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^^ !

như cức

bạn cs thể viết rõ ra đc k

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

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

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

\(\Rightarrow x=5\)

vậy x=5

Vì sao 3^x-1+3^x+3^x+1-1=1052 lại bằng : 3^x-1(1+3+9)=1053 vậy ạ

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

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

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

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

\(3^x=243\)

\(x=5\)

cam un bn

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

1052 x 100 x 10 x1

=105200x10x1

=1052000x1

=1052000

=1052000 Đảm bảo đúng

\(Ai..tích..mk..mk..tích..lại..cho\)

1052000 tk cho minh nhe !

a: Ta có: \(\left(x+1\right)^3-\left(x+2\right)\left(x-1\right)^2-3\left(x-3\right)\left(x+3\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x+2\right)\left(x^2-2x+1\right)-3\left(x^2-9\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-2x^2+x+2x^2-4x+2\right)-3\left(x^2-9\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x-2-3x^2+9=5\)

\(\Leftrightarrow6x=-3\)

hay \(x=-\dfrac{1}{2}\)

b: Ta có: \(\left(x+1\right)^3+\left(x-1\right)^3=\left(x+2\right)^3+\left(x-2\right)^3\)

\(\Leftrightarrow x^3+3x^2+3x+1+x^3-3x^2+3x-1=x^3+6x^2+12x+8+x^3-6x^2+12x-8\)

\(\Leftrightarrow2x^3+6x=2x^3+24x\)

\(\Leftrightarrow x=0\)

c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-1=-10\)

\(\Leftrightarrow12x=-11\)

hay \(x=-\dfrac{11}{12}\)