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\(a,\)\(x^4-4x^3+4x^2=0\)
\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x^2.\left(x^2-2.x.2+2^2\right)=0\)
\(\Leftrightarrow x^2.\left(x-2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b,\)\(x^2+5x+4=0\)
\(\Leftrightarrow x^2+x+4x+4=0\)
\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
\(c,\)\(9x-6x^2-3=0\)
\(\Leftrightarrow-3.\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow2x^2-2x-x+1=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right).\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
\(d,\)\(2x^2+5x+2=0\)
\(\Leftrightarrow2x^2+4x+x+2=0\)
\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\)
a) Ta có: \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{\left(2x+1\right)^2\cdot3}{15}-\frac{5\left(x-1\right)^2}{15}-\frac{7x^2-14x-5}{15}=0\)
\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)-7x^2+14x+5=0\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
\(\Leftrightarrow36x+3=0\)
\(\Leftrightarrow36x=-3\)
\(\Leftrightarrow x=\frac{-3}{36}\)
Vậy: \(x=\frac{-3}{36}\)
b) Ta có: \(\frac{201-x}{99}+\frac{203-x}{97}=\frac{205-x}{95}+3=0\)
\(\Leftrightarrow\frac{201-x}{99}+\frac{203-x}{97}-\frac{205-x}{95}-3=0\)
\(\Leftrightarrow\left(\frac{201-x}{99}+1\right)+\left(\frac{203-x}{97}+1\right)+\left(\frac{205-x}{95}+1\right)=0\)
\(\Leftrightarrow\frac{201-x+99}{99}+\frac{203-x+97}{97}+\frac{205-x+95}{95}=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\ne0\)
nên 300-x=0
\(\Leftrightarrow x=300\)
Vậy: x=300
c) Ta có: \(x^3+x^2+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+1\ge1\ne0\forall x\)(2)
Từ (1) và (2) suy ra x+1=0
hay x=-1
Vậy: x=-1
d) Ta có: \(\left(x-1\right)x\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
Đặt \(x^2+x-1=t\)
\(\Leftrightarrow\left(t+1\right)\left(t-1\right)=24\)
\(\Leftrightarrow t^2-1-24=0\)
\(\Leftrightarrow t^2-25=0\)
\(\Leftrightarrow\left(t-5\right)\left(t+5\right)=0\)
\(\Leftrightarrow\left(x^2+x-1-5\right)\left(x^2+x-1+5\right)=0\)
\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left(x^2+3x-2x-6\right)\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{15}{4}\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\right]\)(3)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}\ne0\forall x\)(4)
Từ (3) và (4) suy ra
\(\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
e) Ta có: \(\left(5x-3\right)-\left(4x-7\right)=0\)
\(\Leftrightarrow5x-3-4x+7=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy: x=-4
f) Ta có: \(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{1}{3}\right\}\)
g) Ta có: \(x^2+6x-16=0\)
\(\Leftrightarrow x^2-2x+8x-16=0\)
\(\Leftrightarrow x\left(x-2\right)+8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-8\right\}\)
h) Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;2\right\}\)
i) Ta có: \(x^2+x-2=0\)
\(\Leftrightarrow x^2-x+2x-2=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;-2\right\}\)
k) Ta có: \(3x^2+7x+2=0\)
\(\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{-1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;\frac{-1}{3}\right\}\)
l) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-2x-10x+5=0\)
\(\Leftrightarrow2x\left(2x-1\right)-5\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{5}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)
f) \(4x^2-12x+9=0\)
<=> \(\left(2x-3\right)^2\) = 0
<=> \(2x-3=0\)
<=> \(2x=3\) <=> \(x=\dfrac{3}{2}\)
Vậy ...............
g) \(3x^2+7x+2=0\)
<=> \(\left(3x^2+6x\right)+\left(x+2\right)=0\)
<=> \(3x\left(x+2\right)+\left(x+2\right)=0\)
<=> \(\left(x+2\right)\left(3x+1\right)=0\)
<=> \(\left[{}\begin{matrix}x=-2\\x=\dfrac{-1}{3}\end{matrix}\right.\)
Vậy ........................
h) \(x^2-4x+1=0\)
<=> \(\left(x^2-4x+4\right)-3=0\)
<=> \(\left(x-2\right)^2=3\)
<=> \(\left[{}\begin{matrix}x+2=\sqrt{3}\\x+2=-\sqrt{3}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{3}-2\\x=-\sqrt{3}-2\end{matrix}\right.\)
Vậy .........................
i) \(2x^2-6x+1=0\)
<=> \(2\left(x^2-3x+2,25\right)-3,5=0\)
<=> \(\left(x-1,5\right)^2=1,75\)
<=> \(\left[{}\begin{matrix}x-1,5=\sqrt{1,75}\\x-1,5=-\sqrt{1,75}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{1,75}+1,5\\x=-\sqrt{1,75}+1,5\end{matrix}\right.\)
Vậy ...................
