căn 6
**** CHO MK NHA