(x+6)2 - (x-6)(x+6) = 0
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![](https://rs.olm.vn/images/avt/0.png?1311)
6 x 1 = 6 6 x 9 = 54
6 x 2 = 12 6 x 8 = 48
6 x 3 = 18 6 x 7 = 42
6 x 4 = 24 0 x 6 = 0
6 x 5 = 30 6 x 0 = 0
6 x 6 = 36 6 x 10 = 60
![](https://rs.olm.vn/images/avt/0.png?1311)
6 x 4 =24
6 x 1 =6
6 x 9 =54
6 x 10 =60
6 x 6 =36
6 x 3 =18
6 x 2 =12
0 x 6 =0
6 x 8 =48
6 x 5 =30
6 x 7 =42
6 x 0 =0
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 6 x 2 = 12 6 x 3 = 18 6 x 8 = 48 6 x 1 = 6
6 x 4 = 24 6 x 5 = 30 6 x 9 = 54 1 x 6 = 6
6 x 6 = 36 6 x 7 = 42 6 x 10 = 60 0 x 6 = 0
b) 6 x 5 = 30 6 x 4 = 24 3 x 6 = 18 2 x 6 = 12
5 x 6 =30 4 x 6 = 24 6 x 3 = 18 6 x 2 =12
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
`x^2 +5x+6=0`
`<=> x^2 + 3x +2x+6=0`
`<=> x(x+3)+2(x+3)=0`
`<=> (x+3)(x+2)=0`
`<=> x+3=0 hoặcx+2=0`
`<=> x=-3 hoặc x=-2`
b)
`x^2 -7x+6=0`
`<=> x^2 -6x-x+6=0`
`<=> x(x-6)-(x-6)=0`
`<=> (x-6)(x-1)=0`
`<=> x-6=0 hoặc x-1=0 `
`<=> x=6 hoặc x=1`
c)
`x^2 +x -12=0`
`<=> x^2 +4x-3x-12=0`
`<=> x(x+4)-3(x+4)=0`
`<=> (x+4)(x-3)=0`
`<=> x+4=0 hoặc x-3=0`
`<=> x=-4 hoặc x=3`
d)
`x^2 -x-6=0`
`<=>x^2 -3x+2x-6=0`
`<=> x(x-3)+2(x-3)=0`
`<=> (x-3)(x+2)=0`
`<=> x-3=0 hoặc x+2=0`
`<=> x=3 hoặc x=-2`
e)
`2x^2 -3x-5=0`
`<=> 2x^2 -5x+2x-5=0`
`<=> x(2x-5)+(2x-5)=0`
`<=> (2x-5)(x+1)=0`
`<=> 2x-5=0 hoặc x+1=0`
`<=> x=5/2 hoặc x=-1`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\)
\(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(2,\)
\(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
\(3,\)
\(x^4-x^2=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(4,\)
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(5,\)
\(x\left(x+6\right)-10\left(x-6\right)=0\)
\(\Leftrightarrow x^2+6x-10x+60=0\)
\(\Leftrightarrow x^2-4x+60=0\)
\(\Leftrightarrow x^2-4x+4+56=0\)
\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)
=> Phương trình vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
2) \(x^3-6x^2+11x-6=0\)
\(\Leftrightarrow\)\(x^3-3x^2-3x^2+9x+2x-6=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(x-2\right)\left(x-1\right)=0\)
bn giải tiếp nha
3) \(x^3-4x^2+x+6=0\)
\(\Leftrightarrow\)\(x^3-3x^2-x^2+3x-2x+6=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(x^2-x-2\right)=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(x-2\right)\left(x+1\right)=0\)
lm tiếp nha
4) \(x^3-3x^2+4=0\)
\(\Leftrightarrow\)\(x^3+x^2-4x^2-4x+4x+4=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\)\( \left(x+1\right)\left(x-2\right)^2=0\)
lm tiếp nha
Mk làm mẫu 1 bài cho nha !
1. <=> (x^3-x^2)+(5x^2-5x)+(6x-6) = 0
<=> (x-1).(x^2+5x+6) = 0
<=> (x-1).[(x^2+2x)+(3x+6)] = 0
<=> (x-1).(x+2).(x+3) = 0
<=> x-1=0 hoặc x+2=0 hoặc x+3=0
<=> x=1 hoặc x=-2 hoặc x=-3
Vậy ..............
Tk mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(x^2+x-2=0\)
\(\Leftrightarrow x^2+2x-x-2=0\)
\(\Leftrightarrow x\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)
2) \(x^2+2x-3=0\)
\(\Leftrightarrow x^2+3x-x-3=0\)
\(\Leftrightarrow x\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}\)
3) \(x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
4) \(x^2+x-6=0\)
\(\Leftrightarrow x^2+3x-2x-6=0\)
\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
5) \(2x^2-x-6=0\)
\(\Leftrightarrow2x^2-4x+3x-6=0\)
\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{3}{2}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Answer:
\(3x^2-4x=0\)
\(\Rightarrow x\left(3x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)
\(\left(x^2-5x\right)+x-5=0\)
\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
\(x^2-5x+6=0\)
\(\Rightarrow x^2-2x-3x+6=0\)
\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)
\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
\(5x\left(x-3\right)-x+3=0\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)
\(x^2-2x+5=0\)
\(\Rightarrow\left(x^2-2x+1\right)+4=0\)
\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)
Vậy không có giá trị \(x\) thoả mãn
\(x^2+x-6=0\)
\(\Rightarrow x^2+3x-2x-6=0\)
\(\Rightarrow x.\left(x+3\right)-2\left(x+3\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) (x+6)(3x-1)+x+6=0
⇔(x+6)(3x-1)+(x+6)=0
⇔(x+6)(3x-1+1)=0
⇔3x(x+6)=0
2) (x+4)(5x+9)-x-4=0
⇔(x+4)(5x+9)-(x+4)=0
⇔(x+4)(5x+9-1)=0
⇔(x+4)(5x+8)=0
3)(1-x)(5x+3)÷(3x-7)(x-1)
=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(X+6)2-(x-6)(x+6)=0
=>x2+12x+36-(x2-36)=0
=>x2+12x+36-x2+36=0
=>12x+72=0
=>12x=-72
=>x=-6
Nhớ bấm tick cho mà!!
Lời giải:
\((x+6)^2-(x-6)(x+6)=0\)
\(\Leftrightarrow (x+6)[(x+6)-(x-6)]=0\)
\(\Leftrightarrow 12(x+6)=0\)
\(\Leftrightarrow x+6=0\Rightarrow x=-6\)
(x+6)2-(x-6)(x+6)=0
=> x2+12x+36-(x2-36)=0
=>x2+12x+36-x2+36=0
=>12x+72=0
=>12x=-72
=>-6