Tìm x:
\(4.3^{x+2}-3^{x-1}=963\)
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TA
20 tháng 7 2021
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`
20 tháng 7 2021
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,
30 tháng 6 2016
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) \(\left(\frac{1}{2}-\frac{1}{6}\right).3^{x+4}-4.3^x=3^{16}-4.3^{13}\)
\(\frac{1}{3}.3^x.4\left(3^4-1\right)=3^{13}.4\left(3^3-1\right)\)
\(3^x.4.\left(3^3-1\right)=3^{13}.4.\left(3^3-1\right)\)
\(3^x=3^{13}\)
\(x=13\)
30 tháng 6 2016
\(\left(\frac{1}{2}-\frac{1}{6}\right).\left(3^x.3^4\right)-4.3^x=3^{16}-4.3^{13}\)
=> \(\frac{1}{3}.3^x.3^4-4.3^x=3^{16}-4.3^{13}\)
=> \(3^x.3^4-4.3^x=\left(3^{16}-4.3^{13}\right):\frac{1}{3}\)
=> \(3^x.3^4-4.3^x=-386339074,3\)
=> \(3^x.\left(3^4-4\right)=-386339074,3\)
=> \(3^x.77=-386339074,3\)
=> \(3^x=-386339074,3:77\)
=> \(3^x=-5017390,575\)
=> x = ... chắc tự ngồi tính đc
NT
2
NN
26 tháng 9 2020
\(3^{x+1}+3^{x+2}+3^{x+3}-4.3^x=315\)
\(\Leftrightarrow3^x.3+3^x.3^2+3^x.3^3-4.3^x=315\)
\(\Leftrightarrow3^x.3+3^x.9+3^x.27-4.3^x=315\)
\(\Leftrightarrow3^x.\left(3+9+27-4\right)=315\)
\(\Leftrightarrow3^x.35=315\)\(\Leftrightarrow3^x=9\)
\(\Leftrightarrow3^x=3^2\)\(\Leftrightarrow x=2\)
Vậy \(x=2\)
B
26 tháng 9 2020
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
AA
2
6 tháng 10 2021
\(3.2^x+12=4.3^2\)
\(\Rightarrow3.2^x+12=4.9\)
\(\Rightarrow3.2^x+12=36\)
\(\Rightarrow3.2^x=36-12\)
\(\Rightarrow3.2^x=24\)
\(\Rightarrow2^x=24\div3\)
\(\Rightarrow2^x=8\)
\(\Rightarrow2^x=2^3\)
Vậy x = 3
NM
1
26 tháng 9 2023
`#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.`
8 tháng 1 2021
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: \(E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
\(\Leftrightarrow\dfrac{1}{3}\cdot E=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E-\dfrac{1}{3}\cdot E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)
\(\Leftrightarrow E\cdot\dfrac{2}{3}=1-\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E=\dfrac{3-\dfrac{3}{3^{101}}}{2}=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)
\(4.3^{x+2}-3^{x-1}=963\)
\(\Rightarrow4.3^{x+2}-3^{x+2-3}=963\)
\(\Rightarrow4.3^{x+2}-\dfrac{1}{3^3}3^{x+2}=963\)
\(\Rightarrow4.3^{x+2}-\dfrac{1}{27}3^{x+2}=963\left(1\right)\)
Đặt \(t=3^{x+2}>0\)
\(\left(1\right)\Rightarrow4t-\dfrac{1}{27}t=963\)
\(\Rightarrow t\left(4-\dfrac{1}{27}\right)=963\)
\(\Rightarrow\dfrac{107}{27}.t=9.107\)
\(\Rightarrow t=3^2\Rightarrow t=3^2:\dfrac{1}{3^3}=3^2.\dfrac{3^3}{1}=3^5\)
\(\Rightarrow3^{x+2}=3^5\Rightarrow x+2=5\Rightarrow x=3\)
4\(\cdot\)3x+2-3x-1=963
=>4\(\cdot\)3xx32-3x:3=963
=>3x\(\cdot\)36-3x\(\cdot\)\(\dfrac{1}{3}\)=963
=>3x\(\cdot\)(36-\(\dfrac{1}{3}\))=963
=>3x\(\cdot\)\(\dfrac{107}{3}\)=963
=>3x=963:\(\dfrac{107}{3}\)=27=33
=>x=3