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b) \(3.2^{x+1}=12\)
\(2^{x+1}=12:3\)
\(2^{x+1}=4\)
\(2^{x+1}=2^2\)
\(x+1=2\)
\(x=2-1\)
\(x=1\)
Vậy \(x=1\)
c) \(2^{x-1}=2^3+2^4-2^3\)
\(2^{x-1}=8+16-8\)
\(2^{x-1}=16\)
\(2^{x-1}=2^4\)
\(x-1=4\)
\(x=5\)
Vậy \(x=5\)
d) \(x^{50}=x\)
\(x^{50}-x=0\)
\(\Rightarrow x\in\left\{0;1\right\}\)
Vậy \(x\in\left\{0;1\right\}\)
\(b.3.2^{x+1}=12\\ \Rightarrow2^{x+1}=4\\ \Rightarrow2^{x+1}=2^2\\ \Rightarrow x=1\\ \)
c) \(2^{x-1}=2^3-2^3+2^4\\ \Rightarrow2^{x-1}=0+16\\ \Rightarrow2^{x-1}=16\\ \Rightarrow2^{x-1}=2^4\\ \Rightarrow x-1=4\\ \Rightarrow x=5\)
d) \(x^{50}=x\\ \Rightarrow x=0;1\)
e) \(2\left(2x-1\right)^4=32\\ \Rightarrow\left(2x-1\right)^4=16\\ \Rightarrow\left(2x-1\right)^4=2^4\\ \Rightarrow2x-1=2\\ \Rightarrow2x=3\\ \Rightarrow x=\frac{3}{2}\)
g) Bí
a) (2^x).4=128
2^x = 128:4
2^x = 32
mà 32=2^5=>x=5
b) ta có: x^15=x
theo quy ước: 0^15=0;1^15=1
=> x=1
4 câu còn lại mai mình sẽ giải nhé
Bài làm
m) (x + 2).(3 - x) = 0;
=> x + 2 = 0 hoặc 3 - x = 0
=> x = -2 hoặc x = 3
Vậy x = -2 hoặc x = 3
d) 511.712 + 511.711
= 511 . ( 712 + 711 )
= 511 . [ 711 . ( 7 + 1 ) ]
= 511 . 711 . 8
= ( 5 . 7 )11 . 8
= 3511 . 8
512.712 + 9.511.711
= 511 ( 5 . 712 + 9 . 1 . 711 )
= 511 [ 711 ( 5 . 7 + 9 . 1 . 1 ) ]
= 511 ( 711 . 44 )
= 511 . 711 . 44
= 3511 . 44
m. \(\left(x+2\right)\left(3-x\right)=0\Leftrightarrow\orbr{\begin{cases}x+2=0\\3-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
d. \(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.\left(7^{12}+7^{11}\right)}{5^{11}.\left(5.7^{12}+9.7^{11}\right)}=\frac{7^{12}+7^{11}}{5.7^{12}+9.7^{11}}=\frac{1}{5.9}=\frac{1}{45}\)
q. \(\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+...+10+11=11\)
\(\Rightarrow\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+...+10=0\)
\(\Rightarrow\left[\left(x-3\right)+\left(x-2\right)+\left(x-1\right)\right]+(1+2+3+...+10)=0\)
\(\Rightarrow\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+55=0\)
\(\Rightarrow x-3+x-2+x-1=-55\)
\(\Rightarrow3x-6=-55\)
\(\Rightarrow3x=-49\)
\(\Rightarrow x=-\frac{49}{3}\)
a) 2x+2x+1+2x+2+2x+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
2x+2x+1+2x+2+2x+3=480
<=>2x.(1+2+22+23)=480
<=>2x.15=480
<=>2x=32=25
<=>x=5
ta co1;2^x+2^x+1+2^x+2+2^x+3=480
=>2^x.1+2^x.2+2^x.2^2+2^x.2^3=480
=>2^x.(1+2+2^2+2^3)=480
=>2^x=480:15=32
=>2^x=2^5
=>x=5
\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
\(2^x+2^x.2+2^x.4+2^x.8=480\)
\(2^x.\left(1+2+4+8\right)=480\)
\(2^x.15=480\)
\(2^x=32\)
\(2^x=2^5\)
\(x=5\)
Vậy \(x=5\)