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\(2x\cdot\dfrac{-4}{9}+2x\cdot\dfrac{-5}{9}=\dfrac{8}{11}\)
\(2x\cdot\left(\dfrac{-4}{9}+\dfrac{-5}{9}\right)=\dfrac{8}{11}\)
\(2x\cdot\left(-1\right)=\dfrac{8}{11}\)
2x = \(\dfrac{8}{11}:\left(-1\right)=\dfrac{-8}{11}\)
x = \(\dfrac{-8}{11}:2=\dfrac{-4}{11}\)
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a)\(\left(x-32\right):16-13=48\)
\(\left(x-32\right):16=48+13\)
\(x+32=61.16\)
\(x+32=976\)
\(x=976-32\)
\(x=944\)
b) \(-2\left(2x-8\right)+\left(4-2x\right)=-72\)
\(-4x+16+4-2x=-72\)
\(-4x-2x=-72-16-4\)
\(-6x=-92\)
\(x=\frac{-92}{-6}=\frac{46}{3}\)
hok tốt!!
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Bài 1: a) \(-2.\left(2x-8\right)+3.\left(4-2x\right)=\left(-72\right)-5.\left(3x-7\right)\)
\(-4x+16+12-6x=-72-15x+35\)
\(-4x-6x+15x=-72+35-16-12\)
\(5x=-65\)
\(x=-\frac{65}{5}\)
\(x=-13\)
b) \(3.\left|2x^2-7\right|=33\)
\(\left|2x^2-7\right|=\frac{33}{3}=11\)
\(\Rightarrow\orbr{\begin{cases}2x^2-7=11\\2x^2-7=-11\end{cases}\Rightarrow\orbr{\begin{cases}2x^2=18\\2x^2=-4\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=9\\x^2=-2\left(vl\right)\end{cases}\Rightarrow}\orbr{\begin{cases}x=\pm3\\\end{cases}}}\)
Bài 2:
Ta có: \(2n+1⋮n-3\)
\(2n-6+7⋮n-3\)
\(2\left(n-3\right)+7⋮n-3\)
Vì \(2\left(n-3\right)⋮n-3\)
Để \(2\left(n-3\right)+7⋮n-3\)
Thì \(7⋮n-3\Rightarrow n-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
n-3 | -1 | 1 | 7 | -7 |
n | 2 | 4 | 10 | -4 |
Vậy.....
hok tốt!!
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\(5.\left(3x+8\right)-7.\left(2x+3\right)=16\)
\(\Rightarrow15x+40-14x+21=16\)
\(\Rightarrow x+19=16\)
\(\Rightarrow x=16-19=-3\)
Vậy x = -3
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1.
a) \(2^x=128\)
\(2^x=2^7\)
\(=>x=7\)
b) \(8^{x-1}=64\)
\(8^{x-1}=8^2\)
\(=>x-1=2\)
\(x=2+1\)
\(=>x=3\)
c) \(3+3^x=30\)
\(3^x=30-3\)
\(3^x=27=3^3\)
\(=>x=3\)
d) \(\left(x+2\right)=64\) -> đề có thiếu không vậy?
e) \(3^2.x=3^5\)
\(x=3^5:3^2\)
\(=>x=3^3=27\)
f) \(\left(2x-1\right)^3=343\)
\(\left(2x-1\right)^3=7^3\)
\(=>2x-1=7\)
\(2x=7+1\)
\(2x=8\)
\(x=8:2\)
\(=>x=4\)
\(#Wendy.Dang\)
a,\(2^x\)=128 b,\(8^{x-1}\)=64 c,3+\(3^x\)=30 d,x+2=64
\(2^7\)=128 \(8^{x-1}\)=\(8^2\) \(3^x\)=30-3 x=64-2
=>x=7 =>x-1=2 \(3^x\)=27 x=62
x=2+1=3 \(3^x\)=\(3^3\)
=>x=3
e,\(3^2\).x=\(3^5\) f,(2x-\(1^3\))=343
x=\(3^5\):\(3^2\) 2x=1+343
x=27 2x=344
x=344:2
x=172
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a) 10-x-5=-5-7-11
=> 5 - x = -23
=> x = 28
b) |x| -3=0
=> |x| = 3
=> x = 3 hoặc x -3
c) ( 7-|x| ) .(2x-4)=0
=> 7 - |x| = 0 hoặc 2x - 4 = 0
=> |x| = 7 hoặc 2x = 4
=> x = 7 hoặc x = - 7 hoặc x = 2
c)2+3x=-15-19
=> 2 + 3x = -34
=> 3x = 36
=> x = 12
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(2x - 15)5 = (2x-15)3
<=> (2x-15) = 0 hoặc (2x-15) = 1
+ TH1: (2x-15)5 = (2x-15)3
05 = 03 = 0
+ TH2: (2x-15)5 = (2x-15)3
15 = 13 = 1
\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\left(2x-15\right)^3\cdot\left[\left(2x-15\right)^2-1\right]=0\)
\(\hept{\begin{cases}2x-15=0\\\left(2x-15\right)^2-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{15}{2}\\\left(2x-15\right)^2=1\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{15}{2}\\\hept{\begin{cases}2x-15=1\\2x-15=-1\end{cases}\Rightarrow\hept{\begin{cases}x=8\\x=7\end{cases}}}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{15}{2}\\\hept{\begin{cases}2x-15=1\\2x-15=-1\end{cases}}\Rightarrow\hept{\begin{cases}x=8\\x=7\end{cases}}\end{cases}}\)
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a) \(2x-8=5\)
\(\Leftrightarrow2x=13\)
\(\Leftrightarrow x=\frac{13}{2}\)
b) \(\left(6x-12\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}6x-12=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}6x=12\\x=-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)
c) \(|2x-1|+3=6^2\)
\(\Leftrightarrow|2x-1|+3=36\)
\(\Leftrightarrow|2x-1|=33\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=33\\2x-1=-33\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=34\\2x=-32\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=17\\x=-16\end{cases}}}\)
a,2x-8=5
2x =5+8
2x =13
x =13:2 k đi rồi tui làm o
x =13/2=6,5
(2x-5)^3=8
=>(2x-5)^3=2^3
=>2x-5 = 2
2x = 2 + 5 = 7
x = 7 : 2 = 7/2