\(a.2x-3=4x+6\) 

\(\Leftrightarrow2x-3-4x-6=0\) 

\(\Leftrightarrow-2x-9=0\)

\(\Leftrightarrow x=\dfrac{9}{2}\)

\(S=\left\{\dfrac{9}{2}\right\}\) 

\(b.x\left(x-1\right)+x\left(x+3\right)=0\) 

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

\(\Leftrightarrow2x^2+2=0\)

\(\Leftrightarrow x\left(2x+2\right)=0\) 

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

\(S=\left\{0,-1\right\}\) 

Mấy câu khác bn gửi lại đc ko tại mik chx hiểu lắm

a: =>-2x=9

=>x=-9/2

c: =>x(x-1+x+3)=0

=>x(2x+2)=0

=>x=0 hoặc x=-1

Bài 2: 

a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

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

\(a,\dfrac{x-3}{4}+\dfrac{2x-1}{3}=-\dfrac{x}{6}\)

\(\Leftrightarrow\dfrac{3\left(x-3\right)+4\left(2x-1\right)+2x}{12}=0\)

\(\Leftrightarrow3x-9+8x-4+2x=0\)

\(\Leftrightarrow13x-13=0\)

\(\Leftrightarrow13x=13\)

\(\Leftrightarrow x=1\)

\(b,\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\-x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)

a, x-3/4+2x-1/3=-x/6

b,(x-3)(2x-1)=(2x-1)(2x+3)

c,6/x-1-4/x-3+8/x^2-4x+3=0

a) \(\left|4-x\right|+2x=3\)

<=> \(\left|4-x\right|=3-2x\)

<=> \(\orbr{\begin{cases}4-x=3-2x\left(x\le4\right)\\x-4=3-2x\left(x>4\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-1\left(tm\right)\\3x=7\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-1\\x=\frac{7}{3}\left(ktm\right)\end{cases}}\)

Vậy x = -1

b) \(\left|x-7\right|+2x+5=6\)

<=> \(\left|x-7\right|=1-2x\)

<=> \(\orbr{\begin{cases}x-7=1-2x\left(đk:x\ge7\right)\\x-7=2x-1\left(đk:x< 7\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}3x=8\\x=-6\left(tm\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{8}{3}\left(ktm\right)\\x=-6\left(tm\right)\end{cases}}\)

Vậy x = -6

c) \(3x-\left|2x+1\right|=2\)

<=> \(\left|2x+1\right|=3x-2\)

<=> \(\orbr{\begin{cases}2x+1=3x-2\left(đk:x\ge-\frac{1}{2}\right)\\2x+1=2-3x\left(đk:x< -\frac{1}{2}\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}x=3\left(tm\right)\\5x=1\end{cases}}\)

<=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{5}\left(ktm\right)\end{cases}}\)

Vậy x = 3

d) \(\left|x+2\right|-x=2\)

<=> \(\left|x+2\right|=x+2\)

<=> \(\orbr{\begin{cases}x+2=x+2\left(đk:x\ge-2\right)\\x+2=-x-2\left(x< -2\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}0x=0\\2x=-4\end{cases}}\)

<=> 0x = 0 (luôn đúng) và x = -2 (ktm)

Vậy x \(\ge\)-2

e) \(\left|x-3\right|=21\)

<=> \(\orbr{\begin{cases}x-3=21\\3-x=21\end{cases}}\)

<=> \(\orbr{\begin{cases}x=24\\x=-18\end{cases}}\)

Vậy x = 24 hoặc x = -18

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

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

<=> \(\orbr{\begin{cases}2x+3=x-3\\2x+3=3-x\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-6\\3x=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-6\\x=0\end{cases}}\)

Vậy x thuộc {-6; 0}

g) Ta có: \(\left|x+\frac{1}{8}\right|\ge0\forall x\)

          \(\left|x+\frac{2}{8}\right|\ge0\forall x\)

    \(\left|x+\frac{5}{8}\right|\ge0\forall x\)

=> VT = \(\left|x+\frac{1}{8}\right|+\left|x+\frac{2}{8}\right|+\left|x+\frac{5}{8}\right|\ge0\forall x\)

=> VP \(\ge0\) => \(4x\ge0\) => \(x\ge0\)

Do đó: \(x+\frac{1}{8}+x+\frac{2}{8}+x+\frac{5}{8}=4x\)

<=> \(3x+1=4x\) <=> \(x=1\left(tm\right)\)

Vậy x = 1

h) \(\left|x-2\right|-\left|2x+3\right|-x=-2\)

<=> \(\left|x-2\right|-\left|2x+3\right|=x-2\)(*)

Lập bảng xét dấu: 

x                     -3/2              2

x - 2        2 - x    |        2 - x    0        x - 2

2x + 3  -2x - 3   0      2x + 3  |          2x + 3

Xét x < -3/2 => pt (*) trở thành: 2 - x + 2x + 3 = x - 2

<=> x + 5 = x - 2 <=> 0x = -7 (vô lí)

