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1, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\) ( Trừ mỗi vế cho 2 ta được phương trình như này nhé ! )
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Do \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên \(x-2010=0\Leftrightarrow x=2010\)
2, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\) \(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\) Do \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\ne0\) nên \(100-x=0\Leftrightarrow x=100\)
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Bài 1 :
Gọi \(A=5+5^2+5^3+...+5^{98}+5^{99}\\ 5A=5^2+5^3+5^4+...+5^{99}+5^{100}\\ 5A-A=\left(5^2+5^3+5^4+...+5^{99}+5^{100}\right)-\left(5+5^2+5^3+...+5^{98}+5^{99}\right)\\ 4A=5^{100}-5\\ A=\dfrac{5^{100}-5}{4}\)
Bài 2:
\(\left(12x-4\right)\cdot8^{2022}=4\cdot8^{2023}\\ 12x-4=4\cdot8^{2023}:8^{2022}\\ 12x-4=4\cdot8\\ 12x-4=32\\ 12x=36\\ x=3\)
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Bài 2:
Ta có: (x-3)(x+4)>0
=>x>3 hoặc x<-4
Bài 3:
a: \(5S=5-5^2+...+5^{99}-5^{100}\)
\(\Leftrightarrow6S=1-5^{100}\)
hay \(S=\dfrac{1-5^{100}}{6}\)
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Lời giải:
a. $(x-3)(y+1)=5=1.5=5.1=(-1)(-5)=(-5)(-1)$
Vì $x-3, y+1$ cũng là số nguyên nên ta có bảng sau:
b.
$A=21+5+(5^2+5^3)+(5^4+5^5)+....+(5^{98}+5^{99})$
$=26+5^2(1+5)+5^4(1+5)+....+5^{98}(1+5)$
$=2+24+(1+5)(5^2+5^4+...+5^{98}$
$=2+24+6(5^2+5^4+....+5^{98})=2+6(4+5^2+5^4+...+5^{98})$
$\Rightarrow A$ chia $6$ dư $2$.
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1.3.5.7........99 = \(\frac{\left(1.3.5.7......99\right)\left(2.4.6......100\right)}{2.4.6......100}\)= \(\frac{1.2.3......99.100}{2^{50}\left(1.2.3.....50\right)}=\frac{51.52.53.......100}{2.2.2......2}=\frac{51}{2}.\frac{52}{2}....\frac{100}{2}\)(ĐPCM)
50 số 2
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A= 1 + 5 + 52 + 5 3 + ... + 5800
5A= 5 + 52 + 53 + .... +5 800 + 5801
5A - A = 5801 - 1
4a = 5801 - 1
5801 - 1 +1 = 5n
⇒ 5801 = 5n ⇒ n = 801