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a) x=3
b) x=1
c) x=1 hoặc -5
d) x=2
e) x=2
g) x=2
h) x=1 hoặc x=0 hoặc x=-1
i) x=-1 hoặc x=0
\(a.4^x=64\)
\(4^x=4^3\)
\(\Rightarrow x=3\)
\(b,3^{x\times4}=81\)
\(3^{x\times4}=3^4\)
\(x\times4=4\)
\(\Rightarrow x=1\)
\(c,\left(2+x\right)^4=81\)
\(\left(2+x\right)^4=3^4\)
\(2+x=3\)
\(x=3-2\)
\(x=1\)
\(d,5^{x\times5}=125\)
\(5^{x\times5}=5^3\)
\(x\times5=3\)
\(x=3:5\)
\(x=\frac{3}{5}\)
1) \(2^x-15=17\)
\(\Leftrightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
2) \(\left(7x-11\right)^3=25\cdot5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=825\)
\(\Leftrightarrow7x-11=\sqrt[3]{825}\)
\(\Leftrightarrow7x=11+\sqrt[3]{825}\)
\(\Rightarrow x=\frac{11+\sqrt[3]{825}}{7}\)
3) \(\left(x+1\right)^{100}-3\left(x+1\right)^{99}=0\)
\(\Leftrightarrow\left(x+1\right)^{99}\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{99}=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
4) \(4x+5\left(x+3\right)=105\)
\(\Leftrightarrow9x+15=105\)
\(\Leftrightarrow9x=90\)
\(\Rightarrow x=10\)
5) \(5\cdot\left(x-2\right)+10\left(x+3\right)=170\)
\(\Leftrightarrow5\left[x-2+2\left(x+3\right)\right]=170\)
\(\Leftrightarrow3x+4=34\)
\(\Leftrightarrow3x=30\)
\(\Rightarrow x=10\)
a, \(390-\left(x-7\right)=13^2:12\)
\(390-\left(x-7\right)=\) \(\dfrac{169}{12}\)
\(x-7=390-\dfrac{169}{12}\)
\(x-7=\dfrac{4511}{12}\)
\(x=\dfrac{4511}{12}+7\)
\(x=\dfrac{4595}{12}\)
Vậy ...
b, \(\left(x-35.2^2\right):7=3^3-24\)
\(\left(x-35.4\right):7=27-24\)
\(\left(x-140\right):7=3\)
\(\Leftrightarrow\left(x-140\right)=3.7\)
\(\Leftrightarrow x-140=21\)
\(\Leftrightarrow x=161\)
Vậy .....
c) \(x-6:2-\left(4^2.3-24\right):2:6=3\)
\(x-3-\left(16.3-24\right):2:6=3\)
\(x-3-\left(48-24\right):2:6=3\)
\(x-3-24:2:6=3\)
\(x-3-2=3\)
\(x=3+2+3\)
\(x=8\)
Vậy ......
d) \(4x-5=5+5^2+5^3+.....+5^{99}\)
Đặt :
\(A=5+5^2+.........+5^{99}\)
\(\Leftrightarrow5A=5^2+5^3+..........+5^{100}\)
\(\Leftrightarrow5A-A=\left(5^2+5^3+......+5^{100}\right)-\left(5+5^2+....+5^{99}\right)\)
\(\Leftrightarrow4A=5^{100}-5\)
\(\Leftrightarrow A=\dfrac{5^{100}-5}{4}\)
\(\Leftrightarrow4x+5=\dfrac{5^{100}-5}{4}\)
Đến đây thì sao nữa nhỉ ?
e) \(\left(2x-1\right)^4=625\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=5\\\left(2x-1\right)^4=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy ....
a) 125 - 5(x + 10) = 12
=> 5(x + 10) = 125 - 12 = 113
=> x + 10 = 113 : 5
=> x + 10 = 113/5
=> x = 113/5 - 10 = 63/5
b) (x - 1)2 = 4 => (x - 1)2 = (\(\pm\)2)2
=> x - 1 = 2 hoặc x - 1 = -2
=> x = 3 hoặc x = -1
c) 5x+1 - 52 . 17 = 52.8
=> 5x . 5 - 52 . 17 = 52 . 8
=> 5x . 5 = 52.8 + 52 . 17
=> 5x . 5 = 52(8 + 17)
=> 5x . 5 = 52 . 52
=> 5x = \(\frac{5^2\cdot5^2}{5}=5^3\)
=> x = 3
d) 84 + (12 - 3.x) : 6 = 85
=> 84 + (12 - 3.x) = 510
=> 12 - 3.x = 426
=> 3.x = -414
=> x = -138
e) (x + 1) + (x + 2) + (x + 3) + ... + (x + 99) + (x + 100) = 5150
=> x + 1 + x + 2 + x + 3 + ... + x + 99 + x + 100 = 5150
=> (x + x + .... + x) + (1 + 2 + 3+ ... + 100) = 5150
=> 100x + 5050 = 5150
=> 100x = 5150 - 5050 = 100
=> x = 1
Bài làm:
\(a,125-5.\left(x+10\right)=12\)
\(\Rightarrow5.\left(x+10\right)=125-12\)
\(\Rightarrow5.\left(x+10\right)=113\)
\(\Rightarrow x+10=\frac{113}{5}\)
\(\Rightarrow x=\frac{63}{5}\)
\(b,\left(x-1\right)^2=4\)
\(\Rightarrow\left(x-1\right)^2=2^2\)
\(\Rightarrow\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
\(c,5^{x+1}-5^2.17=5^2.8\)
\(\Rightarrow5^{x+1}-425=200\)
\(\Rightarrow5^{x+1}=200+425\)
\(\Rightarrow5^{x+1}=625\)
\(\Rightarrow5^{x+1}=5^4\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=3\)
\(d,84+\left(12-3.x\right):6=85\)
\(\left(12-3x\right):6=85-84\)
\(\left(12-3x\right):6=1\)
\(12-3x=6\)
\(3x=6\)
\(x=2\)
\(e,\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+99\right)+\left(x+100\right)=5150\)
\(100.x+\left(1+2+3+...+99+100\right)=5150\)
\(100.x+5050=5150\)
\(100.x=5150-5050\)
\(100.x=100\)
\(x=1\)
Học tốt nhé