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Đặt A = 22009 + 22008 + ... + 21 + 20. Khi đó, M = 22010 - A
Ta có 2A = 22010 + 22009 + ... + 22 + 21.
Suy ra 2A - A = 22010 - 20 = 22010 - 1.
Do đó M = 22010 - A = 22010 - (22010 - 1) = 22010 - 22010 + 1 = = 1.
M=2^2010-(2^2009+2^2008+2^2007+...+2^1+2^0)
M=22010-22009-22008-22007-...-21-20
=>2M=22011-22010-22009-22008-...-22-21
=>2M-M=22011-22010-22009-22008-...-22-21-(22010-22009-22008-22007-...-21-20)
=>M=22011-22010-22009-22008-...-22-21-22010+22009+22008+22007+...+21+20
=22011-22010-22010+20
=22011-2.22010+1
=22011-22011+1
=1
vậy M=1
Bài 1:
a.
$|x+\frac{7}{4}|=\frac{1}{2}$
\(\Leftrightarrow \left[\begin{matrix} x+\frac{7}{4}=\frac{1}{2}\\ x+\frac{7}{4}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-5}{4}\\ x=\frac{-9}{4}\end{matrix}\right.\)
b. $|2x+1|-\frac{2}{5}=\frac{1}{3}$
$|2x+1|=\frac{1}{3}+\frac{2}{5}$
$|2x+1|=\frac{11}{15}$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=\frac{11}{15}\\ 2x+1=\frac{-11}{15}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-2}{15}\\ x=\frac{-13}{15}\end{matrix}\right.\)
c.
$3x(x+\frac{2}{3})=0$
\(\Leftrightarrow \left[\begin{matrix} 3x=0\\ x+\frac{2}{3}=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=\frac{-3}{2}\end{matrix}\right.\)
d.
$x+\frac{1}{3}=\frac{2}{5}-(\frac{-1}{3})=\frac{2}{5}+\frac{1}{3}$
$\Leftrightarrow x=\frac{2}{5}$
Nguyễn Quý Trung:
\(x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)
Bạn bớt 2 vế đi 1/3 thì \(x=\dfrac{2}{5}\)
a) 6x(5x + 3) + 3x(1 – 10x) = 7
⇒ 30x2+18x+3x-30x2=7
⇒21x=7
⇒x=\(\dfrac{7}{21}\)
⇒x= \(\dfrac{1}{3}\)
b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44
⇒15x-63x2-15+63x + 63x2-35x+36x-20=44
⇒79x-35=44
⇒79x=44+35
⇒79x=79
⇒x=1
a: \(\left|3x-2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=4\\3x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b: Ta có: \(\left|5x-3\right|=\left|x-7\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3=x-7\\5x-3=7-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-4\\6x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{3}\end{matrix}\right.\)
2:
a: A(x)=0
=>5x-10-2x-6=0
=>3x-16=0
=>x=16/3
b: B(x)=0
=>5x^2-125=0
=>x^2-25=0
=>x=5 hoặc x=-5
c: C(x)=0
=>2x^2-x-3=0
=>2x^2-3x+2x-3=0
=>(2x-3)(x+1)=0
=>x=3/2 hoặc x=-1
a: =>1/2x-3/4x=-5/6+7/3
=>-1/4x=14/6-5/6=3/2
=>x=-3/2*4=-6
b: =>4/5x-3/2x=1/2+6/5
=>-7/10x=17/10
=>x=-17/7
c: =>6/5x+6/20=6/5-1/3x
=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10
=>x=27/46
d: =>6x+3/2+4/5=1/2-2x
=>8x=1/2-3/2-4/5=-1-4/5=-9/5
=>x=-9/40
\(\dfrac{1}{2}-3x+\left|x-1\right|=0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}-0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}\\ \Rightarrow\left|x-1\right|=\dfrac{1}{2}-3x\\ \Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}-3x\\x-1=-\dfrac{1}{2}+3x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+3x=\dfrac{1}{2}+1\\x-3x=-\dfrac{1}{2}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x=\dfrac{3}{2}\\2x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)
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\(\dfrac{1}{2}\left|2x-1\right|+\left|2x-1\right|=x+1\\ \Rightarrow\left|2x-1\right|\cdot\left(\dfrac{1}{2}+1\right)=x+1\\ \Rightarrow\left|2x-1\right|\cdot\dfrac{3}{2}=x+1\\ \Rightarrow\left|2x-1\right|=x+1:\dfrac{3}{2}\\ \Rightarrow\left|2x-1\right|=x+\dfrac{2}{3}\\ \Rightarrow\left[{}\begin{matrix}2x-1=x+\dfrac{2}{3}\\2x-1=-x-\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-x=\dfrac{2}{3}+1\\2x+x=-\dfrac{2}{3}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\3x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{1}{9}\end{matrix}\right.\)
\(a,\text{Với }x< -2\Rightarrow3-x-x-2=4\\ \Rightarrow-2x=3\Rightarrow x=-\dfrac{3}{2}\left(ktm\right)\\ \text{Với }-2\le x< 3\Rightarrow3-x+x+2=4\\ \Rightarrow0x=-1\Rightarrow x\in\varnothing\\ \text{Với }x\ge3\Rightarrow x-3+x+2=4\\ \Rightarrow2x=5\Rightarrow x=\dfrac{5}{2}\left(ktm\right)\)
Vậy \(x\in\varnothing\)
\(b,\text{Với }x< 2\Rightarrow4-2x+18-6x=21\\ \Rightarrow22-8x=21\Rightarrow x=\dfrac{1}{8}\left(tm\right)\\ \text{Với }2\le x< 3\Rightarrow2x-4+18-6x=21\\ \Rightarrow-4x+14=21\Rightarrow x=-\dfrac{7}{4}\left(ktm\right)\\ \text{Với }x\ge3\Rightarrow2x-4+6x-18=21\\ \Rightarrow8x=43\Rightarrow x=\dfrac{43}{8}\left(tm\right)\)
Vậy \(x\in\left\{\dfrac{1}{8};\dfrac{43}{8}\right\}\)