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5 tháng 12 2016
1, Ta có : \(9^{1006}=\left(3^2\right)^{1006}=3^{2012}\)
Vì \(2011< 2012\)
\(\Rightarrow3^{2011}< 3^{2012}\)
bài 2 bạn tự làm nha
MT
0
16 tháng 3 2017
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.
T
16 tháng 3 2017
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
13 tháng 6 2018
Ta có :
\(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+8}{2010}+\frac{x+7}{2011}\)
\(\Leftrightarrow\)\(\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+8}{2010}+1\right)+\left(\frac{x+7}{2011}+1\right)\)
\(\Leftrightarrow\)\(\frac{x+4+2014}{2014}+\frac{x+3+2015}{2015}=\frac{x+8+2010}{2010}+\frac{x+7+2011}{2011}\)
\(\Leftrightarrow\)\(\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2010}+\frac{x+2018}{2011}\)
\(\Leftrightarrow\)\(\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2010}-\frac{x+2018}{2011}=0\)
\(\Leftrightarrow\)\(\left(x-2018\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)
Vì \(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2010}-\frac{1}{2011}\ne0\)
Nên \(x-2018=0\)
\(\Leftrightarrow\)\(x=2018\)
Vậy \(x=2018\)
Chúc bạn học tốt ~
13 tháng 6 2018
Ta có: \(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+7}{2011}+\frac{x+8}{2010}\)
\(\Rightarrow\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+7}{2011}+1\right)+\left(\frac{x+8}{2010}+1\right)\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2013}=\frac{x+2018}{2011}+\frac{x+2018}{2010}\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2013}-\frac{x+2018}{2011}-\frac{x+2018}{2010}=0\)
\(\Rightarrow\left(x+2018\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
\(\Rightarrow x+2018=0\Rightarrow x=-2018\)
Chúc bn hc tốt! ^_^
11 tháng 10 2020
a, |x-7|=|9-234|
=> |x-7|=|-225|
=> |x-7|=225
=>\(\orbr{\begin{cases}x-7=225\\x-7=-225\end{cases}}\)=>\(\orbr{\begin{cases}x=232\\x=-218\end{cases}}\)
b, \(\left(\frac{x}{8}\right)^2+\frac{3}{16}=\frac{7}{16}\)
=>\(\left(\frac{x}{8}\right)^2=\frac{7}{16}-\frac{3}{16}\)
=>\(\left(\frac{x}{8}\right)^2=\frac{1}{4}\)
=>\(\left(\frac{x}{8}\right)^2=\left(\frac{1}{2}\right)^2\)
=>\(\orbr{\begin{cases}\frac{x}{8}=\frac{1}{2}\\\frac{x}{8}=\frac{-1}{2}\end{cases}}\)=>\(\orbr{\begin{cases}x.2=8\\x.2=-8\end{cases}}\)=>\(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
c, (8-32)2+(x-39)2=10
=>(-1)2+(x-39)2=10
=>(x-39)2=10-1
=>(x-39)2=9
=>(x-39)2=32
=>\(\orbr{\begin{cases}x-39=3\\x-39=-3\end{cases}}\)=>\(\orbr{\begin{cases}x=42\\x=36\end{cases}}\)
11 tháng 10 2020
a)\(\left|x-7\right|=\left|9-234\right|\)
\(\Rightarrow\left|x-7\right|=\left|-225\right|\)
\(\Rightarrow\left|x-7\right|=225\)
\(\Rightarrow x-7=\pm225\)
- \(x-7=225\Rightarrow x=232\)
- \(x-7=-225\Rightarrow x=-218\)
Vậy \(x\in\left\{232;-218\right\}\)
b)\(\left(\frac{x}{8}\right)^2+\frac{3}{16}=\frac{7}{16}\)
\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{7}{16}-\frac{3}{16}\)
\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{4}{16}\)
\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{1}{4}\)
\(\Rightarrow\left(\frac{x}{8}\right)^2=\left(\pm\frac{1}{2}\right)^2\)
\(\Rightarrow\frac{x}{8}=\pm\frac{1}{2}\)
- \(\Rightarrow\frac{x}{8}=\frac{1}{2}\Rightarrow\frac{x}{8}=\frac{4}{8}\Rightarrow x=4\)
- \(\Rightarrow\frac{x}{8}=\frac{-1}{2}\Rightarrow\frac{x}{8}=\frac{-4}{8}\Rightarrow x=-4\)
Vậy \(x\in\left\{\pm4\right\}\)
c)\(\left(8-3^2\right)^2+\left(x-39\right)^2=10\)
\(\Rightarrow\left(8-9\right)^2+\left(x-39\right)^2=10\)
