`5-(x-6)=4(3-2x)`

`<=>5-x+6-4(3-2x)=0`

`<=> 5-x+6-12 +8x=0`

`<=> 7x -1=0`

`<=> 7x=1`

`<=>x=1/7`

Vậy pt đã cho có nghiệm `x=1/7`

__

`3-x(1-3x) =5(1-2x)`

`<=> 3-x+3x^2=5-10x`

`<=> 3-x+3x^2-5+10x=0`

`<=> 3x^2 +9x-2=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9+\sqrt{105}}{6}\\x=\dfrac{-9-\sqrt{105}}{6}\end{matrix}\right.\)

Vậy pt đã cho có tập nghiệm \(S=\left\{\dfrac{-9+\sqrt{105}}{6};\dfrac{-9-\sqrt{106}}{5}\right\}\)

__

`(x-3)(x+4) -2(3x-2)=(x-4)^2`

`<=>x^2+4x-3x-12- 6x +4 =x^2 -8x+16`

`<=>x^2-5x-8=x^2-8x+16`

`<=> x^2 -5x-8-x^2+8x-16=0`

`<=> 3x-24=0`

`<=>3x=24`

`<=>x=8`

Vậy pt đã cho có nghiệm `x=8`

a) 5-(x-6)=4(3-2x)

=> 5 – x + 6 = 12 – 8x

=> -x + 8x = 12 – 5 – 6

=> 7x = 1

=> x=1/7

Vậy phương trình có nghiệm x=1/7

 b) 3 - x ( 1 - 3x)=5(1-2x)

=> 3-x+3x^2=5-10x

=> 3x^2+9x-2= 0

0=105

=> x =\(\dfrac{-9-\sqrt{105}}{6}\)

\(a,4\left(x-3\right)^2-\left(2x-1\right)^2< 10\)

\(\Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-4x+1\right)-10< 0\)
\(\Leftrightarrow4x^2-24x+36-4x^2+4x-1-10< 0\)

\(\Leftrightarrow-20x< -25\)

\(\Leftrightarrow x>\dfrac{5}{4}\)

\(b,x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)\le3\)

\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)\le3\)

\(\Leftrightarrow x^3-25x-\left(x^3+8\right)\le3\)

\(\Leftrightarrow x^3-25x-x^3-8-3\le0\)

\(\Leftrightarrow-25x\le11\)

\(\Leftrightarrow x\ge-\dfrac{11}{25}\)

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

\(x-5=\frac{1}{3\left(x+2\right)}\left(đkxđ:x\ne-2\right)\)

\(< =>3\left(x-5\right)\left(x+2\right)=1\)

\(< =>3\left(x^2-3x-10\right)=1\)

\(< =>x^2-3x-10=\frac{1}{3}\)

\(< =>x^2-3x-\frac{31}{3}=0\)

giải pt bậc 2 dễ r

\(\frac{x}{3}+\frac{x}{4}=\frac{x}{5}-\frac{x}{6}\)

\(< =>\frac{4x+3x}{12}=\frac{6x-5x}{30}\)

\(< =>\frac{7x}{12}=\frac{x}{30}< =>12x=210x\)

\(< =>x\left(210-12\right)=0< =>x=0\)

\(3x\left(2x-3\right)-3\left(3+2x^2\right)=0\)

\(< =>6x^2-9x-9-6x^2=0\)

\(< =>-9x-9=0< =>9x+9=0\)

\(< =>x=-\frac{9}{9}=-1\)

Câu 1 : 

a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)

\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)

\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)

Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)

tương tự 

\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)

\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)

\(< =>95-24x+40=6-4x-15x+5\)

\(< =>-24x+135=-19x+11\)

\(< =>5x=135-11=124\)

\(< =>x=\frac{124}{5}\)

\(\frac{\left(x-2\right).3}{2}+3+\frac{\left(x-3\right).5}{3}+5+\frac{\left(x-5\right).2}{5}+2=10\)

\(< =>\frac{\left(x-2\right).3.15}{30}+\frac{\left(x-3\right).5.10}{30}+\frac{\left(x-5\right).2.6}{30}=10-2-3-5\)

\(< =>\frac{\left(x-2\right).45+\left(x-3\right).50+\left(x-5\right).12}{30}=0\)

\(< =>45x-90+50x-150+12x-60=0\)

\(< =>107x-300=0< =>x=\frac{300}{107}\)

khó quá

khó gì fan gao bạc

khó quá

a) \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x-4\right)\left(x+4\right)\le10\)

\(\Leftrightarrow5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-16\right)\le10\)

\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-240\le10\)

