a: =>-2x=-8

hay x=4

b: =>7x=-21

hay x=-3

c: =>0,25x=-1,5

hay x=-6

d: =>5,3x=6,36

hay x=6/5

e: =>-4x=-12

hay x=3

f: =>-10x=-10

hay x=1

g: =>2x+2-3-2x=0

=>-1=0(vô lý)

h: =>3-3x+4x-3=0

=>x=0

a,

\(3-x=x-5\\ \Leftrightarrow3x-x+5=0\Leftrightarrow2x+5=0\)

\(\Rightarrow x=-\dfrac{5}{2}\)

b, \(\Rightarrow x=-\dfrac{21}{7}=-3\)

c, \(\Leftrightarrow x=\left(0-1,5\right):0,25=-6\)

a. <=> 2x=8 hay x=4

b.<=> x= -21/7 = -3

c. <=> x= -1,5/ 0,25=-6

d. <=> x= -6,36/-5,3=1,2

e.<=> 4x=12 hay x= 3

f. <=> 10x = 10 hay x = 1

g. <=> 2x +2 = 3 + 2x

<=> 2=3 ( vô lí )

h.<=> 3 - 3x + 4x -3 =0

<=> x=0

Bài 1:

a) Ta có: 7x+12=0

\(\Leftrightarrow7x=-12\)

hay \(x=-\frac{12}{7}\)

Vậy: \(x=-\frac{12}{7}\)

b) Ta có: 5x-2=0

\(\Leftrightarrow5x=2\)

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

Vậy: \(x=\frac{2}{5}\)

c) Ta có: 12-6x=0

\(\Leftrightarrow6x=12\)

hay x=2

Vậy: x=2

d) Ta có: -2x+14=0

⇔-2x=-14

hay x=7

Vậy: x=7

Bài 2:

a) Ta có: 3x+1=7x-11

⇔3x+1-7x+11=0

⇔-4x+12=0

⇔-4x=-12

hay x=3

Vậy: x=3

b) Ta có: 2x+x+12=0

⇔3x+12=0

⇔3x=-12

hay x=-4

Vậy: x=-4

c) Ta có: x-5=3-x

⇔x-5-3+x=0

⇔2x-8=0

⇔2x=8

hay x=4

Vậy: x=4

d) Ta có: 7-3x=9-x

⇔7-3x-9+x=0

⇔-2x-2=0

⇔-2x=2

hay x=-1

Vậy: x=-1

e) Ta có: 5-3x=6x+7

⇔5-3x-6x-7=0

⇔-9x-2=0

⇔-9x=2

hay \(x=\frac{-2}{9}\)

Vậy: \(x=\frac{-2}{9}\)

f) Ta có: 11-2x=x-1

⇔11-2x-x+1=0

⇔12-3x=0

⇔3x=12

hay x=4

Vậy: x=4

g) Ta có: 15-8x=9-5

⇔15-8x=4

⇔8x=11

hay \(x=\frac{11}{8}\)

Vậy: \(x=\frac{11}{8}\)

Bài 3:

a) Ta có: 0,25x+1,5=0

⇔0,25x=-1,5

hay x=-6

Vậy: x=-6

b) Ta có: 6,36-5,2x=0

⇔5,2x=6,36

hay \(x=\frac{159}{130}\)

Vậy: \(x=\frac{159}{130}\)

d) \(4x^2-9-x\left(2x-3\right)=0\)

\(\Leftrightarrow4x^2-9-2x^2+3x=0\)

\(\Leftrightarrow2x^2+3x-9=0\)

\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)

e) \(x^3+5x^2+9x=-45\)

\(\Leftrightarrow x^3+5x^2+9x+45=0\)

\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)

\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)

f) \(x^3-6x^2-x+30=0\)

\(\Leftrightarrow\left(x^3-x^2-6x\right)-\left(5x^2-5x-30\right)=0\)

\(\Leftrightarrow x\left(x^2-x-6\right)-5\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+3x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left[x\left(x-2\right)+3\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow x\in\left\{5;-3;2\right\}\)

a)  \(4\left(2x+7\right)^2=9\left(x+3\right)^2\)

\(\Leftrightarrow4\left(4x^2+28x+49\right)=9\left(x^2+6x+9\right)\)

\(\Leftrightarrow16x^2+112x+196=9x^2+54x+81\)

\(\Leftrightarrow7x^2+58x+115=0\)

\(\Leftrightarrow7x^2+35x+23x+115=0\)

\(\Leftrightarrow7x\left(x+5\right)+23\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\7x+23=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-\frac{23}{7}\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-5;-\frac{23}{7}\right\}\)

b) \(2x^3+7x^2+7x+2=0\)

\(\Leftrightarrow2x^3+2x^2+5x^2+5x+2x+2=0\)

\(\Leftrightarrow2x^2\left(x+1\right)+5x\left(x+1\right)+2\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2+5x+2\right)=0\)

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

\(\Leftrightarrow\left(x+1\right)\left[2x\left(x+2\right)+\left(x+2\right)\right]=0\)

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

\(\Leftrightarrow\)\(x+1=0\)

hoặc \(2x+1=0\)

hoặc  \(x+2=0\)

\(\Leftrightarrow\) \(x=-1\)

hoặc    \(x=-\frac{1}{2}\)

hoặc    \(x=-2\)

