a) x.(x-1)=0

\(\Rightarrow\)x=0 hoặc x-1=0

\(\Rightarrow\)x=0+1

\(\Rightarrow\)x=1

vậy x=1 hoặc x=0

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

\(\Rightarrow\)-x = 0 hoặc x+3 = 0

\(\Rightarrow\)x= 0-3

\(\Rightarrow\)x=-3

vậy x=0 hoặc x=-3

c) (2x-4).(x+2)=0

(2x-4)= 0

2x=0+4

2x=4

x=4:2

x=2

hoặc (x+2)=0

x= 0-2

x=-2

vậy x=2 hoặc x=-2

d) (3-x).|x+5|=0

3-x = 0

x= 3-0

x=3

hoặc |x+5|=0

x+ 5=0

x=0-5

x=-5

vậy x=3 hoặc x=-5

e) (|x|+1).( 4-2x) = 0

(|x|+1) =0

|x|= 0-1

|x|=-1

hoặc( 4-2x) = 0

2x=4-0

2x=4

x=4:2

x=2

g) x²+5x=0

x²=0

x=0

hoặc 5x=0

x= 0: 5

x=0

vậy x=0

2)

a) (x+3).(y-5)= 7

(x+3)và (y-5)\(\in\)Ư(7)=\(\left\{1;-1;7;-7\right\}\)

b) xy + 3x - 2y= 11

x( y+3) -2y=11

x(y-3)- 2( y+3) +6 = 11

( y+3) ( x-2) = 5

vì x,y thuộc Z \(\Leftrightarrow\)y+3 và x-2 \(\in\)Z

do đó y+3 và x-2 \(\in\)Ư ( 5)= \(\left\{1;5;-1;-5\right\}\)

\(\in\)\(\in\)

c) xy + 3x - 7y= 21

x( y+3) -7y= 21

x( y+3) - 7( y+3)+21= 21

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

@Lâm Khả Vy siêu quá đi à.

+)   (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

a) \(\frac{1}{3}+\frac{2}{3}:x=-7\)

=> \(\frac{2}{3}:x=-7-\frac{1}{3}\)

=> \(\frac{2}{3}:x=-\frac{22}{3}\)

=> \(x=\frac{2}{3}:\left(-\frac{22}{3}\right)\)

=> \(x=-\frac{1}{11}\)

b) \(\frac{1}{3}x+\frac{2}{5}x=0\)

=> \(\frac{11}{15}x=0\)

=> \(x=0\)

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

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

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

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

=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)

a) \(\frac{1}{3}+\frac{2}{3}:x=-7\)

\(\Rightarrow\frac{2}{3}.\frac{1}{x}=-7-\frac{1}{3}\)

\(\Rightarrow\frac{2}{3x}=\frac{-21-1}{3}\)

\(\Rightarrow\frac{2}{3x}=\frac{-22}{3}\)

\(\Rightarrow-22.3x=6\)

\(\Rightarrow3x=\frac{-6}{22}=\frac{-3}{11}\)

\(\Rightarrow x=\frac{-3}{11}:3=\frac{-3}{11}.\frac{1}{3}\)

\(\Rightarrow x=\frac{-1}{11}\)

b) \(\frac{1}{3}x+\frac{2}{5}x=0\)

\(\Rightarrow x.\left(\frac{1}{3}+\frac{2}{5}\right)=0\)

\(\Rightarrow x=0\)

c) \(\left(2x-3\right).\left(6-2x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-3=0\\6-2x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=3\\2x=6\end{cases}}\)

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

d) \(x:\frac{3}{4}+\frac{1}{4}=\frac{-2}{3}\)

\(\Rightarrow x.\frac{4}{3}=\frac{-2}{3}-\frac{1}{4}\)

\(\Rightarrow x.\frac{4}{3}=\frac{-11}{12}\)

\(\Rightarrow x=\frac{-11}{12}:\frac{4}{3}=\frac{-11}{12}.\frac{3}{4}=\frac{-11}{16}\)

e) \(\frac{3}{4}-\left|x-\frac{2}{3}\right|=\frac{1}{2}\)

\(\Rightarrow\left|x-\frac{2}{3}\right|=\frac{3}{4}-\frac{1}{2}\)

\(\Rightarrow\left|x-\frac{2}{3}\right|=\frac{1}{4}\)

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

\(\Rightarrow\orbr{\begin{cases}x=\frac{11}{12}\\x=\frac{5}{12}\end{cases}}\)

(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)

 20x2−16x−34=10x2+3x−120x2−16x−34=10x2+3x−1

 10x2−19x−33=010x2−19x−33=0

 (10x+11)(x−3)=0

chỉ bt lm con b thoy

..army,,,,,,,,,,

a) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)

\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)

\(\Leftrightarrow3x^2-12x=3x^2-17x+20+2\)

\(\Leftrightarrow3x^2-12x=3x^2-17x+22\left(3x^2-17x\right)\)

\(\Leftrightarrow5x=22\)

\(\Rightarrow x=\frac{22}{5}\)

b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow20x^2-16x-34=10x^2+3x+1\)

\(\Leftrightarrow20x^2-16x-33=10x^2+3x\)

\(\Leftrightarrow20x^2-16x-33=10x^2+3x-3x\)

\(\Leftrightarrow20x^2-16x-33=10x^2\)

\(\Leftrightarrow20x^2-16x-33=10x^2-10x^2\)

\(\Leftrightarrow20x^2-16x-33=0\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)

1) 30x-30x^2-31

2)6x^4-2x^3-15x^2+23x-6

Nhầm 1 chút nhé, Bài 1 câu a) ( x^2 - 5x ) . ( x^2 - x +3) - ( x^3 - x^2 +1) . ( 2x - 1 )