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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

a) \(3\left(x-1\right)^2\cdot3x\left(x-5\right)=0\)

\(\Rightarrow9x\left(x-1\right)^2\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=5\end{matrix}\right.\)

b) \(\left(x+3\right)^2-5x-15=0\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

c) \(2x^5-4x^3+2x=0\)

\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)

\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)

\(\Rightarrow2x\left(x^2-1\right)^2=0\)

\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)

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

\(\text{#}Toru\)

a)

 \(\left(2x-15\right)^5=\left(2x-15\right)^3\\ \Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\\ \Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15-1\right).\left(2d-15+1\right)=0\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right. \)

b) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)

25/7nha

a,  2 x - 15 = 17

⇒ 2 x = 2 = 2 2 ⇒ x = 1

b,  7 x - 11 3 = 2 5 . 5 2 + 200

⇒ 7 x - 11 3 = 1000 = 10 3

⇒ 7x – 11 =10

⇒ 7x = 21

⇒ x = 3

(x-7).(2x-8)=0

suy ra :(x-7)=0   suy ra x=7

           2x-8=0   suy ra x=4

Câu b:

( x-40) .15 =15.3

 (x-40) .15=45

(x-40)      = 45:15

x-40         = 3

x=43

Bài 1: Tìm x biết:

a )  (x-7).(2x-8)=0

=>x-7=0 hoặc 2x-8=0

=>x=7 hoặc x=4

b)   (x-40).15=15.3

=>(x-40).15=45

=>x-40=45:15

=>x-40=3

=>x=3+40

=>x=43

a) ( x - 7 ) . ( 2x - 8 ) = 0 

TH1 : ( x -7 ) = 0 

=>       x      = 0 + 7 

=>       x      = 7 

TH2 : ( 2x -8)=0

=>      2x      = 0 + 8 
=>      2x      = 8 

=>        x      = 8 : 2 

=>        x      = 4 

Vậy...