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b./ \(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)(b)
Mà \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}< 0\)
(b) \(\Leftrightarrow x+2010=0\Leftrightarrow x=-2010\)
a./
\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0.\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)(a)
Mà \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0\)
(a) \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
a) 2x=64:23=64:8=8=23=>x=3
b) 7x=343:7=49=72=> x=2
c) Tương tự như câu trên, với x+1 thì x=1
d) 2x=17+15=32=25=>x=5
e) => \(x\in\left(-1;1\right)\)
g) =>x=0;1
Ta có : x10 = x
=> x10 - x = 0
=> x(x9 - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^9=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
a) 3*2x =48 \(\Leftrightarrow\)2x=16 \(\Leftrightarrow\)x=4 b) (2x +1 )3 =53 \(\Leftrightarrow\)2x +1 =5 \(\Leftrightarrow\)2x =4 \(\Leftrightarrow\)x=2 c) 36+x=36 \(\Leftrightarrow\)x=0 d) 1+x-15=27-1 \(\Leftrightarrow\)x=40
a) \(2^3.2^x=64\Leftrightarrow8.2^x=64\Leftrightarrow2^x=\dfrac{64}{8}=8=2^3\Rightarrow x=3\)
vậy \(x=3\)
b) \(7.7^x=343\Leftrightarrow7^x=\dfrac{343}{7}=49=7^2\Rightarrow x=2\)
vậy \(x=2\)
c) \(7.7^{x+1}=343\Leftrightarrow7^{x+1}=\dfrac{343}{7}=49=7^2\Rightarrow x+1=2\Leftrightarrow x=1\)
vậy \(x=1\)
d) \(2^x-15=17\Leftrightarrow2^x=17+15=32=2^5\Rightarrow x=5\)
vậy \(x=5\)
e) \(x^{10}=1\Leftrightarrow x^{10}=1^{10}\Rightarrow x=1\)
vậy \(x=1\)
\(x^{10}=x\Leftrightarrow x^{10}-x=0\Leftrightarrow x\left(x^9-1\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x^9-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
vậy \(x=0;x=1\)
a) \(2^3.2^x=2^3.2^3\)
=> \(2^x=2^3\)
=> x=3
b) \(7.7^x=343\)
=> \(7.7^x=7.7^2\)
=> \(7^x=7^2\)
=> x=2
c) \(7.7^{x+1}=7.7^2\)
=> \(7^{x+1}=7^2\)
=> x+1=2
=> x=1
d) \(2^x-15=17\)
=> \(2^x=17+15\)
=> \(2^x=32\)
=> \(2^x=2^5\)
=> x=5
e) \(x^{10}=1\)
=> \(x^{10}=1^{10}\)
=> x=1
g) \(x^{10}=x\)
=> x = 0 hoặc 1
+) Với x=0 => \(0^{10}=0\)( t/m )
+) Với x=1 => \(1^{10}=1\)( t/m )
d)19900+(x-15)=3199:3196-12000
1+(x-15)=26
x-15=26-1=25
x=25+15=40
c) (90:15)2+x=26-22.7
36+x=-16320
x=-16320-36=-16356
b) (2x+1)3=125
(2x+1)3=53
2x+1=5
2x=5-1=4=22
x=2
a) ko hiểu đề bài
a) \(3.2^x-3=45\)
\(3\left(2^x-1\right)=45\)
\(2^x-1=15\)
\(2^x=15+1\)
\(2^x=16\)
\(2^x=2^4\)
=>x=4
b) \(\left(2^x+1\right)^3=125\)
\(\left(2^x+1\right)^3=5^3\)
=> \(2^x+1=5\)
Làm tương tự câu a, đc x = 2
c) \(\left(90:15\right)^2+x=2^6-2^{2.7}\)
\(6^2+x=64-16384\)
\(x=-16320-36\)
x=-16356
d) \(1990^0+\left(x-15\right)=3^{199}:3^{196}-1^{2000}\)
1 + x - 15 = 33-1
x= 27-1-1+15
x=40
3x+1+2.3x-1=99
=>3x-1(32+2)=99
=>3x-1.11=99
=>3x-1=99:11=9=32
=.x-1=2
=>x=3
Vậy x=3
b,3.2x+2+3.2x=120
=>3.2x(22+1)=120
=>3.2x.5=120
=>3.2x=120:5
=>2x=24:3=8=23
=>x=3
Vậy x=3
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