j) \(3x^2+4x-4=0\)
<=> \(\left(3x^2+6x\right)-\left(2x+4\right)=0\)
<=> \(3x\left(x+2\right)-2\left(x+2\right)\) = 0
<=> \(\left(x+2\right)\left(3x-2\right)=0\)
<=> \(\left[{}\begin{matrix}x=-2\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ....................................
f) \(4x^2-12x+9=0\)
\(\Rightarrow\left(2x-3\right)^2=0\)
\(\Rightarrow2x-3=0\)
\(\Rightarrow x=\dfrac{3}{2}\)
Vậy..
g) \(3x^2+7x+2=0\)
\(\Rightarrow3x^2+6x+x+2=0\)
\(\Rightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\3x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1}{3}\end{matrix}\right.\)
Vậy..
h) \(x^2-4x+1=0\)
\(\Rightarrow x^2-4x+4-3=0\)
\(\Rightarrow\left(x-2\right)^2-3=0\)
\(\Rightarrow\left(x-2-\sqrt{3}\right)\left(x-2+\sqrt{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2-\sqrt{3}=0\\x-2+\sqrt{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\)
Vậy..
j) \(3x^2+4x-4=0\)
\(\Rightarrow3x^2+6x-2x-4=0\)
\(\Rightarrow3x\left(x+2\right)-2\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(3x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy..
1,x^2-(x+1)(x-1)=0
x^2-x^2+1+0
1=0(vô lý)
2,5x^3+3x^2+3x+1=4x^2
x^3+3x^2+3x+1=0
(x+1)=0
x=-1
3,x^3+x^2=0
x^2(x+1)=0
x=0 or x=-1
4,2x^3-12x^2+18x=0
x^3-6x^2+9x=0
x(x^2-6x+9)=0
x(x-3)^2=0
x=0 or x=3
5,5x^2-4(x^2-2x+1)+20=0
5x^2-4x^2+8x-4+20=0
x^2+8x+16=0
(x+4)^2=0
x=-4
6,5x(x-3)+7x-21=0
5x(x-3)+7(x-3)=0
(5x+7)(x-3)=0
5x-7=0 or x-3=0
x=7/5 or x=3
7,2x^3-50x=0
2x(x^2-25)=0
2x(x-5)(x+5)=0
x=0 or x=5 or x=-5
8,(4x-1)^2-9(x+3)^2=0
(4x-1)^2-3^2*(x+3)^2=0
(4x-1)^2-(3x+9)^2=0
(4x-1-3x-9)(4x-1+3x+9)=0
(x-10)(7x+8)=0
x=10 or x=-8/7
9,3(x-2)^2-x+2=0
3*(x-2)*(x-2)-(x-2)=0
(3x-6)(x-2)-(x-2)=0
(x-2)(3x-6-1)=0
(x-2)(3x-7)=0
x=2 or x=7/3
10,9x^2+6x-8=0
9x^2+12x-6x-8=0
3x(3x-2)+4(3x-2)=0
(3x+4)(3x-2)=0
3x+4=0 or 3x-2=0
x=-4/3 or x=2/3
Dài dữ trời :V Về sau gửi từng bài một thôi, nhìn hoa mắt quá @@
B1: Phân tích thành nhân tử:
a) \(6x^2+9x=3x\left(2x+3\right)\)
b) \(4x^2+8x=4x\left(x+2\right)\)
c) \(5x^2+10x=5x\left(x+2\right)\)
d) \(2x^2-8x=2x\left(x-4\right)\)
e) \(5x-15y=5\left(x-3y\right)\)
f) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
g) \(x^2-2x+1-4y^2=\left(x-1\right)^2-4y^2\)
\(=\left(x-1-2y\right)\left(x-1+2y\right)\)
h) \(x^2-100=\left(x-10\right)\left(x+10\right)\)
i) \(9x^2-18x+9=\left(3x-3\right)^2\)
k) \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)
l) \(x^2+6xy^2+9y^4=\left(x+3y\right)^2\)
m) \(4xy-4x^2-y^2=-\left(4x^2-4xy+y^2\right)\)
\(=-\left(2x-y\right)^2\)
n) \(\left(x-15\right)^2-16=\left(x-15-16\right)\left(x-15+16\right)\)
\(=\left(x-31\right)\left(x+1\right)\)
o) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3+x\right)\)
\(=\left(2+x\right)\left(8+x\right)\)
p) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)
\(=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)
\(=\left(5x-5\right)\left(9x-3\right)\)
Bài 1 :
a ) \(6x^2+9x=3x\left(x+3\right)\)
b ) \(4x^2+8x=4x\left(x+2\right)\)
c ) \(5x^2+10x=5x\left(x+2\right)\)
d ) \(2x^2-8x=2x\left(x-4\right)\)
e ) \(5x-15y=5\left(x-3y\right)\)
f ) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
g ) \(x^2-2x+1-4y^2=\left(x-1\right)^2-\left(2y\right)^2=\left(x-1-2y\right)\left(x-1+2y\right)\)
