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\(3^{x-1}+3^x+3^{x+1}-1=1052\)

\(\Rightarrow3^{x-1}\left(1+3+3^2\right)-1=1052\)

\(\Rightarrow3^{x-1}.13=1053\Rightarrow3^{x-1}=81=>x-1=4\)

\(\Rightarrow x=5\)

vậy x=5

\(3^{x-1}+3^x+3^{x+1}-1=1052\)

\(3^x:3+3^x+3^x\cdot3=1053\)

\(3^x\left(\frac{1}{3}+1+3\right)=1053\)

\(3^x\cdot\frac{13}{3}=1053\)

\(3^x=243\)

\(x=5\)

cam un bn

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1

a, => 3^x.(1+3+3^2)-1 = 1052

=> 3^x.13 = 1052+1 = 1053

=> 3^x = 1053 : 13

=> 3^x = 81 = 3^4

=> x = 4

b, => x^2-49 >=0 ; 81-x^2 >=0 hoặc x^2-49 < = 0 ; 81-x^2 < = 0

=> 49 < = x^2 < = 81

=> -9 < = x < = -7 hoặc 7 < = x < = 9

=> x thuộc {-9;-8;-7;7;8;9}

Tk mk nha

Cảm ơn bạn nhiều nha^^ !

a) 2^x+2^x+1+2^x+2+2^x+3=480

<=> \(2^x+2^x.2+2^x.2^2+2^x.2^3=480\)

<=> \(2^x.\left(1+2+2^2+2^3\right)=480\)

<=>\(2^x=\frac{480}{1+2+2^2+2^3}=32\)

=> x=5

b) (x²-49)*(x²-81)<0 Khi \(\hept{\begin{cases}x^2-49< 0\\x^2-81>0\end{cases}}\) hoặc \(\hept{\begin{cases}x^2-49>0\\x^2-81< 0\end{cases}}\)

TH1 \(\hept{\begin{cases}x^2-49< 0\\x^2-81>0\end{cases}}\)\(\Rightarrow81< x^2< 49\)(Vô lí)

TH2\(\hept{\begin{cases}x^2-49>0\\x^2-81< 0\end{cases}}\) \(\Rightarrow49< x^2< 81\)\(\Leftrightarrow7^2< x^2< 9^2\)Mà x nguyên \(\Rightarrow x=8\)

c) Làm giống câu a

thank

a) 1152-(374+1152)+(-65+374)

=1152-374-1152-65+374

=(1152-1152)+(-374+374)-65

=-65

a, 1152 - ( 374 + 1152 ) + ( -65 + 374 )

= 1152 - 374 - 1152 - 65 + 374 

= ( 1152 - 1152 ) - ( 374 - 374 ) - 65

= 0 - 0 - 65

= -65

b, 13 - 12 + 10 - 9 + 8 - 7 - 6 + 5 - 4 + 3 + 2 - 1

= ( 13 - 12 ) + 11 + ( 10 - 9 ) + ( 8 - 7 ) - ( 6 - 5 ) - ( 4 - 3 ) + ( 2 - 1 )

= 1 + 11 + 1 + 1 - 1 - 1 + 1

= 13

(-5/28+1,75+8/35):(3,9/20)

=9/5 : 0.195

=321/100

=1.605