a) 4y.(x-1) - (1-x) = 4y(x-1) + (x-1) = (4y+1)(x-1)

b) (x-3)³ + 3-x = (x-3)²(x-3) +(x-3) = [(x-3)² +1](x-3) = (x² -6x+10)(x-3)

\(\begin{array}{l} a,\ 4y.(x-1)-(1-x)\\ =3y.(x-1)+1.(x-1)\\ =(3y+1).(x-1)\\ b,\ (x-3)^3+3-x\\ =(x-3)^3-(x-3)\\ =(x-3).[(x-3)^2-1]\end{array}\)

a: =>2x>-6

hay x>-3

e: =>(5-x)/x<0

=>0<x<5

h: \(\Leftrightarrow\dfrac{x+5-x-3}{x+3}< 0\)

\(\Leftrightarrow x+3< 0\)

hay x<-3

g: \(\Leftrightarrow\dfrac{2x+7}{x+4}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{7}{2}\\x< -4\end{matrix}\right.\)

a) x - 1/2 = 3/5

x = 3/5 + 1/2

x = 11/10

b) x - 1/2 = -2/3

x = -2/3 + 1/2

x = -1/6

c) 2/5 - x = 0,25

x = 2/5 - 0,25

x = 2/5 - 1/4

x = 3/20

Chọn D

a) (x-2)³+6(x+1)²-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6x²+12x+6-x³+12=0

\(\Rightarrow\)24x+10=0

\(\Rightarrow\)24x=-10

\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)

b)(x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x-4)(x+4)+3x²

\(\Rightarrow\)x²-25-(x²+6x+9)+3(x²-4x+4)=x²+2x+1-(x²-16)+3x²

\(\Rightarrow\)x²​-25-x²-6x-9+3x²-12x+12=x²+2x+1-x²+16+3x²

\(\Rightarrow\)3x²-18x-22=3x²+2x+17

\(\Rightarrow\)3x²-18x-22-3x²-2x-17=0

\(\Rightarrow\)-20x-39=0

\(\Rightarrow\)-20x=39

\(\Rightarrow\)x=\(-\dfrac{39}{20}\)

c) (2x+3)² +(x-1)(x+1)=5(x+2)²-(x-5)(x+1)+(x+4)

⇒4x²+12x+9+x²-1=5(x²+4x+4)-(x²+x-5x-5)+x+4

⇒5x²+12x+8=5x²+20x+20-x²-x+5x+5+x+4

⇒5x²+12x+8-5x²-20x-20+x²+x-5x-5-x-4=0

⇒x²-13x-21=0

a) \(2\dfrac{3}{4}-x=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{11}{4}-x=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{11}{4}-\dfrac{3}{4}=\dfrac{8}{4}=2\)

b) \(x:\dfrac{5}{6}=-\dfrac{3}{5}\)

\(\Rightarrow x=-\dfrac{3}{5}.\dfrac{5}{6}=-\dfrac{15}{30}=-\dfrac{1}{2}\)

c) \(1\dfrac{1}{3}+\dfrac{2}{3}:x=1\)

\(\Rightarrow\dfrac{2}{3}:x=1-1\dfrac{1}{3}\)

\(\Rightarrow\dfrac{2}{3}:x=-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{3}:-\dfrac{1}{3}\)

\(\Rightarrow x=-2\)

d) \(x-\dfrac{1}{9}=\dfrac{8}{3}\)

\(\Rightarrow x=\dfrac{8}{3}+\dfrac{1}{9}\)

\(\Rightarrow x=\dfrac{25}{9}\)

e) \(\dfrac{1}{2}x+650\%x-x=-6\)

\(\Rightarrow\dfrac{1}{2}x+\dfrac{13}{2}x-x=-6\)

\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{13}{2}-1\right)-6\)

\(\Rightarrow6x=-6\)

\(\Rightarrow x=\dfrac{-6}{6}=-1\)

g) \(2\left(x-\dfrac{1}{2}\right)+3\left(-1+\dfrac{x}{3}\right)=x\left(\dfrac{2}{x}-1\right)\) \(\text{Đ}K:x\ne0\)

\(\Rightarrow2x-1-3+x=2-x\)

\(\Rightarrow3x-4=2-x\)

\(\Rightarrow3x+x=2+4\)

\(\Rightarrow4x=6\)

\(\Rightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)

h) \(x-\dfrac{2}{20}=-\dfrac{5}{2}-x\)

\(\Rightarrow x+x=-\dfrac{5}{2}+\dfrac{2}{20}\)

\(\Rightarrow2x=-\dfrac{12}{5}\)

\(\Rightarrow x=-\dfrac{12}{5}:2=-\dfrac{6}{5}\)

i) \(\left(\dfrac{x}{2}-1\right)^3+2=-\dfrac{11}{8}\)

\(\Rightarrow\left(\dfrac{x}{2}-1\right)^3=-\dfrac{11}{8}-2\)

\(\Rightarrow\dfrac{x}{2}-1=\sqrt[3]{-\dfrac{27}{8}}\)

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

\(\Rightarrow\dfrac{x}{2}=-\dfrac{3}{2}+1\)

