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\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2017}\)
\(\Rightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{x}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2017}\)
\(\Rightarrow-\frac{1}{x+3}=\frac{1}{2017}\)
\(\Rightarrow x+3=-2017\)
\(\Rightarrow x=-2020\)
\(\Leftrightarrow3\left(x^2-4x+4\right)-\dfrac{5}{4}\left(9x^2+6x+1\right)=\dfrac{4}{3}\left(-x^2+4x-3\right)-\dfrac{7}{6}x\left(x-3\right)\)
\(\Leftrightarrow3x^2-12x+12-\dfrac{45}{4}x^2-\dfrac{15}{2}x-\dfrac{5}{4}=-\dfrac{4}{3}x^2+\dfrac{16}{3}x-4-\dfrac{7}{6}x^2+\dfrac{7}{2}x\)
\(\Leftrightarrow x^2\cdot\dfrac{-33}{4}-\dfrac{39}{2}x+\dfrac{43}{4}+\dfrac{5}{2}x^2-\dfrac{53}{6}x+4=0\)
\(\Leftrightarrow x^2\cdot\dfrac{-23}{4}-\dfrac{85}{3}x+\dfrac{59}{4}=0\)
\(\Leftrightarrow12\left(\dfrac{-23}{4}x^2-\dfrac{85}{3}x+\dfrac{59}{4}\right)=0\)
\(\Leftrightarrow-69x^2-340x+177=0\)
\(\Leftrightarrow69x^2+340x-177=0\)
\(\text{Δ}=340^2-4\cdot69\cdot\left(-177\right)=164452\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-170-\sqrt{41113}}{69}\\x_2=\dfrac{-170+\sqrt{41113}}{69}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{1}{2}\left(x^2-4x+4\right)-\dfrac{13}{3}\left(x^2+6x+9\right)=\dfrac{1}{4}\left(x^2-3x+2\right)-2\left(9x^2+3x-2\right)\)
\(\Leftrightarrow x^2\cdot\dfrac{1}{2}-2x+2-\dfrac{13}{3}x^2-26x-39=\dfrac{1}{4}x^2-\dfrac{3}{4}x+\dfrac{1}{2}-18x^2-6x+4\)
\(\Leftrightarrow x^2\cdot\dfrac{167}{12}-\dfrac{85}{4}x-\dfrac{83}{2}=0\)
\(\Leftrightarrow167x^2-255x-498=0\)
\(\text{Δ}=\left(-255\right)^2-4\cdot167\cdot\left(-498\right)=397689\)
Vì Δ>0 nên phương trình có 2 nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{255-\sqrt{397689}}{334}\\x_2=\dfrac{255+\sqrt{397689}}{334}\end{matrix}\right.\)
\(\frac{\left(2-3x\right)^2}{3}-\frac{\left(1+2x\right)^2}{2}=\frac{3}{4}-2\left(x-1\right)\left(x+2\right)+x\left(1+x\right)\)
\(\frac{2^2-12x-3x^2}{3}-\frac{1^2+4x+2x^2}{2}=\frac{3}{4}-\left(x^2+x-2\right)+3x\)
\(\frac{2.\left(4-12x-3x^2\right)}{6}-\frac{3.\left(1+4x+2x^2\right)}{6}=\frac{11}{4}-x^2+2x\)
\(\frac{8-24x-6x^2}{6}-\frac{3+12x+2x^2}{6}=\frac{11}{4}-x^2+2x\)
\(\frac{8-24x-6x^2-3-12x-2x^2}{6}=\frac{11}{4}-x^2+2x\)
\(\frac{5-36x-8x^2}{6}=\frac{11}{4}-x^2+2x\)
Chỗ đây thì mk chịu
=> 1/x - 1/x+1 + 1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 - 1/x = 1/2010
=> -1/x+3 = 1/2010
=> 1/x+3 = 1/-2010
=> x+3 = -2010
=> x = -2010-3 = -2013
k mk nha
1/x - 1/x+1 + 1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 - 1/x = 1/2010
=> -1/x+3 = 1/2010
=> 1/x+3 = 1/-2010
=> x+3 = -2010
=> x = -2010-3 = -2013
Phần a vs phần b tính toán thông thường thôi mà bạn, vs 1 h/s lớp 7 thì ít nhất phải làm được chứ?? :((
a) \(x-\frac{4}{5}=\frac{7}{10}-\frac{3}{4}\)
\(\Leftrightarrow x-\frac{4}{5}=\frac{-1}{20}\)
\(\Leftrightarrow x=\frac{-1}{20}+\frac{4}{5}=\frac{15}{20}=\frac{3}{4}\)
b) \(2\frac{1}{3}-x=\frac{-5}{9}+2x\)
\(\Leftrightarrow2\frac{1}{3}-\frac{-5}{9}=2x+x\)
\(\Leftrightarrow3x=\frac{7}{3}+\frac{5}{9}\)
\(\Leftrightarrow3x=\frac{26}{9}\)
\(\Leftrightarrow x=\frac{26}{9}:3=\frac{26}{27}\)
d) .............................. ( Đề bài)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}\)\(-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(\Leftrightarrow-\frac{1}{x+3}=\frac{1}{2010}\)
\(\Leftrightarrow\frac{1}{-\left(x+3\right)}=\frac{1}{2010}\)\(\Leftrightarrow-\left(x+3\right)=2010\)
\(\Leftrightarrow-x-3=2010\) \(\Leftrightarrow-x=2010+3=2013\)
\(\Leftrightarrow x=-2013\)
Bạn tự kết luận nha!
c)
\(\frac{x+3}{2016}+\frac{x+2}{2017}=\frac{x+1}{2018}+\frac{x}{2019}\\ \Leftrightarrow\frac{x+3}{2016}+1+\frac{x+2}{2017}+1=\frac{x+1}{2018}+1+\frac{x}{2019}+1\\ \Leftrightarrow\frac{x+2019}{2016}+\frac{x+2019}{2017}-\frac{x+2019}{2018}-\frac{x+2019}{2019}=0\\ \Leftrightarrow\left(x+2019\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\\ \Rightarrow x-2019=0\\ \Rightarrow x=2019\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2017}\)
<=> \(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2017}\)
<=> \(\frac{-1}{x+3}=\frac{1}{2017}\)
=> \(x+3=-2017\)
<=> \(x=-2020\)
Vậy...