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a; => \(3^x+3^x.3+3^x.3^2=1053\)
=> \(3^x.\left(1+3+3^2\right)=1053\)
=> \(3^x.13=1053\)
=> \(3^x=81\)
=> \(3^x=3^4\)
=> x=4
b; => (x-1)^2.(x-1)=(x-1)^2
=> (x-1)^2.(x-1)-(x-1)^2=0
=> (x-1)^2.[(x-1)^2-1)=0
\(\hept{\begin{cases}x-1=0\\x-1=1\\x-1=-1\end{cases}}\)
=> \(\hept{\begin{cases}x=1\\x=2\\x=0\end{cases}}\)
a, \(2.x^x=10.3^{12}+8.27^4\)
\(2.x^x=10.3^{12}+8.3^{12}\)
\(2.x^x=3^{12}.\left(10+8\right)\)
\(2.x^x=3^{12}.18\)
\(2.x^x=3^{12}.2.3^3\)
\(2.x^x=3^{15}.2\)
\(x^x=3^{15}\)( Hình như sai đề )
b,\(3^{2x+2}=9^{x+3}\)
\(3^{2x+2}=3^{2x+3}\)
\(a,3x+17x=340\)
\(x\left(17+3\right)=340\)
\(x20=340\)
\(x=340:20=17\)
\(b,\left|2x+1\right|=3\\ \Rightarrow\left[{}\begin{matrix}2x+1=3\\2x+1=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=3-1=2\\2x=-3-1=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(c,3^x+3^{x+1}+3^{x+2}=1053\\ 3^x\left(1+3+9\right)=1053\\ 3^x.13=1053\\ 3^x=1053:13=81=3^4\\ \Rightarrow x=4\)
Đề đúng là 3x+3x+1+3x+2=1053
\(\Rightarrow3^x\left(1+3^1+3^2\right)=1053\)
\(\Rightarrow3^x\cdot13=1053\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
Đề bài phải là :
\(3^x+3^{x+1}3^{x+2}=1053\)
\(3^x+3^x.3+3^x.3^2=1053\)
\(3^x.\left(1+3+3^2\right)=1053\)
\(3^x.\left(1+3+9\right)=1053\)
\(3^x.13=1053\)
\(3^x=1053:13\)
\(3^x=81\)
\(3^x=3^4\)
\(x=4\)
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3x+2=369
=>x+2=69
x=69-2
x=67
2x-5=810
2x-5=230
=>x-5=30
x=30+5
x=35
3x+2+3x=810
3x.32+3x=810
3x.(32+1)=810
3x.10=810
3x=810:10
3x=81
3x=34
=>x=4
5x+1-5x=500
5x.5-5x=500
5x.(5-1)=500
5x.4=500
5x=500:4
5x=125
5x=53
=>x=3
a) 3x+2 = 369
x + 2 = 69
x = 69 - 2
x = 67
b) 2x-5 = 810
2x-5 = 230
x - 5 = 30
x = 30 + 5
x = 35
c) 3x+2 + 3x = 810
3x . 9 + 3x . 1 = 810
3x . ( 9 + 1 ) = 810
3x . 10 = 810
3x = 810 : 10
3x = 81
3x = 34
=> x = 4
d) 5x+1 - 5x = 500
5x . 5 - 5x . 1 = 500
5x . ( 5 - 1 ) = 500
5x . 4 = 500
5x = 500 : 4
5x = 125
5x = 53
=> x = 3
\(-5.\left(x+\frac{1}{5}\right)-\frac{1}{2}.\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(\Rightarrow-5x-1-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(\Rightarrow-5x-\frac{1}{2}x-\frac{3}{2}x=\frac{-5}{6}-\frac{1}{3}+1\)
\(\Rightarrow-7x=\frac{-1}{6}\)
\(\Rightarrow x=\frac{1}{42}\)
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\(3.\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Rightarrow3.\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Rightarrow3x=\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{18}\)
Vậy...