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a) \(\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=\sqrt{16}\) \(\Rightarrow\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=4\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}-\dfrac{1}{3}=-2\\\dfrac{x}{2}-\dfrac{1}{3}=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}=\dfrac{-5}{3}\\\dfrac{x}{2}=\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=\dfrac{14}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{-10}{3}\) hoặc \(x=\dfrac{14}{3}\) thì thỏa mãn đề bài.
b) \(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\) \(\Rightarrow\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\) \(\Rightarrow\dfrac{x+4+2010}{2010}+\dfrac{x+3+2011}{2011}=\dfrac{x+2+2012}{2012}+\dfrac{x+1+2013}{2013}\) \(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\) \(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\) \(\Rightarrow\left(x+2014\right)\times\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\) \(\Rightarrow x+2014=0\) \(\Rightarrow x=-2014\)
Vậy \(x=-2014\) thì thỏa mãn đề bài.
c) \(3^{x+2}+4\times3^{x+1}=7\times3^6\) \(\Rightarrow3^{x+1+1}+4\times3^{x+1}=7\times3^6\) \(\Rightarrow3^{x+1}\times3+4\times3^{x+1}=7\times3^6\) \(\Rightarrow\left(3+4\right)\times3^{x+1}=7\times3^6\) \(\Rightarrow3^{x+1}=3^6\) \(\Rightarrow x+1=6\) \(\Rightarrow x=5\)
Vậy \(x=5\) thì thỏa mãn đề bài.
a)
\(\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=\sqrt{16}\\ \Rightarrow\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=4\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{1}{3}=2\\\dfrac{x}{2}-\dfrac{1}{3}=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{1}{3}+2\\\dfrac{x}{2}=\dfrac{1}{3}-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{7}{3}\\\dfrac{x}{2}=\dfrac{-5}{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{3}.2\\x=\dfrac{-5}{3}.2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{3}\\x=\dfrac{-10}{3}\end{matrix}\right.\)
b)
\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(\Rightarrow\dfrac{x+4}{2010}+1+\dfrac{x+3}{2011}+1=\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\)
\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)
\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)
\(\Rightarrow\left(x+2014\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)
mà \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\)
=> x + 2014 = 0
=> x = -2014
vậy x = -2014
c)\(3^{x+2}+4.3^{x+1}=7.3^6\)
\(\Rightarrow3^{x+1}.3+4.3^{x+1}=7.3^6\\ \Rightarrow3^{x+1}\left(3+4\right)=7.3^6\\ \Rightarrow3^{x+1}.7=7.3^6\\ \Rightarrow3^{x+1}=3^6\\ \Rightarrow x+1=6\\ x=6-1\\ x=5\)
vậy x = 5
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h) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow x=2\)
haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=
bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây
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a: \(\left|x\right|=3+\dfrac{1}{5}=\dfrac{16}{5}\)
mà x<0
nên x=-16/5
b: \(\left|x\right|=-2.1\)
nên \(x\in\varnothing\)
c: \(\left|x-3.5\right|=5\)
=>x-3,5=5 hoặc x-3,5=-5
=>x=8,5 hoặc x=-1,5
d: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=>|x+3/4|=1/2
=>x+3/4=1/2 hoặc x+3/4=-1/2
=>x=-1/4 hoặc x=-5/4
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3. Từ \(\dfrac{x-2}{27}=\dfrac{3}{x-2}\Rightarrow\left(x-2\right)^2=81\)
\(\Rightarrow\left(x-2\right)^2=\left(\pm9\right)^2\\ \Rightarrow\left[{}\begin{matrix}x-2=-9\\x-2=9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=11\end{matrix}\right.\)
Vậy x = -7 hoặc x = 11
4. Từ \(\dfrac{2x+5}{9-2x}=\dfrac{2}{5}\)
\(\Rightarrow5\left(2x+5\right)=2\left(9-2x\right)\\ \Leftrightarrow10x+25=18-4x\\ \Leftrightarrow14x=-7\\ \Rightarrow x=-\dfrac{1}{2}\)
5. Từ \(\dfrac{x-7}{x+8}=\dfrac{x-8}{x+9}\)
\(\Rightarrow\left(x-7\right)\left(x+9\right)=\left(x-8\right)\left(x+8\right)\\ \Leftrightarrow x^2+2x-63=x^2-64\\ \Leftrightarrow2x=-1\\ \Rightarrow x=-\dfrac{1}{2}\)
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1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
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1. Tìm n, biết:
a) \(\dfrac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
\(\Rightarrow\left(-2\right)^n.\left(-2\right)^2=\left(-2\right)^5\)
(-2)n + 2 = (-2)5
n + 2 = 5
n = 5 - 2
n = 3.