Xét -3/2 \(\le\) x < 2 => pt (*) trở thành: 2 - x - 2x - 3 = x - 2

<=> 4x = 1 <=> x = 1/4 ((tm)

Xét x \(\ge\) 2 => pt (*) trở thành x - 2 - 2x - 3 = x - 2

<=> 2x = -3 <=>  x = -3/2 (ktm)

Vậy x = 1/4

i) |2x - 3| - x = |2 - x|

<=> |2x - 3| - |2 - x| = x (*)

Lập bảng xét dấu

x                    3/2               2

2x - 3   3 - 2x   0     2x - 3   |  2x - 3

2 - x     2 - x     |       2 - x    0   x - 2

Xét x < 3/2 => pt (*) trở thành: 3 - 2x - 2 + x =  x

<=> 2x = 1 <=> x = 1//2 ((tm)
Xét \(\frac{3}{2}\le x< 2\)=> pt (*) trở thành: 2x - 3 - 2 + x = x

<=> 2x = 5 <=> x = 5/2 (ktm)

Xét x \(\ge\)2 ==> pt (*) trở thành: 2x - 3 - x + 2 = x

<=> 0x = -5 (vô lí)

Vậy x = 1/2

k) 2|x - 3| - |4x - 1| = 0

<=> 2|x - 3| = |4x - 1|

<=> \(\orbr{\begin{cases}2\left(x-3\right)=4x-1\\2\left(x-3\right)=1-4x\end{cases}}\)

<=> \(\orbr{\begin{cases}2x-6=4x-1\\2x-6=1-4x\end{cases}}\)

<=> \(\orbr{\begin{cases}2x=-5\\6x=7\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-\frac{5}{2}\\x=\frac{7}{6}\end{cases}}\) Vậy ...

Giải pt :

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

\(\Leftrightarrow2x^2+10x-x^2+6x-9-x^2-6=0\)

\(\Leftrightarrow16x-15=0\)

\(\Leftrightarrow x=\frac{15}{16}\)

b) \(6\left(x-3\right)+\left(x-1\right)^2-\left(x+1\right)^2=2x\)

\(\Leftrightarrow2x-18=2x\)

\(\Leftrightarrow-18=0\)( vô lí )

=> x thuộc rỗng

c)d) tương tự

e) \(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)

\(\Leftrightarrow\frac{5x-2}{6}+\frac{9-12x}{6}=\frac{12}{6}-\frac{2x+14}{6}\)

\(\Leftrightarrow5x-2+9-12x=12-2x-14\)

\(\Leftrightarrow-5x+9=0\)

\(\Leftrightarrow x=\frac{9}{5}\)

f) \(\frac{2x-1}{2}=\frac{2x+1}{4}-\frac{1-2x}{8}\)

\(\Leftrightarrow\frac{4\left(2x-1\right)}{8}=\frac{2\left(2x+1\right)}{8}-\frac{1-2x}{8}\)

\(\Leftrightarrow8x-4=4x+2-1+2x\)

\(\Leftrightarrow2x-5=0\)

\(\Leftrightarrow x=\frac{5}{2}\)

Tìm x :

a) \(3x^3-27x=0\)

\(\Leftrightarrow3x\left(x^2-9\right)=0\)

\(\Leftrightarrow3x\left(x-3\right)\left(x+3\right)=0\)

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

b) \(2x^3-12x^2+18x=0\)

\(\Leftrightarrow2x\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow2x\left(x-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

2)  \(x^3-6x^2+11x-6=0\)

\(\Leftrightarrow\)\(x^3-3x^2-3x^2+9x+2x-6=0\)

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

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

bn giải tiếp nha

3)   \(x^3-4x^2+x+6=0\)

\(\Leftrightarrow\)\(x^3-3x^2-x^2+3x-2x+6=0\)

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

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

lm tiếp nha

4)  \(x^3-3x^2+4=0\)

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

\(\Leftrightarrow\)\(\left(x+1\right)\left(x^2-4x+4\right)=0\)

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

lm tiếp nha

Mk làm mẫu 1 bài cho nha !

1. <=> (x^3-x^2)+(5x^2-5x)+(6x-6) = 0

<=> (x-1).(x^2+5x+6) = 0

<=> (x-1).[(x^2+2x)+(3x+6)] = 0

<=> (x-1).(x+2).(x+3) = 0

<=> x-1=0 hoặc x+2=0 hoặc x+3=0

<=> x=1 hoặc x=-2 hoặc x=-3

Vậy ..............