\(\Rightarrow\left(-1\right)^2+\left(x-39\right)^2=10\)
\(\Rightarrow1+\left(x-39\right)^2=10\)
\(\Rightarrow\left(x-39\right)^2=9\)
\(\Rightarrow\left(x-39\right)^2=\left(\pm3\right)^2\)
\(\Rightarrow x-39=\pm3\)
- \(\Rightarrow x-39=3\Rightarrow x=42\)
- \(\Rightarrow x-39=-3\Rightarrow x=36\)
Vậy \(x\in\left\{42;36\right\}\)
I
4 tháng 4 2020
PT đã cho suy ra thành
\(\left(\frac{x^{2010}}{a^2+b^2+c^2+d^2}-\frac{x^{2010}}{a^2}\right)+\left(\frac{y^{2010}}{a^2+b^2+c^2+d^2}-\frac{y^{2010}}{b^2}\right)+\left(\frac{z^{2010}}{a^2+b^2+c^2+d^2}-\frac{z^{2010}}{c^2}\right)\)
\(+\left(\frac{t^{2010}}{a^2+b^2+c^2+d^2}-\frac{t^{2010}}{d^2}\right)=0\)
\(=>x^{2010}\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\right)+\left(tương\right)Tựnha=0\)
Do
\(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\ne0\)
máy cái bạn tự suy ra cx thế
\(=>x^{2010}=y^{2010}=z^{2010}=t^{2010}=0=>x=y=z=t=0\)
ta có
\(T=x^{2011}+y^{2011}+z^{2011}+t^{2011}=0+0+0+0=0\)
4 tháng 4 2020
Ta có:
\(\frac{x^{2010}+y^{2010}+z^{2010}+t^{2010}}{a^2+b^2+c^2+d^2}=\frac{x^{2010}}{a^2}+\frac{y^{2010}}{b^2}+\frac{z^{2010}}{c^2}+\frac{t^{2010}}{d^2}\)
<=> \(x^{2010}\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)+y^{2010}\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\)
\(+z^{2010}\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)+t^{2010}\left(\frac{1}{d^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)=0\)(1)
Lại có: \(x^{2010};y^{2010};z^{2010};t^{2010}\ge0;\forall x,y,z,t\)
và với mọi a; b ; c ; d khác 0 có:
\(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2+d^2}>0\)
\(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2+d^2}>0\);
\(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2+d^2}>0\);
\(\frac{1}{d^2}-\frac{1}{a^2+b^2+c^2+d^2}>0\)
=> \(x^{2010}\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)
\(y^{2010}\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)
\(z^{2010}\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)
\(t^{2010}\left(\frac{1}{d^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)
=> \(x^{2010}\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)+y^{2010}\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\)
\(+z^{2010}\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)+t^{2010}\left(\frac{1}{d^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)
Như vậy (1) xảy ra<=> \(x^{2010}=y^{2010}=z^{2010}=t^{2010}=0\)
<=> x = y = z = t = 0
Thay vào T ta có : T = 0
26 tháng 7 2019
\(a,|x|=2001\)
\(\Rightarrow x=-2001;x=2001\)
\(c,3-\left(x-2\right)=-2x+7\)
\(\Rightarrow3-x+2=-2x+7\)
\(\Rightarrow5-x=-2x+7\)
\(\Rightarrow x=2\)
\(d,\left(\frac{3}{4}\right)+\frac{2}{5}x=\frac{29}{30}\)
\(\Rightarrow\frac{2}{5}x=\frac{13}{60}\)
\(\Rightarrow x=\frac{13}{24}\)
\(e,\left(\frac{3}{7}\right)^5.x=\left(\frac{3}{7}\right)^7\)
\(\Rightarrow x=\left(\frac{3}{7}\right)^2\)
VM
26 tháng 9 2019
\(\frac{16^x}{8}=2^x\)
\(\Rightarrow8=16^x:2^x\)
\(\Rightarrow8=8^x\)
\(\Rightarrow8^x=8^1\)
\(\Rightarrow x=1\)
Vậy \(x=1.\)
Chúc bạn học tốt!
CG
26 tháng 9 2019
16x/8=2x
<=> (24)x/23=2x
<=> 24x/23=2x
<=> 24x-3=2x
=> 4x-3=x
<=> 4x-x=3
<=> 3x=3
<=> x=3:3
<=> x=1
Vậy x=1
Chúc bạn học tốt
\(8^x:2^x=16^{2011}\)
\(\Leftrightarrow\left(2^3\right)^x:2^x=\left(2^4\right)^{2011}\)
\(\Leftrightarrow2^{3x}:2^x=2^{8044}\)
\(\Leftrightarrow2^{2x}=2^{8044}\)
\(\Rightarrow2x=8044\)
\(\Rightarrow x=4022\)
8x : 2x = 162011
(23)x : 2x = (24)2011
23x : 2x = 28044
=> 3x-x=8044
=> 2x=8044
=> x=8044:2
=> x=4022