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

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

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

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}-x=-\frac{1}{2}\\5x=\frac{3}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{10}\end{cases}}\)

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

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}

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 \(x\ge-\frac{15}{7}\)

\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)

\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)

\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)

\(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(x+\dfrac{3}{2}y\right)=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)=\dfrac{9}{4}y^2-x^2\)

\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

\(\left(x-y+z\right)\left(x+y-z\right)=x^2-\left(y-z\right)^2=x^2-y^2-z^2+2yz\)

cảm ơn bạn