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\(4^{x+2}=16=4^2\)
=> x+ 2 = 2
X = 0
\(5^{2x+3}=3125=5^5\)
=> 2x+3 = 5
=> x =1
Mấy phần còn alji tương tự
\(4^{x+2}=16=4^2\)
\(=>x+2=2\) \(=>x=0\)
\(5^{2x+3}=3125=5^5\)
\(=>2x+3=5\) \(=>x=1\)
\(6^{x+2}=216=6^3\)
\(=>x+2=3\) \(=>x=1\)
\(5^{2\left(x+1\right)}=625=5^4\)
\(=>2\left(x+1\right)=4\) \(=>x=1\)
\(3^{4-2x}=3^0\)
\(=>4-2x=0\) \(=>x=2\)
Ta có : \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{9}{10}=\frac{x}{2010}\)
=> \(\frac{1.2.3.....9}{2.3.4....10}=\frac{x}{2010}\)
=> \(\frac{1}{10}=\frac{x}{2010}\)
=> x = 2010/10
=> x = 201
=>1/4:(x-2/3)=2
=>(x-2/3)=1/8
=>x=1/8+2/3=3/24+16/24=19/24
13/4 - 1/4 : ( x - 2/3 )= 5/4
\(\Rightarrow\) 1/4 : ( x- 2/3 ) = 13/4 - 5/4
\(\Rightarrow\) 1/4 : ( x- 2/3)= 2
\(\Rightarrow\) x - 2/3 = 1/4 :2
\(\Rightarrow\) x- 2/3 = 1/8
\(\Rightarrow\) x= 1/8 +2/3 =19/24
Vậy x = 19/24
2+3/4=11/4 ; 2-3/5=7/5
5/7:6=5/42 ; 5+2/7=37/7
Bài 2: a) x-1/6=1/2+1/3
x-1/6=5/6
x=5/6+1/6
x=1
b) x+1/8=1/2-1/3
x+1/8=1/6
x=1/6-1/8
x=1/24
c) x:1/2=2/3.3/4
x:1/2=1/2
x=1/2.1/2
x =1/4
5/4-yx5/6=-1/12
5/4-y = -1/12:5/6
5/4-y = -1/10
y = 5/4-(-1/10)
y = 27/20
\(\frac{5}{4}-y.\frac{5}{6}=\frac{1}{4}-\frac{1}{3}\)
\(\Rightarrow\frac{5}{4}-y.\frac{5}{6}=-\frac{1}{12}\)
\(\Rightarrow y.\frac{5}{6}=\frac{5}{4}-\left(-\frac{1}{12}\right)\)
\(\Rightarrow y.\frac{5}{6}=\frac{5}{4}+\frac{1}{12}\)
\(\Rightarrow y.\frac{5}{6}=\frac{4}{3}\)
\(\Rightarrow y=\frac{4}{3}:\frac{5}{6}\)
\(\Rightarrow y=\frac{8}{5}\)
Vậy \(y=\frac{8}{5}.\)
a)\(2^x.4=128\Leftrightarrow2^x=32\Leftrightarrow2^x=2^5\Rightarrow x=5\)
b)\(\left(2x+1\right)=125\Leftrightarrow2x=126\Leftrightarrow x=13\)
c)\(x^{15}=x\Leftrightarrow\orbr{\begin{cases}x=\pm1\\x=0\end{cases}}\)
d) \(\left(x-5\right)^4=\left(x-5\right)^5\Leftrightarrow\orbr{\begin{cases}x-5=1\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=5\end{cases}}\)
a,
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
b,
2x = 124
x = 62
c,
\(x^{15}-x=0\)
\(x\left(x^{14}-1\right)=0\)
\(\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x^{14}=1\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
d,
\(0=\left(x-5\right)^5-\left(x-5\right)^4\)
\(\left(x-5\right)^4\left(x-5-1\right)=0\)
\(\orbr{\begin{cases}\left(x-5\right)^4=0\\x-6=0\end{cases}}\)
\(\orbr{\begin{cases}x=5\\x=6\end{cases}}\)