h ) \(x^2-100=x^2-10^2=\left(x-10\right)\left(x+10\right)\)
i ) \(9x^2-18x+9=\left(3x-3\right)^2\)
k ) \(x^3-8=\left(x-2\right)\left(x^2+2x+2^2\right)\)
l ) \(x^2+6xy^2+9y^4=\left(x+3y^2\right)^2\)
m ) \(4xy-4x^2-y^2=-\left(2x-y\right)^2\)
n ) \(\left(x-15\right)^2=x^2-30x+15^2\)
o ) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)
p ) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)\)
Bài 2 :
a ) \(3x^3-6x^2+3x^2y-6xy=3x\left(x^2-2x+xy-2y\right)\)
b ) \(x^2-2x+xy-2y=x\left(x-2\right)+y\left(x-2\right)=\left(x-2\right)\left(x+y\right)\)
c ) \(2x+x^2-2y-2xy=......................\)
d ) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
e ) \(x^2+y^2-2xy-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)
f )\(2xy-x^2-y^2+9=-\left(x-y\right)^2+9=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)
câu a, b, c dễ mà. Bạn áp dụng 7 hằng đẳng thúc là làm đc thoii!!
vd: a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=\left(3x+2\right)\left(x-1\right)\left(x+1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)[\left(3x-2\right)-\left(x-1\right)]=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\) (bạn phá ngoặc ra rồi tính là ra bước này)
\(\Leftrightarrow3x+2=0\) hoặc \(x+1=0\) hoặc \(2x-1=0\) ( đến đây bạn chia làm 3 trường hợp r tự tính nhé)
Chúc bạn học tốt!!
d/
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^3+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^3+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
e/
\(\Leftrightarrow x^3+x^2-6x-x^2-x+6=0\)
\(\Leftrightarrow x\left(x^2+x-6\right)-\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)
Câu a mình chắc chắn là đúng vì mình làm rồi.
Chúc bạn học tốt.
b) \(-4x^2-4x-2\) <0 với mọi x
\(=-\left(4x^2+4x+2\right)\)
\(=-\left[\left(2x^2\right)+2.2x.1+1^2+2\right]\)
\(=-\left[\left(2x+1\right)^2+2\right]\)
\(=-\left(2x+1\right)^2-2\)
Nx : \(-\left(2x+1\right)^2\le0\) với mọi x
\(\Rightarrow-\left(2x+1\right)^2-2< 0\) với mọi x
\(\Rightarrow-4x^2-4x-2< 0\) với mọi x
e, Ta có: \(\Delta\)'\(=\left(-6\right)^2-4.5=16>0\)
Suy ra \(\sqrt{\Delta'}=\sqrt{16}=4\)
Vậy phương trình đã cho có 2 nghiệm phân biệt
\(x_1=\dfrac{-b'-\sqrt{\Delta'}}{a}=\dfrac{6-4}{4}=\dfrac{1}{2}\)
\(x_2=\dfrac{-b'+\sqrt{\Delta'}}{a}=\dfrac{6+4}{4}=\dfrac{5}{2}\)
Vậy phương trình đã cho có 2 nghiệm là 1/2;5/2
f,Ta có : a+-b+c=2-5+3=0
Do đó phương trình đã cho có 2 nghiệm \(x_1\)=-1 hoặc \(x_2=\dfrac{-c}{a}=-\dfrac{3}{2}\)
g,Ta có: a+b+c=1+1-2=0
Do phương trình đã cho có 2 nghiệm \(x_1\)=1 hoặc \(x_2=\dfrac{c}{a}=-2\)
h,Ta có a+b+c=1-4+3=0
Do đó phương trình đã cho có 2 nghiệm \(x_1=1\) hoặc \(x_2=\dfrac{c}{a}=3\)
g, \(x^2+x-2=0\)
\(\Rightarrow x^2-x+2x-2=0\)
\(\Rightarrow\left(x^2-x\right)+\left(2x-2\right)=0\)
\(\Rightarrow x.\left(x-1\right)+2.\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right).\left(x+2\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy......
h, \(x^2-4x+3=0\)
\(\Rightarrow x^2-3x-x+3=0\)
\(\Rightarrow\left(x^2-3x\right)-\left(x-3\right)=0\)
\(\Rightarrow x.\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right).\left(x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy......
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