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de thui
nhung bây giờ mk bận
nên ko giải hộ bn được
chúc bn hoc tốt!
hihi
a) 2x . 7 = 224
2x . 7 = 234256
2x =234256:7
2x =đề sai
b) ( 3x + 5 )2 = 289
( 3x + 5 )2 = 172
=>3x + 5=17
3x=17-5
3x=12
x=12:3
x=4
d) 32x +1 .11 = 2673
32x +1 .11 = 19034163
32x +1 =19034163:11
32x +1 =đề sai
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) 70 - 5(x - 3) = 45
5(x - 3) = 70 - 45
5(x - 3) = 25
x - 3 = 25 : 5
x - 3 = 5
x = 8
b) (x + 1) + (x + 2) + (x + 3) = 63
=> (x + x + x) + (1 + 2 + 3) = 63
=> 3x + 6 = 63
=> 3x = 63 - 6
=> 3x = 57
=> x = 57 : 3
=> x = 29
c) 2x - 24 = 1000
=> 2x = 1024
=> 2x = 210
=> x = 10
d) 3x + 4 + 3x = 738
=> 3x . 34 + 3x = 738
=> 3x . 81 + 3x = 738
=> 3x.(81 + 1) = 738
=> 3x . 82 = 738
=> 3x = 738 : 82
=> 3x = 9
=> 3x = 32
=> x = 2
Bài 1 tự làm!
Bài 2:
a, \(\left(3x-4\right)\left(x-1\right)^3=0\Rightarrow\left[{}\begin{matrix}3x-4=0\\\left(x-1\right)^3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=1\end{matrix}\right.\)
b, \(2^{2x-1}:4=8^3\Rightarrow2^{2x-1}:2^2=2^9\)
\(\Rightarrow2x-1-2=9\Rightarrow2x-3=9\Rightarrow2x-12\Rightarrow x=6\)
c, Đề chưa rõ
d, \(\left(x+2\right)^5=2^{10}\Rightarrow\left(x+2\right)^5=4^5\Rightarrow x+2=4\Rightarrow x=2\)
e, \(\left(3x-2^4\right).7^3=2.7^4\Rightarrow3x-2^4=2.7^4:7^3\Rightarrow3x-16=2.7=14\)
\(\Rightarrow3x=14+16=30\Rightarrow x=\dfrac{30}{3}=10\)
f, \(\left(x+1\right)^2=\left(x+1\right)^0\Rightarrow\left(x+1\right)^2=1\) (vì x0 = 1)
\(\Rightarrow x+1=1\Rightarrow x=0\)
a) \(3^{2x-1}=243\)
\(\Leftrightarrow3^{2x-1}=3^5\)
\(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=5+1\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
b) \(\left(3^x\right)^2:3^3=\dfrac{1}{243}\)
\(\Leftrightarrow3^{2x}:3^3=\dfrac{1}{3^5}\)
\(\Leftrightarrow3^{2x}:3^3=3^{-5}\)
\(\Leftrightarrow3^{2x-3}=3^{-5}\)
\(\Leftrightarrow2x-3=-5\)
\(\Leftrightarrow2x=-5+3\)
\(\Leftrightarrow2x=-2\)
\(\Leftrightarrow x=-\dfrac{2}{2}\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
c) \(2^{3x+2}=4^{x+5}\)
\(\Leftrightarrow2^{3x+2}=\left(2^2\right)^{x+5}\)
\(\Leftrightarrow2^{3x+2}=2^{2\left(x+5\right)}\)
\(\Leftrightarrow3x+2=2\left(x+5\right)\)
\(\Leftrightarrow3x+2=2x+10\)
\(\Leftrightarrow3x-2x=10-2\)
\(\Leftrightarrow x=8\)
Vậy \(x=8\)
d) \(3^{x+1}=9^x\)
\(\Leftrightarrow3^{x+1}=\left(3^2\right)^x\)
\(\Leftrightarrow3^{x+1}=3^{2x}\)
\(\Leftrightarrow x+1=2x\)
\(\Leftrightarrow2x-x=1\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
Ta có " (x - 5)7 = (x - 5)4
=> (x - 5)7 - (x - 5)4 = 0
<=> (x - 5)4[(x - 5)3 - 1] = 0
\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^3-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^3=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-5=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=6\end{cases}}\)
a, 3x+3x*34 = 738
<=> 3x(1+34) =738
<=> 3x *82 = 738
<=> 3x =9 =32
<=> x=2
câu b,c tương tự