a) ( 2x - 3 ) - ( x - 5 ) = ( x + 7 ) - ( x + 2 ) 

<=> 2x - 3 - x + 5 = x + 7 - x - 2

<=> x = 3

b)(7x-5)-(6x+4)=(2x+3)-(2x+1)

<=> 7x - 5 - 6x - 4 = 2x + 3 - 2x - 1

<=> x = 11

c)(9x-3)-(8x+5)=(3x+2)

<=> 9x - 3 - 8x - 5 = 3x + 2

<=> -2x = 10

<=> x = -5

d)(x+7)-(2x+3)=(3x+5)-(2x+4)

<=> x + 7 - 2x - 3 = 3x + 5 - 2x - 4

<=> -2x = -3

<=> x = 3/2

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Mài ngu

a) |3x - 5| = 3x - 5 => 3x - 5 > 0 => 3x > 5 => x > 5/3

b) |7 - x| = x - 7 => 7 - x < 0 => - x < - 7 => x > 7

c) |2x - 3| = 3 - 2x => 2x - 3 < 0 => 2x < 3 => x < 3/2

a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)

\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)

\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)

\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)

\(-x^3=27\)

\(x=-3\)

Bài 1:

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

\(6x-9+4-2x=-3\)

\(4x=-2\)

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

b/ \(2x\left(x^2-2\right)+x^2\left(1-2x\right)-x^2=-12\)

\(2x^3-4x+x^2-2x^3-x^2=-12\)

\(-4x=-12\)

\(x=\frac{1}{3}\)

Bài 2:

a/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(6x^2+9x-6x^2-15x+4x+10=8\)

\(-2x=8\)

\(x=-4\)

b/ \(4x\left(x-1\right)-3\left(x^2-5\right)-x^2=\left(x-3\right)-\left(x+4\right)\)

\(4x^2-4x-3x^2+15-x^2=-7\)

\(-4x=-22\)

\(x=\frac{11}{2}\)

c/ \(2\left(3x-1\right)\left(2x+5\right)-6\left(2x-1\right)\left(x+2\right)=-6\)

\(6x-2\left(2x+5\right)-12x+6\left(x+2\right)=-6\)

\(6x-4x-10-12x+6x+12=-6\)

\(-4x=-8\)

\(x=2\)

Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu

a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14) 

=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84) 

=> 156 -  56x = 24x - 324 

=>  24x + 56x = 324 + 156 

=> 80x = 480 

=> x = 480 : 80 =  6 

Vậy x = 6 

b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1 

=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1 

=> (15x - 8x) + (25 + 12) = 11x + 37 

=> 7x + 37 = 11x + 37 

=> 11x - 7x = 0 

=>  x = 0 

a) Phá trị tuyệt đối ra thành 2 trường hợp:

 TH1: |3x - 2| - x = 7

=>  3x - 2 - x =7

=>  2x = 9

=>  x = 4,5

 TH2: |3x - 2| - x = 7

=>   2 - 3x - x = 7

=>  2 - 4x = 7

=> -5 = 4x

=>  x = -1,25

Vậy x = -1,25 hoặc x = 4,5

b) Ta phá trị tuyệt đối:

 TH1: |2x - 3| > 5

=>  2x - 3 > 5

=>  2x > 8

=>  x > 4 (1)

TH2: |2x - 3| > 5 

=>  3 - 2x > 5

=>  2x > -2

=>  x > -1 (2)

Từ (1) và (2) ta suy ra x > 4

HAI Ý CÒN LẠI BẠN CŨNG PHÁ TRỊ TUYỆT ĐỐI RA THÀNH 2 TRƯỜNG HỢP NHA !!!

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