`2^(x+2) . 3^(x+1) . 5^x = 10800`

`=> 2^x. 2^2. 3^x. 3. 5^x=10800`

`=> (2.3.5)^x. (4.3)=10800`

`=> 30^x. 12=10800`

`=> 30^x=10800:12`

`=> 30^x=900`

`=> 30^x=30^2`

`=> x=2`

\(2^{x+2}.3^{x+1}.5^x=10800\\ 2^x.2^2.3^x.3.5^x=10800\\30^x.12=10800\\ 30^x=10800:12\\ 30^x=900\\ 30^x=30^2\\ x=2\)

\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)

......

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)7.2^x=56⇔2^x=8⇔2^x=2³⇔x=3

b)(2x+1)³=9.81⇔(2x+1)³=9³⇔2x+1=9⇔2x=8⇔x=4

c)x³=8²⇔x³=2⁶⇔x=2²⇔x=4

d)4.2^x-3=1⇔4.2^x=4⇔2^x=1⇔2^x=2⁰⇔x=0

e)2.3^x=162⇔3^x=81⇔3^x=3⁴⇔x=4

1/ Ta có: \(5x\left(x-3\right)=\left(x-2\right)\left(5x-1\right)-5\)

\(\Rightarrow5x^2-15x=5x^2-11x+2-5\)

\(\Rightarrow-15x=-11x-3\)

\(\Rightarrow-15x+11x=-3\)

\(\Rightarrow-4x=-3\Rightarrow x=\frac{3}{4}\).

2/ Ta có: \(\left(2x-1\right)\left(x-2\right)-\left(x+3\right)\left(2x-7\right)=3\)

\(\Rightarrow2x^2-5x+2-2x^2+x+21-3=0\)

\(\Rightarrow-4x+20=0\Rightarrow-4x=-20\Rightarrow x=5\).

Chúc bn hc tốt! 

1/ \(5x\left(x-3\right)=\left(x-2\right)\left(5x-1\right)-5\)

<=> \(5x^2-15x=\left(5x^2-x-10x+2\right)-5\)

<=> \(5x^2-15x=5x^2-11x-3\)

<=> \(5x^2-15x-5x^2+11x+3=0\)

<=> \(-4x+3=0\)

<=> \(x=\frac{3}{4}\)

2/ \(\left(2x-1\right)\left(x-2\right)-\left(x+3\right)\left(2x-7\right)=3\)

<=> \(2x^2-4x-x+2-\left(2x^2-7x+6x-21\right)=3\)

<=> \(2x^2-5x+2-2x^2+x+21=3\)

<=> \(-4x+21=3\)

<=> \(4x=17\)

<=> \(x=\frac{17}{4}\)

Cám ơn hai bạn nhiều

a: =>2^4x<2^28

=>4x<28

=>x<7

b: =>5^3x+3<5

=>3x+3<1

=>3x<-2

=>x<-2/3

a) \(16^x< 128^4\)

= (24)^x < (27)⁴

= 2^4x < 2²⁸

= 4x < 28

= x < 7 

Vậy \(x=\left\{0;1;2;3;4;5;6;\right\}\)

\(#Tuyết\)

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