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<=> \(\frac{\left(x+2014\right)}{2011}+1+\frac{\left(x+2013\right)}{2012}+1=\frac{\left(x+2012\right)}{2013}+1+\frac{\left(x+2011\right)}{2014}+1\)
\(\Rightarrow\frac{\left(x+4025\right)}{2011}+\frac{\left(x+4025\right)}{2012}=\frac{\left(x+4025\right)}{2013}+\frac{\left(x+4025\right)}{2014}\)
=> \(\frac{\left(x+4025\right)}{2011}+\frac{\left(x+4025\right)}{2012}-\frac{\left(x+4025\right)}{2013}-\frac{\left(x+4025\right)}{2014}=0\)
=> \(\left(x+4025\right)\left\lbrack\left(\frac{1}{2011}+\frac{1}{2012}\right)-\left(\frac{1}{2013}+\frac{1}{2014}\right)\right\rbrack=0\)
vì \(\left(\frac{1}{2011}+\frac{1}{2012}\right)>\left(\frac{1}{2013}+\frac{1}{2014}\right)\)
=> \(\left\lbrack\left(\frac{1}{2011}+\frac{1}{2012}\right)-\left(\frac{1}{2013}+\frac{1}{2014}\right)\right\rbrack>0\) hay ≠0
=> \(x+4025=0\)
\(x=-4025\)
a) \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
\(\Leftrightarrow\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-1}{17}-5=0\)
\(\Leftrightarrow\frac{x-90-10}{10}+\frac{x-76-2.12}{12}+\frac{x-58-3.14}{14}+\frac{x-36-4.16}{16}+\frac{x-15-5.17}{17}=0\)
\(\Leftrightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
\(\Leftrightarrow x-100=0\Leftrightarrow x=100\)
Vậy \(S=\left\{100\right\}\)
b) \(\frac{x+2011}{2013}+\frac{x+2012}{2012}=\frac{x+2010}{2014}+\frac{x+2013}{2011}\)
\(\Leftrightarrow\frac{x+2011}{2013}+1+\frac{x+2012}{2012}+1=\frac{x+2010}{2014}+1+\frac{x+2013}{2011}+1\)
\(\Leftrightarrow\frac{x+2011+2013}{2013}+\frac{x+2012+2012}{2012}=\frac{x+2010+2014}{2014}+\frac{x+2013+2011}{2011}\)
\(\Leftrightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}-\frac{x+4024}{2014}-\frac{x+4024}{2011}=0\)
\(\Leftrightarrow\left(x+4024\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)=0\)
\(\Leftrightarrow x+4024=0\Leftrightarrow x=-4024\)
Vậy \(S=\left\{-4024\right\}\)
Phương trình a bạn trừ phân thức đầu tiên cho 1, phân thức thứ hai cho 2, phân thức thứ ba cho 3, phân thức thứ tư cho 4, phân thức thứ năm cho 5, vế còn lại trừ đi 15. Tiếp theo bạn đặt x -100 làm nhân tử chung. Cuối cùng tìm được x= 100
\(\Rightarrow\frac{x}{2010}+\frac{x+1}{2011}+\frac{x+2}{2012}+\frac{x+3}{2013}+\frac{x+4}{2014}-5=0\)
\(\left(\frac{x}{2010}-1\right)+\left(\frac{x+1}{2011}-1\right)+\left(\frac{x+2}{2012}-1\right)\)\(+\left(\frac{x+3}{2013}-1\right)+\left(\frac{x+4}{2014}-1\right)=0\)
\(\frac{x-2010}{2010}+\frac{x-2010}{2011}+\frac{x-2010}{2012}+\frac{x-2010}{2013}+\frac{x-2010}{2014}=0\)
\(\left(x-2010\right).\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)=0\)
mà \(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\ne0\Rightarrow x+2010=0\Rightarrow x=-2010\)
Vậy x=-2010
Tìm x:
\(\dfrac{x}{2010}+\dfrac{x+1}{2011}+\dfrac{x+2}{2012}+\dfrac{x+3}{2013}+\dfrac{x+4}{2014}=5\)
\(\dfrac{x}{2010}+\dfrac{x+1}{2011}+\dfrac{x+2}{2012}+\dfrac{x+3}{2013}+\dfrac{x+4}{2014}=5\)
\(\Leftrightarrow\left(\dfrac{x}{2010}-1\right)+\left(\dfrac{x+1}{2011}-1\right)+\left(\dfrac{x+2}{2012}-1\right)+\left(\dfrac{x+3}{2013}-1\right)+\left(\dfrac{x+4}{2014}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-2010}{2010}+\dfrac{x-2010}{2011}+\dfrac{x-2010}{2012}+\dfrac{x-2010}{2013}+\dfrac{x-2010}{2014}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}\right)=0\)
\(\Leftrightarrow x=2010\)
Ta có : \(\dfrac{x+1}{2014}+\dfrac{x+2}{2013}+\dfrac{x+3}{2012}+\dfrac{x+4}{2011}=0\)
\(\Leftrightarrow\left(\dfrac{x+1}{2014}+1\right)+\left(\dfrac{x+2}{2013}+1\right)+\left(\dfrac{x+3}{2012}+1\right)+\left(\dfrac{x+4}{2011}+1\right)=4\)
\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+\dfrac{x+2015}{2012}+\dfrac{x+2015}{2011}=4\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}\right)=4\) \(\Leftrightarrow\left(x+2015\right).0,002=4\) ( mik lấy gần bằng nha )
\(\Leftrightarrow x+2015=2000\Leftrightarrow x=-15\)
Vậy phương trình có nghiệm là x=-15
1.
