c) `3^(x+1)+4.3^3=567`

`3^(x+1)+108 = 567`

`3^x . 3 = 459`

`3^x=153`

`3^x = 3^2 . 17`

`=>` Không có `x` thỏa mãn.

.

`P=(x-2)^2+11/5`

Vì `(x-2)^2 >=0 forall x `

`=> (x-2)^2 + 11/5 >= 11/5 forall x`

`<=> P >=11/5`

`=> P_(min)=11/5 <=> x-2=0 <=>x=2`

a,

 \(\left(5x+3\right)^2=\dfrac{25}{9}\\ \Rightarrow\left[{}\begin{matrix}5x+3=\dfrac{5}{3}\\5x+3=-\dfrac{5}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{4}{15}\\x=-\dfrac{7}{6}\end{matrix}\right.\)

b,

\(\left(-\dfrac{1}{2}x+3\right)^3=-\dfrac{1}{125}\\ \Rightarrow-\dfrac{1}{2}x+3=-\dfrac{1}{5}\\ \Rightarrow x=\dfrac{32}{5}\)

c,

a)\(\left(\frac{1}{2}-\frac{1}{3}\right).6^x+6^{x+2}=6^{15}+6^{18}\)

\(\frac{1}{6}.6^x+6^{x+2}=6^{15}\left(1+6^3\right)\)

\(\frac{1}{6}.6^x\left(1+6^3\right)=6^{15}.217\)

\(6^{x-1}.217=6^{15}.217\)

\(6^{x-1}=6^{15}\)

\(x-1=15\)

\(x=16\)

b) \(\left(\frac{1}{2}-\frac{1}{6}\right).3^{x+4}-4.3^x=3^{16}-4.3^{13}\)

\(\frac{1}{3}.3^x.4\left(3^4-1\right)=3^{13}.4\left(3^3-1\right)\)

\(3^x.4.\left(3^3-1\right)=3^{13}.4.\left(3^3-1\right)\)

\(3^x=3^{13}\)

\(x=13\)

\(\left(\frac{1}{2}-\frac{1}{6}\right).\left(3^x.3^4\right)-4.3^x=3^{16}-4.3^{13}\)

=> \(\frac{1}{3}.3^x.3^4-4.3^x=3^{16}-4.3^{13}\)

=> \(3^x.3^4-4.3^x=\left(3^{16}-4.3^{13}\right):\frac{1}{3}\)

=> \(3^x.3^4-4.3^x=-386339074,3\)

=> \(3^x.\left(3^4-4\right)=-386339074,3\)

=> \(3^x.77=-386339074,3\)

=> \(3^x=-386339074,3:77\)

=> \(3^x=-5017390,575\)

=> x = ... chắc tự ngồi tính đc

cái a tương tự thôi

\(3^{x+1}+3^{x+2}+3^{x+3}-4.3^x=315\)

\(\Leftrightarrow3^x.3+3^x.3^2+3^x.3^3-4.3^x=315\)

\(\Leftrightarrow3^x.3+3^x.9+3^x.27-4.3^x=315\)

\(\Leftrightarrow3^x.\left(3+9+27-4\right)=315\)

\(\Leftrightarrow3^x.35=315\)\(\Leftrightarrow3^x=9\)

\(\Leftrightarrow3^x=3^2\)\(\Leftrightarrow x=2\)

Vậy \(x=2\)

            Bài làm :

Ta có :

\(3^{x+1}+3^{x+2}+3^{x+3}-4.3^x=315\)

\(\Leftrightarrow3^x.3+3^x.3^2+3^x.3^3-4.3^x=315\)

\(\Leftrightarrow3^x.3+3^x.9+3^x.27-4.3^x=315\)

\(\Leftrightarrow3^x.\left(3+9+27-4\right)=315\)

\(\Leftrightarrow3^x.35=315\)

\(\Leftrightarrow3^x=9\)

\(\Leftrightarrow3^x=3^2\)

\(\Leftrightarrow x=2\)

Vậy x=2

\(3.2^x+12=4.3^2\)

\(\Rightarrow3.2^x+12=4.9\)

\(\Rightarrow3.2^x+12=36\)

\(\Rightarrow3.2^x=36-12\)

\(\Rightarrow3.2^x=24\)

\(\Rightarrow2^x=24\div3\)

\(\Rightarrow2^x=8\)

\(\Rightarrow2^x=2^3\)

Vậy x = 3

ohhgfghfg623546785489121751284545

`#3107.\text {DN}`

\(3^{x+2}+4\cdot3^{x+1}+3^{x-1}=6^6\)

`=> 3^x*3^2 + 4*3^x*3 + 3^x * 1/3 = 6^6`

`=>3^x*(3^2 + 12 + 1/3) = 6^6`

`=> 3^x * 64/3 = 6^6`

`=> 3^x = 6^6 \div 64/3`

`=> 3^x = 2187`

`=> 3^x = 3^7`

`=> x = 7`

Vậy, `x = 7.`

Bài 1: 

a) Ta có: \(\dfrac{7^4\cdot3-7^3}{7^4\cdot6-7^3\cdot2}\)

\(=\dfrac{7^3\cdot\left(7\cdot3-1\right)}{7^3\cdot2\left(7\cdot3-1\right)}\)

\(=\dfrac{1}{2}\)

c) Ta có: \(E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)

\(\Leftrightarrow\dfrac{1}{3}\cdot E=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\)

\(\Leftrightarrow E-\dfrac{1}{3}\cdot E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)

\(\Leftrightarrow E\cdot\dfrac{2}{3}=1-\dfrac{1}{3^{101}}\)

\(\Leftrightarrow E=\dfrac{3-\dfrac{3}{3^{101}}}{2}=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)