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NV
20 tháng 7 2016
a/ \(\frac{5x-4}{3-2x}=\frac{7+4x}{x+2}\) (ĐK: \(x\ne\frac{3}{2};x\ne-2\))
\(\Rightarrow\left(x+2\right)\left(5x-4\right)=\left(7+4x\right)\left(3-2x\right)\)
\(\Rightarrow5x^2-4x+10x-8=21-14x+12x-8x^2\)
\(\Rightarrow13x^2+8x-29=0\)
\(\Rightarrow13\left(x^2+\frac{8}{13}x-\frac{29}{13}\right)=0\)
\(\Rightarrow13\left[x^2+2.\frac{4}{13}.x+\left(\frac{4}{13}\right)^2-\left(\frac{4}{13}\right)^2-\frac{29}{13}\right]=0\)
\(\Rightarrow13\left[\left(x+\frac{4}{13}\right)^2-\frac{393}{169}\right]=0\)
\(\Rightarrow13\left(x+\frac{4}{13}\right)^2-\frac{393}{13}=0\)
\(\Rightarrow\left(x+\frac{4}{13}\right)^2=\frac{393}{169}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{4}{13}=\sqrt{\frac{393}{169}}=\frac{\sqrt{393}}{13}\Rightarrow x=\frac{-4+\sqrt{393}}{13}\\x+\frac{4}{3}=-\sqrt{\frac{393}{169}}=-\frac{\sqrt{393}}{13}\Rightarrow x=\frac{-4-\sqrt{393}}{13}\end{cases}}\)
Vậy biểu thức có 2 nghiệm \(x=\left\{\frac{-4+\sqrt{393}}{13};\frac{-4-\sqrt{393}}{13}\right\}\)
b/ \(\frac{x-1}{99}+\frac{x-2}{98}-\frac{x-3}{97}-\frac{x-4}{96}=0\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1-\left(\frac{x-3}{97}-1\right)-\left(\frac{x-4}{96}-1\right)=0\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}-\frac{x-100}{97}-\frac{x-100}{96}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
=> x - 100 = 0 => x = 100
Vậy x = 100
NM
1
LL
28 tháng 11 2016
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\Leftrightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=0\)
\(\Leftrightarrow\frac{x+10}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)\(\Rightarrow x+100=0\Leftrightarrow x=-100\)
25 tháng 9 2016
\(\frac{x-1}{99}-\frac{x+1}{101}+\frac{x-2}{98}-\frac{x+2}{102}+\frac{x-3}{97}-\frac{x+3}{103}+\frac{x-4}{96}-\frac{x+4}{104}=0\)
\(\Rightarrow\frac{x-1}{99}-1-\frac{x+1}{101}+1+\frac{x-2}{98}-1-\frac{x+2}{102}+1+\frac{x-3}{97}-1-\frac{x+3}{103}+1+\frac{x-4}{96}-1-\frac{x+4}{104}+1=0\)
\(\Rightarrow\frac{x-100}{99}-\frac{x-100}{101}+\frac{x-100}{98}-\frac{x-100}{102}+\frac{x-100}{97}-\frac{x-100}{103}+\frac{x-100}{96}-\frac{x-100}{104}=0\)
\(\Rightarrow\left(x-100\right).\left(\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\right)=0\)
Vì \(\frac{1}{99}>\frac{1}{101};\frac{1}{98}>\frac{1}{102};\frac{1}{97}>\frac{1}{103};\frac{1}{96}>\frac{1}{104}\)
\(\Rightarrow\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy \(x=100\)
TN
3
7 tháng 10 2016
\(\frac{x+1}{98}+\frac{x+2}{97}=\frac{x+3}{96}+\frac{x+4}{95}\)
=> \(\left(\frac{x+1}{98}+1\right)+\left(\frac{x+2}{97}+1\right)=\left(\frac{x+3}{96}+1\right)+\left(\frac{x+4}{95}+1\right)\)
=> \(\frac{x+99}{98}+\frac{x+99}{97}-\frac{x+99}{96}-\frac{x+99}{95}=0\)
=> \(\left(x+99\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
=> \(x+99=0\) (Vì: \(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\) )
=>\(x=-99\)
9 tháng 10 2016
Ta có :
\(\frac{x+1}{98}+\frac{x+2}{97}=\frac{x+3}{96}+\frac{x+4}{95}\)