b) \(\dfrac{8}{2^n}=2\)
\(\Rightarrow\dfrac{2^3}{2^n}=2\)
\(\Rightarrow\) 2n . 2 = 23
n + 1 = 3
n = 3 - 1
n = 2.
c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
2n - 1 = 3
2n = 3 + 1
2n = 4
n = 4 : 2
n = 2.
2. Tính:
a) \(\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left[\left(\dfrac{1}{2}\right)^2\right]^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{2}\right)^4\)
\(=\left(\dfrac{1}{2}\right)^7\)
\(=\dfrac{1}{128}\)
b) 273 : 93
= (33)3 : (32)3
= 39 : 36
= 33
= 27
c) 1252 : 253
= (53)2 : (52)3
= 56 : 56
= 1
d) \(\dfrac{27^2.8^5}{6^6.32^3}\)
\(=\dfrac{\left(3^3\right)^2.\left(2^3\right)^5}{6^6.\left(2^5\right)^3}\)
\(=\dfrac{3^6.2^{15}}{6^6.2^{15}}\)
\(=\dfrac{3^6}{6^6}\)
\(=\dfrac{1}{64}.\)
B2 :
b) 27\(^3\): 9\(^3\)= (27:9)\(^3\)= 3\(^3\)
c) 125\(^2\): 25\(^3\)= 15625 : 15625 = 1
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b: =>(3x-1)(3x+1)(2x+3)=0
hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)
c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)
=>2x-1/3=19/12 hoặc 2x-1/3=-19/12
=>2x=23/12 hoặc 2x=-15/12=-5/4
=>x=23/24 hoặc x=-5/8
d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)
=>-5/6x=-3/2
=>x=3/2:5/6=3/2*6/5=18/10=9/5
e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4
=>2/5x=5/4 hoặc 2/5x=-1/4
=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8
f: =>14x-21=9x+6
=>5x=27
=>x=27/5
h: =>(2/3)^2x+1=(2/3)^27
=>2x+1=27
=>x=13
i: =>5^3x*(2+5^2)=3375
=>5^3x=125
=>3x=3
=>x=1
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Số thập phân hữu hạn là mấy số thập phân không có dấu .... ở đuôi ý bạn ạ.
Còn bài này mình không hiểu rõ đề bài mấy bạn ạ
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a: Đặt A=0
=>-2/3x=5/9
hay x=-5/6
b: Đặt B(x)=0
=>(x-2/5)(x+2/5)=0
=>x=2/5 hoặc x=-2/5
c: Đặt C(X)=0
\(\Leftrightarrow x^3\cdot\dfrac{1}{2}=-\dfrac{4}{27}\)
\(\Leftrightarrow x^3=-\dfrac{8}{27}\)
hay x=-2/3
n=3
m=2
\(\dfrac{8}{27}=\dfrac{2^n}{3\text{x}3^m}\)
=>\(\dfrac{2^3}{3^3}=\dfrac{2^n}{3\text{x}3^m}\)
n=3và \(3\text{x}3^2=3^{1+2}\)=>m=2
Vậyn=3,m=2