7^x + 7^2x+3=344

7^x + 7^2x . 7^3 =344

7^x + 7^2 . 7^x . 7^3 =344

7^x . (7^2+1) . 7^3 = 344

7^x . 50 . 343 =344

làm tiếp ........

\(\dfrac{1}{6}+x=\dfrac{5}{12}\)
\(=>x=\dfrac{5}{12}-\dfrac{2}{12}=\dfrac{1}{4}\)
\(\dfrac{3}{4}+\dfrac{1}{4}x=-\dfrac{1}{2}\)
\(=>\dfrac{1}{4}x=-\dfrac{5}{4}\)
\(=>x=-\dfrac{5}{4}.4=-5\)
\(7^{2x}+7^{2x+3}=344\)
\(< =>49^x+49^x.343=344\)
\(=>x=?\)

\(7^{2x}+7^{2x+3}=344\)

\(7^{2x}+7^{2x}.7^3=344\)

\(7^{2x}+7^{2x}.343=344\)

\(7^{2x}.1+7^{2x}.343=344\)

\(7^{2x}.\left(1+343\right)=344\)

\(7^{2x}.344=344\)

\(7^{2x}=344:344\)

\(7^{2x}=1\)

\(\Rightarrow7^{2x}=7^0\)

\(2x=0\)

\(x=0:2\)

\(x=0\)

\(7^{2x}+7^{2x+3}=344\\\Rightarrow7^{2x}+7^{2x}\cdot7^3=344\\\Rightarrow7^{2x}\cdot(1+7^3)=344\\\Rightarrow7^{2x}\cdot(1+343)=344\\\Rightarrow7^{2x}\cdot344=344\\\Rightarrow7^{2x}=1\\\Rightarrow2x=0\\\Rightarrow x=0\)

7^2x + 7^2x . 7³ = 344

=> 7^2x.(1 + 343) = 344

=> 7^2x . 344 = 344

=> 7^2x = 344 : 344

=> 7^2x = 1

=> 7^2x = 7⁰

=> 2x = 0

=> x=  0

7^2x+7^2x.7³=344

=>7^2x(1+7³)=344

=>7^2x=1

=>7^2x=7⁰

=>2x=0

=>x=0

Vậy x=0.

7^2x + 7^2x . 7³ = 344

=> 7^2x. (1 + 7³) = 344

=> 7^2x.344 = 344

=> 7^2x = 1

=> 7^2x = 7⁰

=> 2x = 0

=> x = 0

Vậy x = 0

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bác nhắn linh tinh cái gì thế ???

Nếu x = 0 => \(7^0+7^{2.0+3}=344\)

=> \(344=344\) (luôn đúng)

Vậy x = 0 là nghiệm của pt

Nếu x > 0 => VP= \(344\equiv1\left(mod7\right)\)

VT = \(7^x+7^{2x+3}⋮7\)

=> VT \(\ne\)VP => x > 0 không là nghiệm của pt

Nếu x < 0 =>  \(7^x+7^{2x+3}< 344\)

=> x < 0 không là nghiệm của pt

Vậy x = 0 

giúp mik vs

\(7^{2x}+7^{2x+3}=344\)

\(\Leftrightarrow7^{2x}+7^{2x}.343=344\)

\(\Leftrightarrow7^{2x}\left(1+343\right)=344\)

\(\Leftrightarrow7^{2x}.344=344\)

\(\Leftrightarrow7^{2x}=1\)

\(\Leftrightarrow7^{2x}=7^0\)

\(\Leftrightarrow2x=0\)

\(\Leftrightarrow x=0\)

Vậy \(x=0\)

\(\Leftrightarrow7^{2x}\left(1+7^3\right)=344\)

\(\Leftrightarrow7^{2x}.344=344\)

\(\Leftrightarrow7^{2x}=1\)

\(\Rightarrow x=0\)

\((2x-7)^3=8(7-2x)^2\)

⇔ \((2x-7)^3=8(2x-7)^2\)       (*)

\(TH1: (2x-7)^2=0\)

Khi đó: \(2x-7=0\)  ⇔ \(x=\dfrac{7}{2} \)

\(TH2:\left(2x-7\right)^2\ne0\) 

Khi đó: (*) ⇔ \(2x-7=8\) (chia 2 vế cho \((2x-7)^2\))

⇔ \(x=\dfrac{15}{2} \) 

Vậy \(x=\dfrac{15}{2}\); \(x=\dfrac{7}{2}\)