<=> B=\(3^{24}-\left\lbrack\left(3^3\right)^4+1\right\rbrack\left\lbrack\left(3^2\right)^6-1\right\rbrack\)
\(B=3^{24}-\left(3^{12}+1\right)\left(3^{12}-1\right)\)
\(B=3^{24}-3^{24}+1\)
\(B=1\)
2.
xét vế đầu tiên
\(2011\cdot2013+2012\cdot2014\)
\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)
\(=2012^2-1+2013^2-1\)
\(=2012^2+2013^2-2\)
=> \(2011\cdot2013+2012\cdot2014=2012^2+2013^2-2\)
\(\Leftrightarrow\frac{x-1}{2011}+1+\frac{x-2}{2012}+1+\frac{x-3}{2013}+1+\frac{x-4}{2014}+1-\left(x+2010\right)=0\)
\(\Leftrightarrow\frac{x+2010}{2011}+\frac{x+2010}{2012}+\frac{x+2010}{2013}+\frac{x+2010}{2014}-\left(x+2010\right)=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}-1\right)=0\)
\(\Leftrightarrow x=-2010\)
x+1/2013+1+x+2/2012+1=x+3/2011+1+x+4/2010+1
x+1+2013/2013+x+2+2012/2012=x+3+2011/2011+x+4+2010/2010
x+2014/2013+x+2014/2012-x+2014/2011-x+2014/2010=0
(x+2014)(1/2013+1/2012-1/2011-1/2010)=0
x+2014=0
x=-2014
\(\frac{x+1}{2013}+1+\frac{x+2}{2012}+1=\frac{x+3}{2011}+1+\frac{x+4}{2010}+1\)
\(\Rightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}=\frac{x+2014}{2011}+\frac{x+2014}{2010}\)
\(\Rightarrow\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
\(\Rightarrow\left(x+2014\right)=0\)
\(\Rightarrow x=-2014\)
\(\frac{x+2011}{2013}+\frac{x+2012}{2012}=\frac{x+2010}{2014}+\frac{x+2013}{2011}\)
\(\left(\frac{x+2011}{2013}+1\right)+\left(\frac{x+2012}{2012}+1\right)=\left(\frac{x+2010}{2014}+1\right)+\left(\frac{x+2013}{2011}+1\right)\)
\(\frac{x+4024}{2013}+\frac{x+4024}{2012}=\frac{x+4024}{2014}+\frac{x+4024}{2011}\)
\((x+4024)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)=0\)
Vì \(\left(\frac{1}{2013} + \frac{1}{2012} - \frac{1}{2014} - \frac{1}{2011}\right) \neq 0\), nên \(x+4024 = 0\).
\(x = -4024\) [1]
\(\frac{\left(x+2011\right)}{2013}+\frac{\left(x+2012\right)}{2012}=\frac{\left(x+2010\right)}{2014}+\frac{\left(x+2013\right)}{2011}\) ( đoạn này/ là? đúng ko nhỉ)
\(\Rightarrow\frac{\left(x+2011\right)}{2013}+\frac{\left(x+2012\right)}{2012}+2=\frac{\left(x+2010\right)}{2014}+\frac{\left(x+2013\right)}{2011}+2\)
\(\Rightarrow\left(\frac{\left(x+2011\right)}{2013}+1\right)+\left(\frac{\left(x+2012\right)}{2012}+1\right)=\left(\frac{\left(x+2010\right)}{2014}+1\right)+\left(\frac{\left(x+2013\right)}{2011}+1\right)\)
\(\frac{\left(x+4024\right)}{2013}+\frac{\left(x+4024\right)}{2012}=\frac{\left(x+4024\right)}{2014}+\frac{\left(x+4024\right)}{2011}\)
\(\Rightarrow\frac{\left(x+4024\right)}{2013}+\frac{\left(x+4024\right)}{2012}-\frac{\left(x+4024\right)}{2014}-\frac{\left(x+4024\right)}{2011}=0\)
\(\Rightarrow\left(x+4024\right)\left\lbrack\left(\frac{1}{2023}+\frac{1}{2012}\right)-\left(\frac{1}{2014}+\frac{1}{2011}\right)\right\rbrack=0\)
vì \(\left\lbrack\left(\frac{1}{2023}+\frac{1}{2012}\right)-\left(\frac{1}{2014}+\frac{1}{2011}\right)\right\rbrack\) ≠0
=>x+4024=0
=>x=-4024
\(\frac{x+2011}{2013}+\frac{x+2012}{2012}=\frac{x+2010}{2014}+\frac{x+2013}{2011}\)
\(\Rightarrow\left(\frac{x+2011}{2013}+1\right)+\left(\frac{x+2012}{2012}+1\right)=\left(\frac{x+2010}{2014}+1\right)+\left(\frac{x+2013}{2011}+1\right)\)
\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}-\frac{x+4024}{2014}-\frac{x+4024}{2011}=0\)
\(\Rightarrow\left(x+4024\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)=0\)
\(\Rightarrow x+4024=0\)
\(\Rightarrow x=-4024\)