+)   (5x-1). (2x+3)-3. (3x-1)=0

10x^2+15x-2x-3 - 9x+3=0

10x^2 +8x=0

2x(5x+4)=0

=> x=0 hoặc x= -4/5

+)    x^3 (2x-3)-x^2 (4x^2-6x+2)=0

2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0

-2x^4 + 3x^3-2x^2=0

x^2(-2x^2+x-2)=0

-2x^2(x-1)^2=0

=> x=0 hoặc x=1

+)   x (x-1)-x^2+2x=5

x^2 -x -x^2+2x=5

x=5

+)     8 (x-2)-2 (3x-4)=25

8x - 16-6x+8=25

2x=33

x=33/2

1) -3x²+5x=0

-x(3x-5)=0

suy ra hoặc x=0 hoặc 3x-5=0. giải ra ta có nghiệm phương trình là 0 và 3/5

2) x²+3x-2x-6=0

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

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

suy ra hoặc x-2=0 hoặc x+3=0. giải ra ta có nghiệm là 2 và -3

3) x²+6x-x-6=0

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

(x-1)(x+6)=0. vậy nghiệm là 1 và -6

4) x²+2x-3x-6=0

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

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

vậy nghiệm là -2 và 3

5) x(x-6)-4(x-6)=0

(x-4)(x-6)=0. vậy nghiệm là 4 và 6

6)x(x-8)-3(x-8)=0

(x-3)(x-8)=0

suy ra nghiệm là 3 và 8

7) x²-5x-24=0

x²-8x+3x-24=0

x(x-8)+3(x-8)=0

(x+3)(x-8)=0

vậy nghiệm là -3 và 8

câu 1:  -3x² + 5x = 0

suy ra -x(3x-5)=0

sung ra x = 0 hoặc 3x-5=0 suy ra 3x = 5 suy ra x = 5/3

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

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

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

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

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) 2x – (3 – 5x) = 4( x +3)

<=>2x-3+5x=4x+12

<=>2x-3+5x-4x-12=0

<=>3x-15=0

<=>x=5

2) 5(2x-3) - 4(5x-7) =19 - 2(x+11)

<=>10x-15-20x+28=19-2x-22

<=>10x-15-20x+28-19+2x+22=0

<=>-8x+16=0

<=>x=2

tham khảo

1) 2x – (3 – 5x) = 4( x +3)

<=>2x-3+5x=4x+12

<=>2x-3+5x-4x-12=0

<=>3x-15=0

<=>x=5

2) 5(2x-3) - 4(5x-7) =19 - 2(x+11)

<=>10x-15-20x+28=19-2x-22

<=>10x-15-20x+28-19+2x+22=0

<=>-8x+16=0

<=>x=2

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

\(\Rightarrow x+2-2=0\)

\(\Rightarrow x=0\)

\(\left(x+3\right)+1=7\)

\(\Rightarrow x+3+1=7\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=3\)

\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)

\(\Rightarrow3x=12\)

\(\Rightarrow x=4\)

\(\left(5x+4\right)-1=13\)

\(\Rightarrow5x+4-1=13\)

\(\Rightarrow5x+3=13\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

\(\left(4x-8\right)-3=5\)

\(\Rightarrow4x-8-3=5\)

\(\Rightarrow4x-11=5\)

\(\Rightarrow4x=16\)

\(\Rightarrow x=4\)

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

\(\Rightarrow8-2x-4=2\)

\(\Rightarrow4-2x=2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

\(7+\left(5x+2\right)=14\)

\(\Rightarrow7+5x+2=14\)

\(\Rightarrow9+5x=14\)

\(\Rightarrow5x=5\)

\(\Rightarrow x=1\)

\(5-\left(3x-11\right)=1\)

\(\Rightarrow5-3x+11=1\)

\(\Rightarrow16-3x=1\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

1,\(5x^2=13x\Leftrightarrow5x^2-13x=0\Leftrightarrow x\left(5x-13\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{13}{5}\end{cases}}\)

2,\(\left(5x^2+3x-2\right)^2=\left(4x^2-3x-2\right)^2\Leftrightarrow\orbr{\begin{cases}5x^2+3x-2=4x^2-3x-2\\5x^2+3x-2=-4x+3x+2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+6x=0\\9x^2-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\left(x+6\right)=0\\\left(3x\right)^2=2^2\end{cases}\Leftrightarrow}}\orbr{\begin{cases}x=0or-6\\x=-\frac{2}{3}or\frac{2}{3}\end{cases}}\)

3,\(x^3+27+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2+3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+3x+9+x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2+4x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x^2+4x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x\left(x+4\right)=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=0or-4\end{cases}}\)

4,\(5x\left(x-2000\right)-x+2000=0\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)

\(\Leftrightarrow\left(x-2000\right)\left(5x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=2000\\x=\frac{1}{5}\end{cases}}\)

5,\(5x\left(x-2\right)-x+2=0\Leftrightarrow5x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(5x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x-2=0\\5x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=\frac{1}{5}\end{cases}}\)

6,\(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(4x-8\right)=0\Leftrightarrow\orbr{\begin{cases}x+1=0\\4x-8=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