\(\Rightarrow\) \(\left(\frac{x+1}{98}+1\right)+\left(\frac{x+2}{97}+1\right)=\left(\frac{x+3}{96}+1\right)+\left(\frac{x+4}{95}+1\right)\)
\(\Rightarrow\frac{x+99}{98}+\frac{x+99}{97}=\frac{x+99}{96}+\frac{x+99}{95}\)
\(\Rightarrow\frac{x+99}{98}+\frac{x+99}{97}-\frac{x+99}{96}-\frac{x+99}{95}=0\)
\(\Rightarrow\left(x+99\right).\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Vì \(\frac{1}{96}+\frac{1}{97}< \frac{1}{96}+\frac{1}{95}\)
\(\Rightarrow\) \(\frac{1}{96}+\frac{1}{97}< \frac{1}{96}+\frac{1}{95}\ne0\)
Nên \(x+99=0\)
\(\Rightarrow x=0-99\)
\(\Rightarrow x=-99\)
Vậy : \(x=-99\)
NQ
2
2 tháng 10 2021
a: Ta có: \(\left(\dfrac{3}{2}x-\dfrac{1}{5}\right)^2\cdot\left(x^2+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=\dfrac{1}{5}\)
hay \(x=\dfrac{1}{5}:\dfrac{3}{2}=\dfrac{2}{15}\)
b: Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)
\(\Leftrightarrow x+100=0\)
hay x=-100
24 tháng 6 2018
\(a)\) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=-4+4\)
\(\Leftrightarrow\)\(\frac{x+1+99}{99}+\frac{x+2+98}{98}+\frac{x+3+97}{97}+\frac{x+4+96}{96}=0\)
\(\Leftrightarrow\)\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)
Nên \(x+100=0\)
\(\Rightarrow\)\(x=-100\)
Vậy \(x=-100\)
Chúc bạn học tốt ~
24 tháng 6 2018
\(b)\) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\Leftrightarrow\)\(1-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=1-\frac{2008}{2009}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{2009}\)
\(\Leftrightarrow\)\(x+1=2009\)
\(\Leftrightarrow\)\(x=2009-1\)
\(\Leftrightarrow\)\(x=2008\)
Vậy \(x=2008\)
Chúc bạn học tốt ~
27 tháng 6 2017
Ta có:
\(\frac{x+1}{98}+1+\frac{x+2}{97}+1=\frac{x+3}{96}+1+\frac{x+4}{95}+1\)
\(\frac{x+1}{98}+\frac{98}{98}+\frac{x+2}{97}+\frac{97}{97}=\frac{x+3}{96}+\frac{96}{96}+\frac{x+4}{95}+\frac{95}{95}\)
\(\frac{x+99}{98}+\frac{x+99}{97}=\frac{x+99}{96}+\frac{x+99}{95}\)
\(\frac{x+99}{98}+\frac{x+99}{97}-\frac{x+99}{96}-\frac{x+99}{95}=0\)
\(\left(x+99\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Vì: \(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)nên x+99=0
=> x=-99
26 tháng 8 2018
a) \(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
Vì 1/99 + 1/98 - 1/97 - 1/96 khác 0
=> x + 100 = 0 => x = -100
b) \(\frac{x-3}{47}+\frac{x-2}{48}=\frac{x-1}{49}+1\)
\(\Rightarrow\frac{x-3}{47}-1+\frac{x-2}{48}-1=\frac{x-1}{49}+1-2\)
\(\Rightarrow\frac{x-50}{47}+\frac{x-50}{48}-\frac{x-50}{49}=0\)
\(\Rightarrow\left(x-50\right)\left(\frac{1}{47}+\frac{1}{48}-\frac{1}{49}\right)=0\)
Vì 1/47 + 1/48 - 1/49 khác 0
Nên x -50 = 0 => x = 50
\(E=\frac{1+x^2+x^4+...+x^{98}}{1+x^4+x^8+...+x^{96}}\)
\(=\frac{\left(1+x^4+x^8+...+x^{96}\right)+\left(x^2+x^6+x^{10}...+x^{94}+x^{98}\right)}{1+x^4+x^8+...+x^{96}}\)
\(=\frac{\left(1+x^4+x^8+...+x^{96}\right)+x^2\left(1+x^4+x^8...+x^{96}\right)}{1+x^4+x^8+...+x^{96}}\)
\(=\frac{\left(1+x^4+x^8+...+x^{96}\right)\left(1+x^2\right)}{1+x^4+x^8+...+x^{96}}=1+x^2\)
#alibaba nguyễn
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