\(\Rightarrow x=-\dfrac{1}{2}.2=-1\)

k) \(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}x\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\\dfrac{3}{2}x=\dfrac{3}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}.3=-\dfrac{3}{2}\\x=\dfrac{3}{4}:\dfrac{3}{2}=\dfrac{1}{2}\end{matrix}\right.\)

1) Ta có: \(\dfrac{3x+7}{11}=\dfrac{5x-7}{9}\)

\(\Leftrightarrow9\left(3x+7\right)=11\left(5x-7\right)\)

\(\Leftrightarrow27x+63=55x-77\)

\(\Leftrightarrow27x-55x=-77-63\)

\(\Leftrightarrow-28x=-140\)

hay x=5

Vậy: S={5}

1, 3x+7/11= 5x-7/9

3x-5x=-7/11-7/9

-2x=-140/99

x= 70/99

vậy ...

2, x-7/4=9/-7+x

x-x=7/4-9/7

x không có gt thỏa mãn 

3, x-1/-2=8/1-x 

x+x =-1/2+8/1 

2x =15/2 

x =15/4 

vậy ...

Ít thôi -..-

a) ( 3x + 2 )( 2x + 9 )  - ( x + 3 )( 6x + 1 ) = ( x + 1 )² - ( x + 2 )( x - 2 )

<=> 6x² + 31x + 18 - ( 6x² + 19x + 3 ) = x² + 2x + 1 - ( x² - 4 )

<=> 6x² + 31x + 18 - 6x² - 19x - 3 = x² + 2x + 1 - x² + 4

<=> 12x + 15 = 2x + 5

<=> 12x - 2x = 5 - 15

<=> 10x = -10

<=> x = -1

b) ( 2x + 3 )( x - 4 ) + ( x - 5 )( x - 2 ) = ( 3x - 5 )( x - 4 )

<=> 2x² - 5x - 12 + x² - 7x + 10 = 3x² - 17x + 20

<=> 3x² - 12x - 2 = 3x² - 17x + 20

<=> 3x² - 12x - 3x² + 17x = 20 + 2

<=> 5x = 22

<=> x = 22/5

c) ( x + 2 )³ - ( x - 2 )³ - 12x( x - 1 ) = -8

<=> x³ + 6x² + 12x + 8 - ( x³ - 6x² + 12x - 8 ) - 12x² + 12x = -8

<=>  x³ + 6x² + 12x + 8 - x³ + 6x² - 12x + 8 - 12x² + 12x = -8

<=> 12x + 16 = -8

<=> 12x = -24

<=> x = -2

d) ( 3x - 1 )² - 5( x + 1 ) + 6x - 3.2x + 1 - ( x - 1 )² = 16

<=> 9x² - 6x + 1 - 5x - 5 + 6x - 6x + 1 - ( x² - 2x + 1 ) = 16

<=> 9x² - 11x - 3 - x² + 2x - 1 = 16

<=> 8x² - 9x - 4 = 16

<=> 8x² - 9x - 4 - 16 = 0

<=> 8x² - 9x - 20 = 0

( Đến đây bạn có hai sự lựa chọn : 1 là vô nghiệm

                                                         2 là nghiệm vô tỉ =) )

a) (3x + 2)(2x + 9) - (x + 3)(6x + 1) = (x + 1)² - (x + 2)(x - 2)

=> 3x(2x + 9) + 2(2x + 9) - x(6x + 1) - 3(6x + 1) = x² + 2x + 1 - x(x - 2) - 2(x - 2)

=> 6x² + 27x + 4x + 18 - 6x² - x - 18x - 3 = x² + 2x + 1 - x² + 2x - 2x + 4

=> (6x² - 6x²) + (27x + 4x - x - 18x) + (18 - 3) = (x² - x²) + (2x + 2x - 2x) + (1 + 4)

=> 12x + 15 = 2x + 5

=> 12x + 15  - 2x - 5 = 0

=> 10x + 10 = 0

=> 10x = -10 => x = -1

b) (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)

=> 2x(x - 4) + 3(x - 4) + x(x - 2) - 5(x - 2) = 3x(x - 4) - 5(x - 4)

=> 2x² - 8x + 3x - 12 + x² - 2x - 5x + 10 = 3x² - 12x - 5x + 20

=> (2x² + x²) + (-8x + 3x - 2x - 5x) + (-12 + 10) = 3x² - 17x + 20

=> 3x² - 12x - 2 = 3x² - 17x + 20

=> 3x² - 12x - 2 - 3x² + 17x - 20 = 0

=> (3x² - 3x²) + (-12x + 17x) + (-2 - 20) = 0

=> 5x - 22 = 0

=> 5x = 22 => x = 22/5

c) (x + 2)³ - (x - 2)³ - 12x(x - 1) = -8

=> x³ + 6x² + 12x + 8 - (x³  - 6x² + 12x - 8) - 12x² + 12x = -8

=> x³ + 6x² + 12x + 8 -x³ + 6x² - 12x + 8 - 12x² + 12x = -8

=> (x³ - x³) + (6x² + 6x² - 12x²) + (12x - 12x + 12x) + (8 + 8) = -8

=> 12x + 16 = -8

=> 12x = -24

=> x = -2

Còn bài cuối làm nốt