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\(\frac{x+1}{2019}+\frac{x+2}{2018}=\frac{x+3}{2017}+\frac{x+4}{2016}\)
\(\Leftrightarrow\left(\frac{x+1}{2019}-1\right)+\left(\frac{x+2}{2018}-1\right)=\left(\frac{x+3}{2017}-1\right)+\left(\frac{x+4}{2016}-1\right)\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}=\frac{x+2020}{2017}+\frac{x+2020}{2016}\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)
\(\Leftrightarrow x+2020=0:\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)\)
\(\Leftrightarrow x+2020=0\)
Còn lại tự làm :V
Lộn chỗ này , thay chút nha !
\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)=\left(\frac{x+3}{2017}+1\right)+\left(\frac{x+4}{2016}+1\right)\)
Sorry =))
\(\frac{x-1}{2019}+\frac{x-2}{2018}-\frac{x-3}{2017}=\frac{x-4}{2016}\)
\(\Leftrightarrow\frac{x-1}{2019}+\frac{x-2}{2018}-\frac{x-3}{2017}-\frac{x-4}{2016}=0\)
\(\Leftrightarrow\frac{x-1}{2019}-1+\frac{x-2}{2018}-1-\frac{x-3}{2017}+1-\frac{x-4}{2016}+1=0\)
\(\Leftrightarrow\frac{x-2020}{2019}+\frac{x-2020}{2018}-\frac{x-2020}{2017}-\frac{x-2020}{2016}=0\)
\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)
\(\Leftrightarrow x-2020=0\Leftrightarrow x=2020\)
\(\frac{x-1}{2019}+\frac{x-2}{2018}-\frac{x-3}{2017}=\frac{x-4}{2016}\)
\(\frac{x-1}{2019}+\frac{x-2}{2018}=\frac{x-3}{2017}+\frac{x-4}{2016}\)
\(\frac{x-1}{2019}+\frac{x-2}{2018}-2=\frac{x-3}{2017}+\frac{x-4}{2016}-2\)
\(\left(\frac{x-1}{2019}-1\right)+\left(\frac{x-2}{2018}-1\right)=\left(\frac{x-3}{2017}-1\right)+\left(\frac{x-4}{2016}-1\right)\)
\(\frac{x-1-2019}{2019}+\frac{x-2-2018}{2018}=\frac{x-3-2017}{2017}+\frac{x-4-2016}{2016}\)
\(\frac{x-2020}{2019}+\frac{x-2020}{2018}=\frac{x-2020}{2017}+\frac{x-2020}{2016}\)
\(\frac{x-2020}{2019}+\frac{x-2020}{2018}-\frac{x-2020}{2017}-\frac{x-2020}{2016}=0\)
\(\left(x-2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)
\(\Rightarrow x-2020=0\)
Vậy \(x=2020\)
a,\(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\) (1)
<=> \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)
<=> \(\left(x+1\right)\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)
=> x+1=0 (vì \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\ne0\))
<=> x=-1
Vậy pt (1) có tập nghiệm S\(=\left\{-1\right\}\)
b, \(\frac{x+6}{2015}+\frac{x+5}{2016}+\frac{x+4}{2017}=\frac{x+3}{2018}+\frac{x+2}{2019}+\frac{x+1}{2010}\)(2)
<=> \(\frac{x+6}{2015}+1+\frac{x+5}{2016}+1+\frac{x+4}{2017}+1=\frac{x+3}{2018}+1+\frac{x+2}{2019}+1+\frac{x+1}{2020}+1\)
<=> \(\frac{x+2021}{2015}+\frac{x+2021}{2016}+\frac{x+2021}{2017}-\frac{x+2021}{2018}-\frac{x+2021}{2019}-\frac{x+2021}{2020}=0\)
<=> \(\left(x+2021\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
=> x+2021=0(vì \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))
<=> x=-2021
Vậy pt (2) có tập nghiệm S=\(\left\{-2021\right\}\)
c,\(\frac{x+6}{2016}+\frac{x+7}{2017}+\frac{x+8}{2018}=\frac{x+9}{2019}+\frac{x+10}{2020}+1\) (3)