Tk mk nha

2.  x3−6x2+11x−6=0

⇔x3−3x2−3x2+9x+2x−6=0

⇔(x−3)(x2−3x+2)=0

⇔(x−3)(x−2)(x−1)=0

bn giải tiếp nha

3)   x3−4x2+x+6=0

⇔x3−3x2−x2+3x−2x+6=0

⇔(x−3)(x2−x−2)=0

⇔(x−3)(x−2)(x+1)=0

lm tiếp nha

4)  x3−3x2+4=0

⇔x3+x2−4x2−4x+4x+4=0

⇔(x+1)(x2−4x+4)=0

⇔(x+1)(x−2)2=0

lm tiếp nha

a) 3x(x - 1) + 2(x - 1) = 0

<=> (3x + 2)(x - 1) = 0

<=> \(\orbr{\begin{cases}3x+2=0\\x-1=0\end{cases}}\)

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

Vậy S = {-2/3; 1}

b) x² - 1 - (x + 5)(2 - x) = 0

<=> x² - 1 - 2x + x² - 10 + 5x = 0

<=> 2x² + 3x - 11 = 0

<=> 2(x² + 3/2x + 9/16 - 97/16) = 0

<=> (x + 3/4)² - 97/16 = 0

<=> \(\orbr{\begin{cases}x+\frac{3}{4}=\frac{\sqrt{97}}{4}\\x+\frac{3}{4}=-\frac{\sqrt{97}}{4}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{\sqrt{97}-3}{4}\\x=-\frac{\sqrt{97}-3}{4}\end{cases}}\)

Vậy S = {\(\frac{\sqrt{97}-3}{4}\); \(-\frac{\sqrt{97}-3}{4}\)

d) x(2x - 3) - 4x + 6 = 0

<=> x(2x - 3) - 2(2x - 3) = 0

<=> (x - 2)(2x - 3) = 0

<=> \(\orbr{\begin{cases}x-2=0\\2x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\\x=\frac{3}{2}\end{cases}}\)

Vậy  S = {2; 3/2}

e)  x³ - 1 = x(x - 1)

<=> (x - 1)(x² + x + 1) - x(x - 1) = 0

<=> (x - 1)(x² + x +  1 - x) = 0

<=> (x - 1)(x² + 1) = 0

<=> x - 1 = 0

<=> x = 1

Vậy S = {1}

f) (2x - 5)² - x² - 4x - 4 = 0

<=> (2x - 5)² - (x + 2)² = 0

<=> (2x - 5 - x - 2)(2x - 5 + x + 2) = 0

<=> (x - 7)(3x - 3) = 0

<=> \(\orbr{\begin{cases}x-7=0\\3x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=7\\x=1\end{cases}}\)

Vậy S = {7; 1}

h) (x - 2)(x² + 3x - 2) - x³ + 8 = 0

<=> (x - 2)(x² + 3x - 2) - (x- 2)(x² + 2x + 4) = 0

<=> (x - 2)(x² + 3x - 2 - x² - 2x - 4) = 0

<=> (x - 2)(x - 6) = 0

<=> \(\orbr{\begin{cases}x-2=0\\x-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\\x=6\end{cases}}\)

Vậy S = {2; 6}

\(a,3x\left(x-1\right)+2\left(x-1\right)=0\)

\(3x.x-3x+2x-2=0\)

\(2x-2=0\)

\(2x=2\)

\(x=1\)

\(a,\left(x-2\right)\left(x-3\right)-3\left(4x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2-5x+6-12x+6=x^2-8x+16\\ \Leftrightarrow-9x-4=0\\ \Leftrightarrow x=-\dfrac{4}{9}\)

\(b,\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\\ \Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\\ \Leftrightarrow10x=1\\ \Leftrightarrow x=\dfrac{1}{10}\)

\(c,x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\\ \Leftrightarrow30x-12x+12+5x+40=210+10x-10\\ \Leftrightarrow13x=148\\ \Leftrightarrow x=\dfrac{148}{13}\)

\(d,\left(2x+5\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)

\(e,x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

\(g,2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)

\(h,\left(x+\dfrac{1}{x}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\left(x\ne0\right)\)

Đặt \(x+\dfrac{1}{x}=t\), pt trở thành:

\(t^2+2t-8=0\\ \Leftrightarrow\left(t-2\right)\left(t+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1-2x=0\\x^2+1+4x=0\left(1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\Delta\left(1\right)=16-4=12>0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\)

Tick plzz

a: Ta có: \(\left(x-2\right)\left(x+3\right)-3\left(4x-2\right)=\left(x-4\right)^2\)

\(\Leftrightarrow x^2+x-6-12x+6-x^2+8x-16=0\)

\(\Leftrightarrow-3x=16\)

hay \(x=-\dfrac{16}{3}\)

b: Ta có: \(\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\)

\(\Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\)

\(\Leftrightarrow-14x+7+4x-6=0\)

\(\Leftrightarrow10x=1\)

hay \(x=\dfrac{1}{10}\)

c: Ta có: \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)

\(\Leftrightarrow30x-12x+30+5x+40=210+10x-10\)

\(\Leftrightarrow23x+70=10x+200\)

\(\Leftrightarrow x=10\)

roois vãi

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