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

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

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

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

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

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

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

a) \(\Rightarrow\left(x-2\right)\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

b) \(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

c) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)

d) \(\Rightarrow\left(x-7\right)\left(3x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(a,\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\\ c,\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\\ d,\Leftrightarrow\left(x-7\right)\left(3x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)

Bài 1

1.(x-3)(x+2)-x(x-7)=15

\(\Leftrightarrow x^2+2x-3x-6-x^2+7x=15\)

\(\Leftrightarrow-6+6x=15\)

\(\Leftrightarrow6x=15+6\) =21

\(\Rightarrow x=\dfrac{21}{6}=3,5\)

2.(x-5)(x+5)+x(3-x)=20

\(\Leftrightarrow x^2-25+3x-x^2=20\)

\(\Leftrightarrow-25+3x=20\)

\(\Leftrightarrow3x=20+25=45\)

\(\Rightarrow x=\dfrac{45}{3}=15\)

3.(x-7)²-x(2+x)=-7

\(\Leftrightarrow x^2-14x+49-2x-x^2=-7\)

\(\Leftrightarrow-16x+49=-7\)

\(\Leftrightarrow-16x=-7-49=-56\)

\(\Rightarrow x=\dfrac{-56}{-16}=\dfrac{7}{2}=3,5\)

Tiếp bài 1

4.(x-4)²-(x+4)(x-4)=-16

\(\Leftrightarrow x^2-8x+16-x^2-16=-16\)

\(\Leftrightarrow-8x=-16\)

\(\Rightarrow x=\dfrac{-16}{-8}=2\)

5.(x-5)(x+5)-x(2-3x)=4x²-7

\(\Leftrightarrow x^2-25-2x+3x^2=4x^2-7\)

\(\Leftrightarrow4x^2-25-2x+3x^2=4x^2-7\)

\(\Leftrightarrow4x^2-4x^2-2x=-7+25\)

\(\Leftrightarrow-2x=18\)

\(\Rightarrow x=\dfrac{18}{-2}=-9\)

B2

1. 2x(x-2)+3x-6=0

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

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

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

2.3x(x-5)-5x+25=0

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

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

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

3. x(x+7)-x-7=0

=>x(x+7)-(x+7)=0

=>(x+7)(x-1)=0

\(\Rightarrow\left[{}\begin{matrix}x+7=0\Leftrightarrow x=-7\\x-1=0\Leftrightarrow x=1\end{matrix}\right.\)

\(\left(4-3x\right)\left(10x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)

\(\left(7-2x\right)\left(4+8x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)

rồi thực hiện đến hết ... 

Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>

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

\(2x^2-7x+3=4x^2+4x-3\)

\(2x^2-7x+3-4x^2-4x+3=0\)

\(-2x^2-11x+6=0\)

\(2x^2+11x-6=0\)

\(2x^2+12x-x-6=0\)

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

\(\left(x+6\right)\left(2x-1\right)=0\)

\(x+6=0\Leftrightarrow x=-6\)

\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

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

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

\(x=0\)

\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Tự lm tiếp nha 

chiu lop 3 ma

1/(2.x-5)+17=6

=> 2x - 5 = -11

=> 2x = -6

=> x = 3

vậy_

2/10-2.(4-3x)=-4 

=> 2(4 - 3x) = 14

=> 4 - 3x = 7

=> 3x = -3

=> x = -1

3/-12+3.(-x+7)=-18

=> 3(-x+7) = -6

=> -x+7 = -2

=> -x = -9

=> x = 9

4/24:(3.x-2)=-3

=> 3x - 2 = -8

=> 3x = -6

=> x = -2

5/-45:5.(-3-2.x)=3

=> 5(-3 - 2x) = -15

=> -3 - 2x = -3

=> - 2x = 0

=> x = 0

6/x.(x+7)=0

=> x = 0 hoặc x + 7 = 0

=> x = 0 hoặc x = -7

7/(x+12).(x-3)=0

=> x + 12 = 0 hoặc x - 3 = 0

=> x = -12 hoặc x = 3

8/(-x+5).(3-x)=0

=> -x + 5 = 0 hoặc 3  - x = 0

=> x = 5 hoặc x = 3

9/x.(2+x).(7-x)=0

=> x = 0 hoặc 2  + x = 0 hoặc  7 - x = 0

=> x =  0 hoặc x = -2  hoặc x = 7

10/(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

em tường anh vô LIÊM SỈ KO CÓ THẬT CƠ

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)

=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)

\(3x\left(x-2020\right)-x+2020=0\)

\(3x\left(x-2020\right)-\left(x-2020\right)=0\)

\(\left(3x-1\right)\left(x-2020\right)=0\)

\(\orbr{\begin{cases}x=\frac{1}{3}\left(TM\right)\\x=2020\left(TM\right)\end{cases}}\)

