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2x-1+2x+2x+1=224
2x:2+2x+2x.2=224
2x.\(\frac{1}{2}\)+2x+2x.2=224
\(2^x.\left(\frac{1}{2}+1+2\right)=224\)
\(2^x.\frac{7}{2}=224\)
\(2^x=224:\frac{7}{2}\)
\(2^x=64\)
2x=26
=> x=6
Vậy x=6
|x+3|=|x-5|
\(\Rightarrow\orbr{\begin{cases}x+3=x-5\\x+3=-x+5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-x=-5-3\Rightarrow0x=-8\left(vl\right)\\x+x=5-3\Rightarrow2x=2\Rightarrow x=2:2=1\end{cases}}\)
Vậy x=1
a) \(\left(x-1\right)^3=125\)
\(\Leftrightarrow\left(x-1\right)^3=5^3\)
\(\Leftrightarrow x-1=5\)
\(\Leftrightarrow x=5+1\)
\(\Leftrightarrow x=6\)
Vậy \(x=6\)
b) \(2^{x+2}-2^x=96\)
\(\Leftrightarrow\left(2^2-1\right)\cdot2^x=96\)
\(\Leftrightarrow\left(4-1\right)\cdot2^x=96\)
\(\Leftrightarrow3\cdot2^x=96\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
c) \(\left(2x+1\right)^3=343\)
\(\Leftrightarrow\left(2x+1\right)^3=7^3\)
\(\Leftrightarrow2x+1=7\)
\(\Leftrightarrow2x=7-1\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
d) \(720:\left[41-\left(2x-5\right)\right]=2^3\cdot5\)
\(\Leftrightarrow720:\left[41-\left(2x-5\right)\right]=2^3\cdot5\left(đk:x\ne23\right)\)
\(\Leftrightarrow720:\left(41-2x+5\right)=8\cdot5\)
\(\Leftrightarrow720:\left(46-2x\right)=40\)
\(\Leftrightarrow\dfrac{720}{46-2x}=40\)
\(\Leftrightarrow\dfrac{720}{2\left(23-x\right)}=40\)
\(\Leftrightarrow\dfrac{360}{23-x}=40\)
\(\Leftrightarrow360=40\left(23-x\right)\)
\(\Leftrightarrow9=23-x\)
\(\Leftrightarrow x=23-9\)
\(\Leftrightarrow x=14\left(đk:x\ne23\right)\)
\(\Leftrightarrow x=14\)
Vậy \(x=14\)
e) \(2^x\cdot7=224\)
\(\Leftrightarrow7\cdot2^x=224\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
f) \(\left(3x+5\right)^2=289\)
\(\Leftrightarrow3x+5=\pm17\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+5=17\\3x+5=-17\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{22}{3}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{22}{3};x_2=4\)
a)\(\left(x-1\right)^3=125\Leftrightarrow\left(x-1\right)^3=5^3\Leftrightarrow x-1=5\Leftrightarrow x=6\)b)\(2^{x+2}-2^x=96\Leftrightarrow2^x.2^2-2^x=96\Leftrightarrow2^x\left(2^2-1\right)=96\Leftrightarrow2^x.3=96\Leftrightarrow2^x=32\Leftrightarrow x=5\)c)\(\left(2x-1\right)^3=343\Leftrightarrow\left(2x-1\right)^3=7^3\Leftrightarrow2x-1=7\Rightarrow2x=8\Rightarrow x=4\)d)\(720:\left[41-\left(2x-5\right)\right]=2^3.5\)
\(720:\left[41-\left(2x-5\right)\right]=40\Leftrightarrow\left[41-\left(2x-5\right)\right]=720:40=18\)
\(\Leftrightarrow41-2x+5=18\Leftrightarrow36-2x=18\Leftrightarrow2x=18\Leftrightarrow x=9\)
e)\(2^x.7=224\Leftrightarrow2^x=224:7=32\Leftrightarrow2^x=2^5\Leftrightarrow x=5\)
f) \(\left(3x+5\right)^2=289\Leftrightarrow\left(3x+5\right)=17^2\Leftrightarrow3x+5=17\Leftrightarrow3x=12\Leftrightarrow x=4\)
a, 2x=224:7
2x=32
2x=25
=> x=5
b, \(\left(3^x+5\right)^2=289\)
\(\left(3^x+5\right)^2=17^2\)
\(\Rightarrow3^x+5=17 \)
\(3^x=17-5\)
\(3^x=12\)
a,2x.7=224
=> 2x=32
Mà : 25=32
=> 2x=25
=> x=5
b,(3x+5)2=289
Ta có : 172=289
=> (3x+5)2=172
=> 3x+5=17
=> 3x=12
=> sai đề :v
a,(3x+5)2=289
=>(3x+5)2=172
=>3x+5=17
=>3x=12
=>x=4
b,x50=x
=>x50-x=0
=>x=1
c,2x.7=224
=>2x=224:7
=>2x=32
=>2x=25
=>x=5
#Koo#
2x + 2x+1 + 2x+2 =224
<=>2x(1+2+4)=224
<=>2x.7=224
<=>2x=32
<=>x=5
vậy x=5
\(=>2^x+2^x.2^1+2^x.2^2=224\)
=> \(2^x+2^x.2+2^x.4=224\)
=> x = 5
a. ( x - 1 )2 = 25
<=> \(\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}}\)
<=>\(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
b. 2x + 2 - 2x = 96
<=> 2x.4-2x=96
<=> 2x.3=96
<=> 2x=32=25
<=> x=5
c. 2x . 7 = 224
<=> 2x=32=25
<=> x=5
d. ( 7x - 11 )3 = 25 . 52 = 200 (xem lại đề)
e. 9 < 3x < 81
<=> 32<3x<34
<=> 2<x<4
\(2^{x+3}-2^x=224\)
=>\(2^x\cdot8-2^x=224\)
=>\(7\cdot2^x=7\cdot32\)
=>\(2^x=32=2^5\)
=>x=5
2x+3 - 2x = 224
2x+3 - 2x = 28- 25
=> x+3 - x = 8 - 5
3 = 3
=> pt luôn bằng 3 với mọi x