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Câu 1:
\(\dfrac{128}{\left(n-3\right)^3}=2\)
\(\Leftrightarrow\left(n-3\right)^3=64\)
=>n-3=4
hay n=7
Câu 2:
\(\Leftrightarrow x^3+12=57\cdot4=228\)
\(\Leftrightarrow x^3=216\)
hay x=6
a) (x^3+12):4=60+(-3)
(x^3+12).1/4=57
x^3+12=228
x^3=216
x^3=6^3
=> x=6
b) 2^x+1.3=96
2^x+1.3=2^5.3
2^x+1=2^5.3:3
2^x+1=2^5
=> x+1=5
x=4
a) x = 17
b) x = 23
c) x = 63
d) x = 2499
e) x = -3
j) x = 3
g) x = 7
h) x = 3
c/ 2x - 1 = \(5^{98}:5^{96}\)
2x - 1 = \(5^2\) = 25
2x = 25 + 1 = 26
x = 26 : 2
x = 13
d/ 7x + 3 = \(3^5.2^3.9\)
7x + 3 = \(3^5.3^2.8=3^7.8=2187.8\)
7x + 3 = \(17496\)
7x = 17496 - 3 = 17493
x = 17493 : 7
x = 2499
e/\(2^{2x+6}=1\)
\(2^{2x+6}=2^0\)
2x + 6 = 0
2x = 0 - 6 = - 6
x = - 6 : 2
x = - 3
j/ \(2^x=8\)
\(2^x=2^3\)
x = 3
g/ \(2^x:2^3=16\)
\(2^{x-3}=2^4\)
x - 3 = 4
x = 4 + 3
x = 7
h/ \(2^x+2^{x+1}+2^{x+2}=56\)
\(2^x\left(1+2+2^2\right)\) = 56
\(2^x.7=56\)
\(2^x=56:7\)
\(2^x=8\)
\(2^x=2^3\)
x = 3
Bài a, b thiên phong giải r, mk chỉ làm những bài còn lại thôi. Chúc bạn học tốt!!!
a) 2x + 2x+1 = 96
=> 2x(1 + 2) = 96
=> 2x.3 = 96
=> 2x = 96 : 3
=> 2x = 32
=> 2x = 25
=. x = 5
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) \(\left(3x+5\right)^2=289\Leftrightarrow\left(3x+5\right)=17^2\Leftrightarrow3x+5=17\Leftrightarrow3x=12\Leftrightarrow x=4\)