a: Ta có: \(7x+25=144\)

\(\Leftrightarrow7x=119\)

hay x=17

b: Ta có: \(33-12x=9\)

\(\Leftrightarrow12x=24\)

hay x=2

c: Ta có: \(128-3\left(x+4\right)=23\)

\(\Leftrightarrow3\left(x+4\right)=105\)

\(\Leftrightarrow x+4=35\)

hay x=31

d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)

\(\Leftrightarrow5\left(726-3x\right)=2175\)

\(\Leftrightarrow726-3x=435\)

\(\Leftrightarrow3x=291\)

hay x=97

e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)

\(\Leftrightarrow41-\left(2x+5\right)=18\)

\(\Leftrightarrow2x+5=23\)

\(\Leftrightarrow2x=18\)

hay x=9

Bn cần bài nào vậy

f: Ta có: \(10-4x+120:8=16+1\)

\(\Leftrightarrow4x=17-25=-8\)

hay x=-2

g: Ta có: \(x+9x+7x+5x=2244\)

\(\Leftrightarrow22x=2244\)

hay x=102

h: Ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)

\(\Leftrightarrow100x+5050=5750\)

\(\Leftrightarrow100x=700\)

hay x=7

a) (7x - 11)³ = 2⁵ x 5² + 200

     (7x - 11)³ = 800 + 200

     (7x - 11)³ = 1000

     (7x - 11)³ = 10³

=>     7x - 11 = 10

=>            7x = 10 + 11

=>             7x = 21

=>                x =  3

b) \(3\frac{1}{3}x+16\frac{3}{4}=-13,25\)

                     \(3\frac{1}{3}x=-13,25-16\frac{3}{4}\)

                     \(\frac{10}{3}x=-30\)

                             \(x=-9\)

a, x= 3

b, -9

đáp số 

a=3

b=-9

hok tốt

Bài 3: 

a: x=-15

b: =>2x=18

hay x=9

\(a,3\frac{1}{3}x+16\frac{3}{4}=-13,25\)

       \(\frac{10}{3}x+\frac{67}{4}=-13,25\)

                    \(\frac{10}{3}x=-13,25-\frac{67}{4}\)

                    \(\frac{10}{3}x=-30\)

                            \(x=\left(-30\right):\frac{10}{3}\)

                            \(x=-9\)

\(b,\left(7x-11\right)^3=2^5.5^2+200\)

    \(\left(7x-11\right)^3=32.25+200\)

     \(\left(7x-11\right)^3=1000\)

 \(\Rightarrow\left(7x-11\right)^3=10^3\)

 \(\Rightarrow7x-11=10\)

\(\Rightarrow7x=10+11\)

\(\Rightarrow7x=21\)

\(\Rightarrow x=3\)

A) (10/3)x+67/4=-53/4<=>(10/3)x=-53/4-67/4=-30<=>x=-30:(10/3)=-9 b) (7x-11)^3=1000=10^3<=>7x-11=10=>7x=21=>x=3

​

Cảm ơn bạn nhiều

9) Ta có: \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)

\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)

\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3-4-3x^2-10x+x+3=0\)

\(\Leftrightarrow-4x=9\)

hay \(x=-\dfrac{9}{4}\)

10) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3-7x}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}

11) Ta có: \(\dfrac{5+9x}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)

\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{9x+5}{\left(x-4\right)\left(x+5\right)}\)

Suy ra: \(2x^2-9x+4+3x^2+12x-x-4-9x-5=0\)

\(\Leftrightarrow5x^2-7x=0\)

\(\Leftrightarrow x\left(5x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)

12) Ta có: \(\dfrac{2x}{2x-1}+\dfrac{x}{2x+1}=1+\dfrac{4}{\left(2x-1\right)\left(2x+1\right)}\)

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

Suy ra: \(4x^2+2x+2x^2-x-4x^2-3=0\)

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

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

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

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

13) Ta có: \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\)

\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)

Suy ra: \(x^2+2x-x+2-2=0\)

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

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

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

......

Bài 1: Tìm x

a) Ta có: \(\left(7x-11\right)^3=2^5\cdot5^2+200\)

\(\Leftrightarrow\left(7x-11\right)^3=32\cdot25+200=1000\)

\(\Leftrightarrow7x-11=10\)

\(\Leftrightarrow7x=21\)

hay x=3

Vậy: x=3

b) Ta có: \(720:\left(41-\left(2x-5\right)\right)=2^3\cdot5\)

\(\Leftrightarrow720:\left(41-2x+5\right)=8\cdot5=40\)

\(\Leftrightarrow-2x+46=18\)

\(\Leftrightarrow-2x=18-46=-28\)

hay x=14

Vậy: x=14

c) Ta có: \(2\cdot3^x=10\cdot3^{12}+8\cdot27^4\)

\(\Leftrightarrow2\cdot3^x=10\cdot3^{12}+8\cdot3^{12}\)

\(\Leftrightarrow2\cdot3^x=18\cdot3^{12}\)

\(\Leftrightarrow3^x=\dfrac{18\cdot3^{12}}{2}=9\cdot3^{12}\)

\(\Leftrightarrow3^x=3^2\cdot3^{12}=3^{14}\)

hay x=14

Vậy: x=14

A (7x-11)³=2⁵.5²+200

    (7x-11)³=800+200

    (7x-11)³=1000

    (7x-11)³=10³

    7x-11=10

     7x=10+11

     7x=21

       x=21:7

       x=3

B 720:(41-(2x-5))=2³.5

   720:(41-(2x-5))=40

             41-(2x-5)=720:40

             41-(2x-5)=18

                    2x-5=41-18

                    2x-5=23

                       2x=23+5

                       2x=28

                         x=28:2

                         x=14

 C 2.3^x=10.3¹²+8.27⁴

    2.3^x=10.3¹²+8.(3^3.4)

      2.3^x=10.3¹²+8.3¹²

      2.3^x=3¹².(10+8)

     2.3^x=3¹².18

        3^x=( 3¹².18):2

        3^x=3¹².9

        3^x=3¹² .3²

        3^x=3¹⁴

          3^x=14

        x=14