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Nguyễn Anh Dũng
Giới thiệu về bản thân
![](https://rs.olm.vn/images/medal_mam_non.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
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![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_win_1.png)
![](https://rs.olm.vn/images/medal_tan_binh.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_win_1.png)
![](https://rs.olm.vn/images/medal_chuyen_can.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_win_1.png)
![](https://rs.olm.vn/images/medal_cao_thu.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_win_1.png)
![](https://rs.olm.vn/images/medal_thong_thai.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_win_1.png)
![](https://rs.olm.vn/images/medal_kien_tuong.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_win_1.png)
![](https://rs.olm.vn/images/medal_dai_kien_tuong.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_ngoi_sao.png)
![](https://rs.olm.vn/images/medal_win_1.png)
a)
=(3x+1)(x-1)
⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)
⇒TH2:x-1=0⇒x=1
Vậy xϵ{-\(\dfrac{1}{3}\);1}
b)
x(x-9)=0
⇒TH1:x=0
⇒TH2:x-9=0⇒x=9
Vậy xϵ{0;9}
a)
=(3x+1)(x-1)
⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)
⇒TH2:x-1=0⇒x=1
Vậy xϵ{-\(\dfrac{1}{3}\);1}
b)
x(x-9)=0
⇒TH1:x=0
⇒TH2:x-9=0⇒x=9
Vậy xϵ{0;9}
a)
=(3x+1)(x-1)
⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)
⇒TH2:x-1=0⇒x=1
Vậy xϵ{-\(\dfrac{1}{3}\);1}
b)
x(x-9)=0
⇒TH1:x=0
⇒TH2:x-9=0⇒x=9
Vậy xϵ{0;9}
a)
=(3x+1)(x-1)
⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)
⇒TH2:x-1=0⇒x=1
Vậy xϵ{-\(\dfrac{1}{3}\);1}
b)
x(x-9)=0
⇒TH1:x=0
⇒TH2:x-9=0⇒x=9
Vậy xϵ{0;9}
a)
=(3x+1)(x-1)
⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)
⇒TH2:x-1=0⇒x=1
Vậy xϵ{-\(\dfrac{1}{3}\);1}
b)
x(x-9)=0
⇒TH1:x=0
⇒TH2:x-9=0⇒x=9
Vậy xϵ{0;9}
a)
=(3x+1)(x-1)
⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)
⇒TH2:x-1=0⇒x=1
Vậy xϵ{-\(\dfrac{1}{3}\);1}
b)
x(x-9)=0
⇒TH1:x=0
⇒TH2:x-9=0⇒x=9
Vậy xϵ{0;9}
a)
=(3x+1)(x-1)
⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)
⇒TH2:x-1=0⇒x=1
Vậy xϵ{-\(\dfrac{1}{3}\);1}
b)
x(x-9)=0
⇒TH1:x=0
⇒TH2:x-9=0⇒x=9
Vậy xϵ{0;9}
a)x2+25-10x
=(x+5)2
b)-8y3+x3
=(-2y+x)(4y2+2xy+x3)