\(\Leftrightarrow\left(5x^3-5x^3\right)-\left(30x^2-15x^2-15x^2\right)-\left(45x-15x\right)+5-240\le10\)

\(\Leftrightarrow30x-235\le10\)

\(\Leftrightarrow30x\le10+235\)

\(\Leftrightarrow30x\le245\)

\(\Leftrightarrow30x:30\le245:30\)

\(\Leftrightarrow x\le\dfrac{49}{6}\)

Vậy nghiệm của bất phương trình là: \(x\le\dfrac{49}{6}\)

b) \(\left(3x-2\right)\left(9x^2+6x+4\right)+27x\left(\dfrac{1}{3}-x\right)\left(\dfrac{1}{2}+x\right)\ge1\)

\(\Leftrightarrow27x^3-8+27x\left(\dfrac{1}{9}-x^2\right)\ge1\)

\(\Leftrightarrow27x^3-8+3x-27x^3\ge1\)

\(\Leftrightarrow\left(27x^3-27x^3\right)-8+3x\ge1\)

\(\Leftrightarrow-8+3x\ge1\)

\(\Leftrightarrow3x\ge1+8\)

\(\Leftrightarrow3x\ge9\)

\(\Leftrightarrow3x:3\ge9:3\)

\(\Leftrightarrow x\ge3\)

Vậy nghiệm của bất phương trình là \(x\ge3\)

a: =>5x(x^2-6x+9)-5(x^3-3x^2+3x-1)+15(x^2-16)<=10

=>5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-240<=10

=>30x-235<=10

=>30x<=245

=>x<=49/6

b: =>27x^3-8+27x(1/9-x^2)>=1

=>27x^3-8+3x-27x^3>=1

=>3x>=9

=>x>=3

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

la`thu'hai nga`y 19 nhe

a, \(x-5=\frac{1}{3}\left(x+2\right)\)

\(\Leftrightarrow\frac{3x-15}{3}=\frac{x+2}{3}\Leftrightarrow\frac{3x-15-x-2}{3}=0\)

\(\Leftrightarrow2x-17=0\Leftrightarrow x=\frac{17}{2}\)

b, \(\frac{x}{3}+\frac{x}{4}=\frac{1}{5}-\frac{x}{6}\)

\(\Leftrightarrow\frac{2x}{6}+\frac{x}{6}=\frac{4}{20}-\frac{5x}{20}\Leftrightarrow\frac{x}{2}=\frac{4-5x}{20}\)

\(\Leftrightarrow\frac{10x}{20}-\frac{4-5x}{20}=0\Leftrightarrow15x-4=0\Leftrightarrow x=\frac{4}{15}\)

a, x - 5 = \(\frac{1}{3}\).(x + 2) 

<=> x - 5 = \(\frac{1}{3}\)x + \(\frac{2}{3}\)

<=> x - 5 - \(\frac{1}{3}\)x - \(\frac{2}{3}\)= 0

<=>\(\frac{2}{3}\)x  - \(\frac{17}{3}\)= 0

<=>x = \(\frac{17}{2}\)

a) Ta có: \(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow7-2x-4=-x-4\)

\(\Leftrightarrow-2x+3+x+4=0\)

\(\Leftrightarrow-x+7=0\)

\(\Leftrightarrow-x=-7\)

hay x=7

Vậy: S={7}

b) Ta có: \(\dfrac{2+x}{5}-0.5x=\dfrac{1-2x}{4}+0.25\)

\(\Leftrightarrow\dfrac{4\left(2+x\right)}{20}-\dfrac{0.5x\cdot20}{20}=\dfrac{5\left(1-2x\right)}{20}+\dfrac{20\cdot0.25}{20}\)

\(\Leftrightarrow4\left(2+x\right)-10x=5\left(1-2x\right)+5\)

\(\Leftrightarrow8+4x-10x=5-10x+5\)

\(\Leftrightarrow-6x+8=-10x+10\)

\(\Leftrightarrow-6x+8+10x-10=0\)

\(\Leftrightarrow4x-2=0\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

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

d) Ta có: \(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-59}{1}+\dfrac{x-58}{2}+\dfrac{x-57}{3}\)

\(\Leftrightarrow\dfrac{x-1}{59}-1+\dfrac{x-2}{58}-1+\dfrac{x-3}{57}-1=\dfrac{x-59}{1}-1+\dfrac{x-58}{2}-1+\dfrac{x-57}{3}-1\)

\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}=\dfrac{x-60}{1}+\dfrac{x-60}{2}+\dfrac{x-60}{3}\)

\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}\right)-\left(x-60\right)\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)

mà \(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\ne0\)

nên x-60=0

hay x=60

Vậy: S={60}