7,\(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(2x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x-4=0\\2x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}\)

tí làm nửa kia 

8,\(x^2-6x+8=0\Leftrightarrow x^2-6x+9-1=0\Leftrightarrow\left(x-3\right)^2-1^2=0\)

\(\Leftrightarrow\left(x-3-1\right)\left(x-3+1\right)=0\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=2\end{cases}}\)

9,\(9x^2+6x-8=0\Leftrightarrow9x^2+6x+1-9=0\Leftrightarrow\left(3x+1\right)^2-3^2=0\)

\(\Leftrightarrow\left(3x+1-3\right)\left(3x+1+3\right)=0\Leftrightarrow\left(3x-2\right)\left(3x+4\right)=0\Leftrightarrow\orbr{\begin{cases}3x-2=0\\3x+4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{4}{3}\end{cases}}\)

10,\(x^3+x^2+x+1=0\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2+1=0\end{cases}\Leftrightarrow}x=-1\)

11,\(x^3-x^2-x+1=0\Leftrightarrow\left(x-1\right)\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

12,\(\left(5-2x\right)\left(2x+7\right)=4x^2-25\Leftrightarrow\left(5-2x\right)\left(2x+7\right)-4x^2+25=0\)

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

\(\Leftrightarrow\left(5-2x\right)\left(2x+7-5-2x\right)=0\Leftrightarrow\left(5-2x\right).2=0\Leftrightarrow5-2x=0\Leftrightarrow x=\frac{5}{2}\)

13,\(x\left(2x-1\right)+\frac{1}{3}.\frac{2}{3}x=0\Leftrightarrow x\left(2x-1\right)+\frac{2}{9}x=0\)

\(\Leftrightarrow x\left(2x-1+\frac{2}{9}\right)=0\Leftrightarrow x\left(2x-\frac{7}{9}\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\2x=\frac{7}{9}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{7}{18}\end{cases}}\)

14,\(4\left(2x+7\right)-9\left(x+3\right)^2=0\Leftrightarrow8x+28-9x^2-54x-81=0\)

\(\Leftrightarrow-9x^2+\left(8x-54x\right)+\left(28-81\right)=0\Leftrightarrow-9x^2-46x-53=0\)

\(\Leftrightarrow9x^2+46x+53=0\)Ta có : \(\Delta'=\frac{2116}{4}-477=529-477=52\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-23+\sqrt{52}}{9}\\x=\frac{-23-\sqrt{52}}{9}\end{cases}}\)

a)

\(\left(4x-10\right)\cdot\left(24+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-\frac{24}{5}\end{matrix}\right.\)

Vậy \(S=\left\{\frac{5}{2};-\frac{24}{5}\right\}\)

b)

\(\left(2x-5\right)\left(3x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)

Vậy \(S=\left\{\frac{5}{2};\frac{2}{3}\right\}\)

c)

\(\left(2x-1\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{3}\end{matrix}\right.\)

Vậy \(S=\left\{\frac{1}{2};-\frac{1}{3}\right\}\)

d)

\(x\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy \(S=\left\{0;\frac{1}{2}\right\}\)

e) \(\left(5x+3\right)\left(x^2+4\right)\left(x-1\right)=0\)

Do \(x^2\ge0\) Nên \(x^2+4>0\)

\(\left(5x+3\right)\left(x^2+4\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{5}\\x=1\end{matrix}\right.\)

Vậy \(S=\left\{-\frac{3}{5};1\right\}\)

....... Còn lại cứ cho mỗi thừa số = 0 rồi tìm x như bình thường thôi bạn

1. (4x - 10)(24 + 5x) = 0

\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-24}{5}\end{matrix}\right.\)

Vậy S = {\(\frac{5}{2}\); \(\frac{-24}{5}\)}

2. (2x - 5)(3x - 2) = 0

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)

Vậy S = {\(\frac{5}{2}\); \(\frac{2}{3}\)}

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{3}\end{matrix}\right.\)

Vậy S = {\(\frac{1}{2}\); \(\frac{-1}{3}\)}

4. x(x² - 1) = 0

\(\Leftrightarrow\) x(x - 1)(x + 1) = 0

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

Vậy S = {0; 1; -1}

5. (5x + 3)(x² + 4)(x - 1) = 0

VÌ x² + 4 > 0 với mọi x nên

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

Vậy S = {\(\frac{-3}{5}\); 1}

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)

Vậy S = {1; -2; -3}

7. (x - 1)(x + 5)(-3x + 8) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\\-3x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\\x=\frac{8}{3}\end{matrix}\right.\)

Vậy S = {1; -5; \(\frac{8}{3}\)}

Chúc bn học tốt!!

2:

a: =>x-1=0 hoặc 3x+1=0

=>x=1 hoặc x=-1/3

b: =>x-5=0 hoặc 7-x=0

=>x=5 hoặc x=7

c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)

d: =>x=0 hoặc x^2-1=0

=>\(x\in\left\{0;1;-1\right\}\)

Bạn tách ra từng câu thoi nhe .

Bạn gõ ct lại dưới dạng latex (kí hiệu này gọi là latex Σ ) nhé