 Vậy tập nghiệm của phương trình là \(S=\left\{-1;-\frac{1}{2};-2\right\}\)

c) \(x^4+x^2+6x-8=0\)

\(\Leftrightarrow x^4-x^3+x^3-x^2+2x^2-2x+8x-8=0\)

\(\Leftrightarrow x^3\left(x-1\right)+x^2\left(x-1\right)+2x\left(x-1\right)+8\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+2x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2-x^2-2x+4x+8\right)=0\)

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

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

\(\Leftrightarrow\)\(x-1=0\)

hoặc   \(x+2=0\)

hoặc   \(x^2-x+4=0\)

\(\Leftrightarrow\)\(x=1\)(tm)

hoặc   \(x=-2\)(tm)

hoặc  \(\left(x-\frac{1}{2}\right)^2+\frac{15}{4}=0\)(ktm)

Vậy tập nghiệm của phương trình là \(S=\left\{1;-2\right\}\)

d) \(\left(x-1\right)^3+\left(2x+3\right)^3=27x^3+8\)

\(\Leftrightarrow x^3-3x^2+3x-1+8x^3+36x^2+54x+27=27x^3+8\)

\(\Leftrightarrow9x^3+33x^2+57x+26=27x^3+8\)

\(\Leftrightarrow18x^3-33x^2-57x-18=0\)

\(\Leftrightarrow18x^3-54x^2+21x^2-63x+6x-18=0\)

\(\Leftrightarrow18x^2\left(x-3\right)+21x\left(x-3\right)+6\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(18x^2+21x+6\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(18x^2+9x+12x+6\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[9x\left(2x+1\right)+6\left(2x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)\left(9x+6\right)=0\)

\(\Leftrightarrow\)\(x-3=0\)

hoặc  \(2x+1=0\)

hoặc  \(9x+6=0\)

\(\Leftrightarrow\)\(x=3\)

hoặc  \(x=-\frac{1}{2}\)

hoặc \(x=-\frac{2}{3}\)

Vậy tập nghiệm của phương trình là \(S=\left\{3;-\frac{1}{2};-\frac{2}{3}\right\}\)

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

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

Vậy....

hk tốt

^^

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

    x\(^2\)- x -x\(^{^2}\)-2x +x+2=0

     -2x+2=0

      -2x=0+2

       -2x=2

         x=-1

Vậy x bằng -1

a.

a). 3. |9 - 2x| - 17 = 16

3. |9 - 2x| = 16 + 17

3. |9 - 2x| = 33

|9 - 2x| = 33 : 3

|9 - 2x| = 11

=> 9 - 2x = 11

2x = 9 - 11

2x = -2

x = - 2 : 2

x = - 1

hay 9 - 2x = - 11

2x = 9 - (- 11)

2x = 9 + 11

2x = 20

x = 20 : 2

x = 10

Vậy x = -1; x = 10

a) 3.| 9 - 2x | -17 = 16

    3. | 9 - 2x |      = 16 + 17 = 33

        | 9 - 2x |      = 33 : 3 = 11

  \(\Rightarrow\)9 - 2x = 11                hoặc                9 - 2x = -11

               2x  = 9 - 11                                       2x = 9 - ( - 11 )

               2x  = -2                                            2x  = 20

                  x = -2 : 2                                           x = 20 : 2

                   x = -1                                               x  = 10

b). 3 - 4 |5 - 6x| = 7

4 |5 - 6x| = 3 - 7

4 |5 - 6x| = - 4

|5 - 6x| = - 4 : 4

|5 - 6x| = -1

Mà |5 - 6x| luôn lớn hơn 0 với mọi x

Do đó, x không tìm được giá trị

a) \(\left(4x^2-25\right)\left(2x^2-7x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x^2-25=0\left(1\right)\\2x^2-7x-9=0\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x^2=\frac{25}{4}\Leftrightarrow x=\pm\frac{5}{2}\)

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

\(\Leftrightarrow2x\left(x+1\right)-9\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{9}{2}\end{matrix}\right.\)

Vậy....

b) \(\left(2x^2-3\right)^2-4\left(x-1\right)^2=0\)

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

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

\(\Leftrightarrow\left(2x^2-2x-1\right)\left(2x^2+2x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2-2x-1=0\left(3\right)\\2x^2+2x-5=0\left(4\right)\end{matrix}\right.\)

\(\left(3\right)\Delta=2^2-4\cdot2\cdot\left(-1\right)=12\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2-\sqrt{12}}{4}=\frac{1-\sqrt{3}}{2}\\x=\frac{2+\sqrt{12}}{4}=\frac{1+\sqrt{3}}{2}\end{matrix}\right.\)

\(\left(4\right)\Delta=2^2-4\cdot2\cdot\left(-5\right)=44\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2-\sqrt{44}}{4}=\frac{-1-\sqrt{11}}{2}\\x=\frac{-2+\sqrt{44}}{4}=\frac{-1+\sqrt{11}}{2}\end{matrix}\right.\)

Vậy...

c) \(x^3+5x^2+7x+3=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)

Vậy...

d) \(x^3-6x^2+11x-6=0\)

\(\Leftrightarrow x^3-2x^2-4x^2+8x+3x-6=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=3\end{matrix}\right.\)

Vậy...