\(b,4-9x^2=0\)

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

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

\(\orbr{\begin{cases}2-3x=0\\2+3x=0\end{cases}\orbr{\begin{cases}x=\frac{2}{3}\left(TM\right)\\x=-\frac{2}{3}\left(TM\right)\end{cases}}}\)

\(c,x^2-x+\frac{1}{4}=0\)

\(x^2-x+\left(\frac{1}{2}\right)^2=0\)

\(\left(x-\frac{1}{2}\right)^2=0\)

\(x-\frac{1}{2}=0\)

\(x=\frac{1}{2}\)

\(d,x\left(x-3\right)+\left(x-3\right)=0\)

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

\(\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}\orbr{\begin{cases}x=3\left(TM\right)\\x=-1\left(TM\right)\end{cases}}}\)

\(e,9x\left(x-7\right)-x+7=0\)

\(9x\left(x-7\right)-\left(x-7\right)=0\)

\(\left(9x-1\right)\left(x-7\right)=0\)

\(\orbr{\begin{cases}9x-1=0\\x-7=0\end{cases}\orbr{\begin{cases}x=\frac{1}{9}\left(TM\right)\\x=7\left(TM\right)\end{cases}}}\)

a) 3x(x - 2020) - x + 2020 = 0 

<=> 3x(x - 2020) - (x - 2020) = 0

<=> (3x - 1)(x - 2020) = 0

<=> \(\orbr{\begin{cases}3x-1=0\\x-2020=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2020\end{cases}}\)

Vậy tập nghiệm phương trình là \(S=\left\{\frac{1}{3};2020\right\}\)

b) \(4-9x^2=0\)

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

<=> \(\orbr{\begin{cases}2-3x=0\\2+3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{2}{3}\end{cases}}\)

Vậy \(x\in\left\{\frac{2}{3};-\frac{2}{3}\right\}\)là nghiệm phương trình 

c) \(x^2-x+\frac{1}{4}=0\)

<=> \(\left(x-\frac{1}{2}\right)^2=0\)

<=> \(x-\frac{1}{2}=0\)

<=> \(x=\frac{1}{2}\)

d) x(x - 3) + (x - 3) = 0

<=> (x + 1)(x - 3) = 0

<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)

Vậy \(x\in\left\{-1;3\right\}\)là nghiệm phương trình

e) 9x(x - 7) - x + 7 = 0

<=> (9x - 1)(x - 7) = 0

<=> \(\orbr{\begin{cases}9x-1=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{9}\\x=7\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{9};7\right\}\)là nghiệm phương trình

1, \(3x\left(x-2020\right)-x+2020=0\)

\(\Leftrightarrow\)  \(3x\left(x-2020\right)-\left(x-2020\right)=0\)

\(\Leftrightarrow\left(x-2020\right)\left(3x-1\right)=0\)  

  \(\Leftrightarrow\)\(\orbr{\begin{cases}x-2020=0\\3x-1=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2020\\3x=1\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2020\\x=\frac{1}{3}\end{cases}}\)

Vậy phương trình có nghiệm x=2020 hoặc x=\(\frac{1}{3}\)

2, \(4-9x^2=0\)

\(\Leftrightarrow4=9x^2\)

\(\Leftrightarrow\frac{4}{9}=x^2\)

\(\Leftrightarrow x=\pm\frac{2}{3}\)

Vậy phương trình có nghiệm x=\(\pm\frac{2}{3}\)

3, \(x^2-x+\frac{1}{4}=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\frac{1}{2}=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy phương trình có nghiệm x=\(\frac{1}{2}\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

Vậy phương trình có nghiệm x=3 hoặc x= -1

5, \(9x\left(x-7\right)-x+7=0\)

\(\Leftrightarrow9x\left(x-7\right)-\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(9x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\9x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\9x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=\frac{1}{9}\end{cases}}\)

Vậy phương trình có nghiệm x=7 hoặc x=\(\frac{1}{9}\)

a) \(\left(x-4\right)^2-\left(x-4\right)=0\)

\(\left(x-4\right)\left(x-4-1\right)=0\)

\(\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

b) \(5x^2\left(x-7\right)+7\left(x-7\right)=0\)

\(\left(x-7\right)\left(5x^2+7\right)=0\)

\(\left[{}\begin{matrix}x-7=0\\5x^2+7=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=7\\x^2=\dfrac{-7}{5}\end{matrix}\right.\)

\(x=7\)

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

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

\(\left[{}\begin{matrix}x=3\\x=\pm1\end{matrix}\right.\)

a) (x - 4)^2=(x - 4)

(x - 4) (x -4)=(x -4 )

(x - 4) (x - 4)-(x - 4)=0

(x-4) (x-4-1)=0

(x-4) (x-5)=0

TH1:x-4=0                          TH2:x-5=0

            x=4                                      x=5