<=> \(\frac{x+6}{2016}-1+\frac{x+7}{2017}-1+\frac{x+8}{2018}-1=\frac{x+9}{2019}-1+\frac{x+10}{2020}-1+1-1\)
<=> \(\frac{x-2010}{2016}+\frac{x-2010}{2017}+\frac{x-2010}{2018}=\frac{x-2010}{2019}+\frac{x-2010}{2020}\)
<=> \(\frac{x-2010}{2016}+\frac{x-2010}{2017}+\frac{x-2010}{2018}-\frac{x-2010}{2019}-\frac{x-2010}{2020}=0\)
<=> \(\left(x-2010\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
=> x-2010=0 (vì \(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))
<=> x=2010
Vậy pt (3) có tập nghiệm S=\(\left\{2010\right\}\)
d, \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\) (4)
<=>\(\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=15-1-2-3-4-5\)
<=> \(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
<=> (x-100)(\(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\))=0
=> x -100=0(vì \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\))
<=> x=100
Vậy pt (4) có tập nghiệm S=\(\left\{100\right\}\)
a) \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)
\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)
\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=0-1\)
\(\Rightarrow x=-1\)
Vậy \(x=-1.\)
Mình chỉ làm câu a) thôi nhé.
Chúc bạn học tốt!
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)
\(\frac{x+3}{2016}+\frac{x+2}{2017}=\frac{x+1}{2018}+\frac{x}{2019}\\ \Leftrightarrow\frac{x+3}{2016}+1+\frac{x+2}{2017}+1=\frac{x+1}{2018}+1+\frac{x}{2019}+1\\ \Leftrightarrow\frac{x+2019}{2016}+\frac{x+2019}{2017}-\frac{x+2019}{2018}-\frac{x+2019}{2019}=0\\ \Leftrightarrow\left(x+2019\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\\ \Rightarrow x-2019=0\\ \Rightarrow x=2019\)
x+4/2015 + x+3/2016 = x+2/2017 + x+1/2018
=> 1 + x+4/2015 + 1 + x+3/2016 = 1 + x+2/2017 + 1 + x+1/2018
=> x+2019/2015 + x+2019/2016 = x+2019/2017 + x+2019/2018
=> x+2019/2015 + x+2019/2016 - x+2019/2017 - x+2019/2018 = 0
=> (x + 2019).(1/2015 + 1/2016 - 1/2017 - 1/2018) = 0
Vì 1/2015 > 1/2017; 1/2016 > 1/2018
=> 1/2015 + 1/2016 - 1/2017 - 1/2018 khác 0
=> x + 2019 = 0
=> x = -2019
Bài giải
a, \(\frac{x+5}{2017}-\frac{x+5}{2018}+\frac{x+5}{2019}-\frac{x+5}{2020}=0\)
\(\left(x+5\right)\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}\right)=0\)
Do \(\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}\right)\ne0\)
\(\Rightarrow\text{ }x+5=0\)
\(x=0-5\)
\(=-5\)
a)\(\frac{1}{4}-\frac{1}{3}x=\frac{2}{5}-\frac{3}{2}x\)
\(\Leftrightarrow\)\(\frac{15-20x}{60}=\frac{24-90x}{60}\)
\(\Leftrightarrow15-20x=24-90x\)
\(\Leftrightarrow-20x+90x=24-15\)
\(\Leftrightarrow70x=9\)
\(\Leftrightarrow x=\frac{9}{70}\)
c) (1/2-1/6)*3^x+4-4*3^x=3^16-4*3^13
=1/3*3^x*3^4-4*3^x=3^13*3^3-4*3^13
=27*3^x-4*3^x=3^13*(27-4)
=3^x*(27-4)=3^13*(27-4)
=>x=13
Câu a đề thiếu vế phải rồi bạn
b: \(\Leftrightarrow x\cdot0+1=0\)
=>0x+1=0(vô lý)
đề thiếu hả bn
Hình như bạn ghi thiếu đề rồi. Để tìm đc x trong đẳng thức này thì ta phải có kết quả của biểu thức trên chứ đề cộc lốc thế này ko giải đc đâu