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TA CÓ: \(\frac{x}{2013}+\frac{x-1}{2012}=\frac{x-2}{2011}+\frac{x-3}{2010}\)
\(\Rightarrow\frac{x}{2013}+\frac{x-1}{2012}-\frac{x-2}{2011}-\frac{x-3}{2010}=0\)
\(\frac{x}{2013}-1+\frac{x-1}{2012}-1-\frac{x-2}{2011}+1-\frac{x-3}{2010}+1=0\)
\(\left(\frac{x}{2013}-1\right)+\left(\frac{x-1}{2012}-1\right)-\left(\frac{x-2}{2011}-1\right)-\left(\frac{x-3}{2010}-1\right)=0\)
\(\frac{x-2013}{2013}+\frac{x-2013}{2012}-\frac{x-2013}{2011}-\frac{x-2013}{2010}=0\)
\(\left(x-2013\right).\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
MÀ \(\frac{1}{2013}< \frac{1}{2011};\frac{1}{2012}< \frac{1}{2010}\)
\(\Rightarrow\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\ne0\)
\(\Rightarrow x-2013=0\)
\(x=2013\)
VẬY X= 2013
CHÚC BN NĂM MỚI VUI VẺ NHA!!!!!!!
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=> x + 1 là Ước của 63
=> x + 1 thuộc {+1; +3; +7; +9; +21; +63}
với x + 1 = +1
=> x = 0 hoặc x = -2
với x + 1 = +3
=> x = 2 hoặc x = -4
với x + 1 = +7
=> x = 6 hoặc x = -8
với x + 1 = +9
=> x = 8 hoặc x = -10
với x + 1 = +21
=> x = 20 hoặc x = -22
với x + 1 = +63
=> x = 62 hoặc x = -64
mà x + 1 nhỏ hơn 63
=> x thuộc {0; -2; 2; -4; 6; -8; 8; -10; 20; -22}
\(x\in\left\{2;5;6;8;11;13;14;17;20;23;26;27;29;32;34;35;38;41;44;47;48;50;53;55;56;59;\right\}\)
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( x + 1 ) + ( x + 2 ) + ( x + 3 ) + .... + ( x + 100 ) = 5750
( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750
100 chữ số x
100 . x + 5050 = 5750
100 . x = 5750 - 5050
100 . x = 700
x = 700 : 100
x = 7
(x+x+...+x)+(1+2+3+...+100)=5750
100x+5050=5750
100x=5750-5050
100x=700
x=700/100
x=7
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a)231-(x-6)=103
x-6=231-103=128
x=128+6=134
b)x+10-25=0
x+10=0+25=25
x=25-10=15
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Câu 1 bị sai đề bài.
Câu 2:
\(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}=\frac{2012-1}{2012}+\frac{2013-1}{2013}+\frac{2011+1+1}{2011}\)
\(=1-\frac{1}{2012}+1-\frac{1}{2013}+1+\frac{1}{2011}+\frac{1}{2011}\)
Vì:
\(\frac{1}{2011}>\frac{1}{2012};\frac{1}{2011}>\frac{1}{2013}\Rightarrow\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}>0\)
\(\Rightarrow\)\(\frac{2012-1}{2012}+\frac{2013-1}{2013}+\frac{2011+1+1}{2011}>3\)
\(\Rightarrow\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}>3\)
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\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2001}{2003}\)
\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\right)=\frac{1}{2}\cdot\frac{2001}{2003}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2003}\)
\(\Rightarrow x+1=2003\)
\(x=2002\)
Vậy x = 2002
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Từ đề bài ta sẽ có: \(\frac{a}{2011}+\frac{b}{2012}+\frac{c}{2013}=\frac{a+b+c}{6036}.\)
Suy ra a + b + c = 6036 : 3 = 2012
Ta có: \(\frac{a}{2011}+\frac{b}{2012}+\frac{c}{2013}=\frac{2012}{6036}.\)
tới đây thì mình bí rồi! Bạn tự giải nhé! Ai thấy đúng nhớ tk cho mình
như thế vậy thì tớ cg nghĩ ra rồi, dù sao thì cg cảm ơn bạn đã trả lời câu hỏi của mk
2^x+2012=2013
2^x=1
Vậy k có số nguyên x thỏa mãn !
2^x=2013-2012
2^x=1
x=0