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loki
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Nguyễn Lê Phước Thịnh
5 tháng 10 2021 lúc 20:56

\(x^3-3x^2+3x-1=-8\)

\(\Leftrightarrow x-1=-2\)

hay x=-1

Đỗ khánh linh
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Nguyễn Lê Phước Thịnh
9 tháng 2 2022 lúc 15:32

Bài 2: 

a: =>x-45=-20

hay x=25

b: =>35+30-5x=-12+112

=>65-5x=100

=>5x=-35

hay x=-7

c: =>x-124=1000

hay x=1124

d: \(\Leftrightarrow46-\left(3x-2\right)^3=-18\)

\(\Leftrightarrow\left(3x-2\right)^3=64\)

=>3x-2=4

hay x=2

Đỗ khánh linh
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Nguyễn Lê Phước Thịnh
12 tháng 2 2022 lúc 14:07

Bài 2: 

a: =>x-45=-20

hay x=25

b: =>35+30-5x=-12+112=100

=>65-5x=100

=>5x=-35

hay x=-7

c: =>x-124=1000

hay x=1124

d: \(\Leftrightarrow46-\left(3x-2\right)^3=-18\)

\(\Leftrightarrow\left(3x-2\right)^3=64\)

=>3x-2=4

hay x=2

Nguyễn Thanh Nhung
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Phong
1 tháng 9 2023 lúc 13:30

c) \(x^2-9=2\cdot\left(x+3\right)^2\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-2\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(x+3\right)\left[x-3-2\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-3-2x-6\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(-x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)

b) \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

d) \(x^2-8x+3x-24=0\)

\(\Leftrightarrow\left(x^2-8x\right)+\left(3x-24\right)=0\)

\(\Leftrightarrow x\left(x-8\right)+3\left(x-8\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=8\end{matrix}\right.\)

Nguyễn Đức Trí
1 tháng 9 2023 lúc 13:38

a) \(x^2-9=2\left(x+3\right)^2\)

\(\Leftrightarrow\left(x+3\right)\left(x-3\right)=2\left(x+3\right)^2\)

\(\Leftrightarrow2\left(x+3\right)^2-\left(x+3\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left[2\left(x+3\right)-\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left[2x+6-x+3\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+9\right)=0\)

\(\)\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+9=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)

b) \(x^2-8x+3x-24=0\)

\(\Leftrightarrow\left(x-8\right)x+3\left(x-8\right)=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

c) \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Mai Anh Nguyễn
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Darlingg🥝
7 tháng 11 2021 lúc 15:38

a) 15x2-3x=0

=>3x(5x-1)=0

=>2 TH

=>*3x=0                   *5x-1=0

=>x=0                        =>5x=1=>x=1/5

vậy x=0 hoặc x=1/5

b) (3x-2) (x+3)+ (x2-9)=0

=>(3x-2)(x+3)+(x-3)(x+3)=0

=>(x+3).(3x-2+x-3)=0

=>(x+3).(4x-5)=0

=> 2 TH

*x+3=0=>x=0-3=>x=-3

*4x-5=0=>4x=5=>x=5/4

vậy x=-3 hoặc x=5/4

c) (x-1)3- (x+1) (2-3x)=-3

\(\Rightarrow\left(x-1\right)^3-\left(x+1\right)\left(2-3x\right)+3=0\)

\(\Rightarrow\left(x^3-3x^2+3x-1\right)-\left(2x-3x^2+2-3x\right)+3=0\)

\(\Rightarrow x^3-3x^2+3x-1-2x+3x^2-2+3x+3=0\)

\(\Rightarrow x^3-3x^2+3x^2+3x-2x+3x-1-2+3=0\)

\(\Rightarrow x^3+4x=0\)

\(\Rightarrow x\left(x^2+4\right)=0\)

=> 2 TH

*x=0

*x^2+4=0

vì: x^2>0

do đó:x^2+4>0

=> x^2+4 ko có gt nào x t/m y/cầu đề bài

vậy x=0

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Hiếu Tuấn
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Nguyễn Lê Phước Thịnh
5 tháng 11 2023 lúc 19:56

