a, \(5\left|2x-1\right|-3=7\Leftrightarrow5\left|2x-1\right|=10\Leftrightarrow\left|2x-1\right|=2\)

TH1 : \(2x-1=2\Leftrightarrow x=\frac{3}{2}\)

TH2 : \(2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)

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

\(\Leftrightarrow\left(x-2\right)\left(2x+3-x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow x=-1;x=2\)

c, \(\frac{2x-3}{2}< \frac{1-3x}{-5}\Leftrightarrow\frac{2x-3}{2}+\frac{1-3x}{5}< 0\)

\(\Leftrightarrow\frac{10x-15+2-6x}{10}< 0\Rightarrow4x-13< 0\Leftrightarrow x< \frac{13}{4}\)

mk chỉ giải đc có bài 1 thui nha bn

\(\frac{4}{x-2}+\frac{1}{x+3}=0\)

ĐKXĐ: x ≠ 2 và x ≠ -3

QĐKM:

⇔(x+3)4 + (x-2)1 = 0

⇔4x + 12 + x - 2 = 0

⇔4x + x = -12 + 2

⇔5x = -10

⇔x= -2

S={-2}

a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0

=>-5x-4=0

=>x=-4/5

b: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81

=>83x=83

=>x=1

Cảm ơn nhìu ạ :3

Cách giải

a, 2x - x (3x + 1 ) < 15 - 3x(x + 2)

<=> 2x - 3x² - x < 15 - 3x² - 6x

<=> 7x < 15

<=> x < 15/7 Vậy Tập nghiệm của BPT là : { x / x < 15/7 }

b , BPT <=> 2(1 - 2x ) - 16 < 1 - 5x + 8x

    <=> -7x < 15

   <=> x > -15/7 Vậy tập nghiệm của BPT là : { x / x > -15/7 }

a) 2x-x(3x+1) < 15-3x(x+2)

<=> 2x-3x²-x < 15-3x²-6x

<=> 2x-3x²-x+3x²+6x < 15

<=> 7x < 15

<=> x < 15/7

Vậy tập nghiệm của bất phương trình là x < 15/7

b) \(\frac{1-2x}{4}-2\le\frac{1-5x}{8}+x\)

Quy đồng mẫu ta được :

\(\frac{2-4x}{8}-\frac{16}{8}\le\frac{1-5x}{8}+\frac{8x}{8}\)

Khử mẫu

=> \(2-4x-16\le1-5x+8x\)

<=> \(-4x+5x-8x\le1-2+16\)

<=> \(-7x\le15\)

<=> \(x\ge-\frac{15}{7}\)

Vậy tập nghiệm của bất phương trình là \(x\ge-\frac{15}{7}\)

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

a: 3x-5>15-x

=>4x>20

hay x>5

b: \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)

=>3x²+x>3x²-12

=>x>-12

Đây là giải bất phương trình nhé bạn

a) Ta có: \(3\left(1-2x\right)< 4\left(5-\frac{3x}{2}\right)\)

\(\Leftrightarrow3-6x< 20-6x\)

\(\Leftrightarrow3-6x-20+6x< 0\)

hay -17<0(vô lý)

Vậy: \(S=\varnothing\)

b) Ta có: \(4-\left(x-3\right)^2-\left(2x-1\right)^2>12x\)

\(\Leftrightarrow4-\left(x^2-6x+9\right)-\left(4x^2-4x+1\right)-12x>0\)

\(\Leftrightarrow4-x^2+6x-9-4x^2+4x-1-12x>0\)

\(\Leftrightarrow-5x^2-2x-6>0\)

\(\Leftrightarrow-5\left(x^2+\frac{2}{5}x+\frac{6}{5}\right)>0\)

\(\Leftrightarrow x^2+\frac{2}{5}x+\frac{6}{5}< 0\)

\(\Leftrightarrow x^2+2\cdot x\cdot\frac{2}{10}+\frac{4}{100}+\frac{29}{25}< 0\)

\(\Leftrightarrow\left(x+\frac{1}{5}\right)^2+\frac{29}{25}< 0\)(vô lý)

Vậy: \(S=\varnothing\)

a: 3x-5>15-x

=>3x+x>15+5

=>4x>20

=>x>5

b: \(3\left(x-2\right)\left(x+2\right)<3x^2+x\)

=>\(3\left(x^2-4\right)<3x^2+x\)

=>\(3x^2-12-3x^2-x<0\)

=>-x-12<0

=>x+12>0

=>x>-12

c: \(\left(2x+1\right)^2+3x\left(1-x\right)\le\left(x+2\right)^2\)

=>\(4x^2+4x+1+3x-3x^2\le x^2+4x+4\)

=>\(x^2+7x+1\le x^2+4x+4\)

=>7x+1<=4x+4

=>7x-4x<=4-1

=>3x<=3

=>x<=1

d: \(\frac{5x-20}{3}-\frac{2x^2+x}{2}>\frac{x\left(1-3x\right)}{3}-\frac{5x}{4}\)

=>\(\frac{4\left(5x-20\right)-6\left(2x^2+x\right)}{12}>\frac{4x\left(1-3x\right)-15x}{12}\)

=>\(4\left(5x-20\right)-6\left(2x^2+x\right)>4x\left(1-3x\right)-15x\)

=>\(20x-80-12x^2-6x>4x-12x^2-15x\)

=>14x-80>-11x

=>25x>80

=>\(x>\frac{80}{25}=\frac{16}{5}\)

e: 4-2x<=3x-6

=>-2x-3x<=-6-4

=>-5x<=-10

=>x>=2

f: \(\left(x+4\right)\left(5x-1\right)>5x^2+16x+2\)

=>\(5x^2-x+20x-4>5x^2+16x+2\)

=>19x-4>16x+2

=>3x>6

=>x>2

g: \(x\left(2x-1\right)-8<5-2x\left(1-x\right)\)

=>\(2x^2-x-8<5-2x+2x^2\)

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

=>-x+2x<5+8

=>x<13

h: \(\frac{3x-1}{4}-\frac{3\left(x-2\right)}{8}-1>\frac{5-3x}{2}\)

=>\(\frac{2\left(3x-1\right)}{8}-\frac{3\left(x-2\right)}{8}-\frac88>\frac{4\left(5-3x\right)}{8}\)

=>2(3x-1)-3(x-2)-8>4(5-3x)

=>6x-2-3x+6-8>20-12x

=>3x-4>20-12x

=>15x>24

=>\(x>\frac{24}{15}\)

=>x>1,6