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\(2^{225}=\left(2^3\right)^{75}=8^{75}< 9^{75}=\left(3^2\right)^{75}=3^{150}\)
\(2^{2009}+2^{2008}+.......+2+1=b\)
\(\Rightarrow2b=2^{2010}+2^{2009}+.........+2^2+2\)
\(\Rightarrow2b-b=2^{2010}-1\Rightarrow b=2^{2010}-1\)
\(\Rightarrow A=2^{2010}-b=2^{2010}-\left(2^{2010}-1\right)=1\)
Lời giải:
$\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}$
$\Leftrightarrow \frac{x-1}{2011}+\frac{x-2}{2010}=\frac{x-3}{2009}+\frac{x-4}{2008}$
$\Leftrightarrow \frac{x-1}{2011}-1+\frac{x-2}{2010}-1=\frac{x-3}{2009}-1+\frac{x-4}{2008}-1$
$\Leftrightarrow \frac{x-2012}{2011}+\frac{x-2012}{2010}=\frac{x-2012}{2009}+\frac{x-2012}{2008}$
$\Leftrightarrow (x-2012)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0$
Dễ thấy $\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}< 0$
Do đó $x-2012=0\Rightarrow x=2012$
\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}.\)
\(\Rightarrow\frac{x-1}{2011}+\frac{x-2}{2010}=\frac{x-4}{2008}+\frac{x-3}{2009}\)
\(\Rightarrow\left(\frac{x-1}{2011}-1\right)+\left(\frac{x-2}{2010}-1\right)=\left(\frac{x-4}{2008}-1\right)+\left(\frac{x-3}{2009}-1\right)\)
\(\Rightarrow\left(\frac{x-1-2011}{2011}\right)+\left(\frac{x-2-2010}{2010}\right)=\left(\frac{x-4-2008}{2008}\right)+\left(\frac{x-3-2009}{2009}\right)\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}=\frac{x-2012}{2008}+\frac{x-2012}{2009}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2008}-\frac{x-2012}{2009}=0\)
\(\Rightarrow\left(x-2012\right).\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)
Vì \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2008}-\frac{1}{2009}\ne0.\)
\(\Rightarrow x-2012=0\)
\(\Rightarrow x=0+2012\)
\(\Rightarrow x=2012\)
Vậy \(x=2012.\)
Chúc bạn học tốt!
a: \(\left(-\dfrac{3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
\(\Leftrightarrow\left(-\dfrac{3}{4}\right)^{3x-1}=\left(-\dfrac{3}{4}\right)^{-4}\)
=>3x-1=-4
=>3x=-3
hay x=-1
b: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=-1\\x-7=1\end{matrix}\right.\Leftrightarrow x\in\left\{7;6;8\right\}\)
c: \(\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2=0\)
=>x-1/2=0 và y+1/2=0
=>x=1/2 và y=-1/2
=20122011-2012.20122010+2012.20122009-.......................-2012.20122-1
còn lại tự làm nhá
Bài 1
\(a,\left|x\right|=-\left|-\frac{5}{7}\right|=>x\in\varnothing\)
\(b,\left|x+4,3\right|-\left|-2,8\right|=0\)
\(=>\left|x+4,3\right|-2,8=0\)
\(=>\left|x+4,3\right|=0+2,8=2,8\)
\(=>x+4,3=\pm2,8\)
\(=>\hept{\begin{cases}x+4,3=2,8\\x+4,3=-2,8\end{cases}=>\hept{\begin{cases}x=-1,5\\x=-7,1\end{cases}}}\)
\(c,\left|x\right|+x=\frac{2}{3}\)
\(=>\hept{\begin{cases}x+x=\frac{2}{3}\\-x+x=\frac{2}{3}\end{cases}}=>\hept{\begin{cases}x=\frac{1}{3}\\x=-\frac{1}{3}\end{cases}}\)
Bài 1 :
a) \(-3+\left(-4\right)-\left(-3\right)+\left(2+7-10\right)=-3-4+3+2+7-10=-5\)
b) \(3-\left(-3+2-7\right)+\left(-4\right)=3+3-2+7-4=7\)
c) \(7+\left(-2-3+7\right)-\left(-2\right)=7-2-3+7+2=17\)
d) \(-\left(-3\right)-\left(-2+3-8\right)+\left(-6\right)=3+2-3+8-6=4\)
Bài 2 :
a) \(x^2-2x-\left(3x-2x\right)=x^2-2x-3x+2x=x^2-3x\)
b) \(-\left(x^2+3x^2\right)-\left(-5x^2+3x\right)=-x^2-3x^2+5x^2-3x=x^2-3x\)
c) \(\left(x-y\right)-\left(x+3y+1\right)=x-y-x-3y-1=-4y-1\)
Bài 1:
a, -3+ (-4) - (-3) + (2 + 7 - 10)
= -3 - 4 + 3 + 2 + 7 - 10
= 5 - 10
= -5.
b, 3 - (-3 + 2 - 7) + (-4)
= 3 + 3 - 2 + 7 - 4
= 11 - 4
= 7
c, 7 + (-2 - 3 + 7) - (-2)
= 7 - 2 - 3 + 7 + 2
= 9 + 2
= 11.
d, - (-3) - (-2 + 3 - 8) + (-6)
= 3 + 2 - 3 + 8 - 6
= 10 - 6
= 4.
Mình chỉ làm bài 1 thôi nhé.
Chúc bạn học tốt!
a, Đặt \(A=2^{2010}+2^{2009}+2^{2008}+...+2^1+2^0\)
\(\Rightarrow2A=2^{2011}+2^{2010}+2^{2009}+...+2^2+2^1\)
\(\Rightarrow2A-A=2^{2011}-2^0\)
\(\Rightarrow A=2^{2011}-1\)
b,\(7^{x+2}+2.7^{x-1}=345\)
\(7^{x-1}.\left(7^3+2\right)=345\)
\(\Rightarrow7^{x-1}.345=345\)
\(\Rightarrow7^{x-1}=345:345=1\)
\(\Rightarrow7^{x-1}=7^0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
Thanks bạn nhen . Hi hi.