2: \(3x\left(x-4\right)+2x-8=0\)

=>\(3x\left(x-4\right)+2\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(3x+2\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)

3: 4x(x-3)+x2-9=0

=>\(4x\left(x-3\right)+\left(x+3\right)\left(x-3\right)=0\)

=>\(\left(x-3\right)\left(4x+x+3\right)=0\)

=>\(\left(x-3\right)\left(5x+3\right)=0\)

=>\(\left[{}\begin{matrix}x-3=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{5}\end{matrix}\right.\)

4: \(x\left(x-1\right)-x^2+3x=0\)

=>\(x^2-x-x^2+3x=0\)

=>2x=0

=>x=0

5: \(x\left(2x-1\right)-2x^2+5x=16\)

=>\(2x^2-x-2x^2+5x=16\)

=>4x=16

=>x=4

Vũ Ngọc Thuận
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Trần Ngọc Mỹ
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Monkey D. Luffy
17 tháng 11 2021 lúc 9:59

\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)

ILoveMath
17 tháng 11 2021 lúc 10:00

\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)

i love Vietnam
17 tháng 11 2021 lúc 10:11

1) \(x^2-9x=0\Rightarrow x\left(x-9\right)=0\Rightarrow x=0;9\)

2) \(x\left(x-4\right)-x^2=7\Rightarrow-4x=7\Rightarrow x=-\dfrac{7}{4}\)

3) \(3x+2\left(x-5\right)=5\Rightarrow5x-10=5\Rightarrow5x=15\Rightarrow x=3\)

4) \(25x^2-1=0\Rightarrow x^2=\dfrac{1}{25}\Rightarrow x=\pm\dfrac{1}{5}\)

5) \(3x\left(x-2\right)-5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(3x-5\right)=0\Rightarrow x=2;\dfrac{5}{3}\)

6) \(3x\left(x-7\right)+4\left(x-7\right)\Rightarrow\left(3x+4\right)\left(x-7\right)=0\Rightarrow x=-\dfrac{4}{3};7\)

7) \(4x^2-9=0\Rightarrow x^2=\dfrac{9}{4}\Rightarrow x=\pm\dfrac{3}{2}\)

8) \(10x\left(x-4\right)+2x-8=0\Rightarrow2\left(x-4\right)\left(5x+1\right)=0\Rightarrow x=4;-\dfrac{1}{5}\)

9) \(x\left(2x-5\right)-2x^2=0\Rightarrow x\left(2x-5-2x=0\right)\Rightarrow x=0\)

10) \(2x^2-4x=0\Rightarrow2x\left(x-2\right)=0\Rightarrow x=0;2\)

11) \(2x\left(3-4x\right)+3\left(4x-3\right)=0\Rightarrow2x\left(4x-3\right)-3\left(4x-3\right)=0\Rightarrow\left(4x-3\right)\left(2x-3\right)=0\Rightarrow x=\dfrac{3}{4};\dfrac{3}{2}\)

12) \(2x\left(x-5\right)-2x^2=3\Rightarrow-10x=3\Rightarrow x=-\dfrac{3}{10}\)

Du thương thán
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No Name
8 tháng 12 2019 lúc 20:55

\(2x\left(x^2-25\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x=0\\x^2-25=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

\(2x\left(3x-5\right)+\left(3x-5\right)=0\)

\(\left(2x+1\right)\left(3x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x-5=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{3}\end{cases}}\)

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No Name
8 tháng 12 2019 lúc 20:58

\(9\left(3x-2\right)-x\left(2-3x\right)=0\)

\(9\left(3x-2\right)+x\left(3x-2\right)=0\)

\(\left(9+x\right)\left(3x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}9+x=0\\3x-2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-9\\x=\frac{2}{3}\end{cases}}\)

\(\left(2x-1\right)^2=25\)

\(\Rightarrow\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

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Hiếu Mình Là
8 tháng 12 2019 lúc 21:04

\(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x^2-25\right)=0\)

\(\Leftrightarrow x^2-25=0\)

\(\Leftrightarrow x^2=25\)

\(\Leftrightarrow x=\pm5\)

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