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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!
Bài này dễ mà!
\(P\left(2019\right)=2019^3\left(a+2016\right)+2019\left(b+2017\right)+2018=2020\)
\(\Rightarrow2019^3\left(a+2016\right)+2019\left(b+2017\right)=2\)
Có: \(P\left(-2019\right)=-2019^3\left(a+2016\right)-2019\left(b+2017\right)+2018\)
\(=-\left[2019^3\left(a+2016\right)+2019\left(b+2017\right)\right]+2018\)
\(=-2+2018=2016\)
b) để \(\left(x-7\right)^{x+2015}-\left(x-7\right)^{x+2016}=0\)
thì \(\left(x-7\right)^{x+2015}=\left(x-7\right)^{x+2016}\)
mà \(x+2015
nên \(x-7=x-7\Rightarrow x=7\)
bài a)
|2x+3|=x+2
2x+3=x+2 hoặc -(2x+3)=x+2
2x-x=2-3 -2x-3=x+2
1x=-1 -2x-x=2+3
x=-1 -3x =5
x=\(\frac{-5}{3}\)
Ta có : \(\frac{x-1}{2017}+\frac{x-2}{2018}-\frac{x-3}{2019}=\frac{x-4}{2020}\)
\(\Rightarrow\frac{x-1}{2017}+\frac{x-2}{2018}=\frac{x-4}{2020}+\frac{x-3}{2019}\)
\(\Rightarrow1+\frac{x-1}{2017}+1+\frac{x-2}{2018}=1+\frac{x-4}{2020}+1+\frac{x-3}{2019}\)
\(\Rightarrow\frac{2016+x}{2017}+\frac{2016+x}{2018}=\frac{2016+x}{2020}+\frac{2016+x}{2019}\)
\(\Rightarrow\frac{2016+x}{2017}+\frac{2016+x}{2018}-\frac{2016+x}{2019}-\frac{2016+x}{2020}=0\)
\(\Rightarrow\left(2016+x\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
\(\text{Mà :
}\)\(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\)
\(\text{Nên : }\) \(2016+x=0\)
\(\Rightarrow x=-2016\)
\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}+\frac{x+4}{2017}=-4\)
=> \(\left[\frac{x+1}{2020}+1\right]+\left[\frac{x+2}{2019}+1\right]+\left[\frac{x+3}{2018}+1\right]+\left[\frac{x+4}{2017}+1\right]=-4\)
=> \(\left[\frac{x+1}{2020}+\frac{2020}{2020}\right]+\left[\frac{x+2}{2019}+\frac{2019}{2019}\right]+\left[\frac{x+3}{2018}+\frac{2018}{2018}\right]+\left[\frac{x+4}{2017}+\frac{2017}{2017}\right]=-4\)
=> \(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}+\frac{x+2021}{2017}=-4\)
=> \(\left[x+2021\right]\left[\frac{1}{2000}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}\right]=-4\)
Do \(\frac{1}{2020}>\frac{1}{2019}>\frac{1}{2018}>\frac{1}{2017}\)nên \(\frac{1}{2000}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}\ne0\)
Do đó : x + 2021 = -4 => x = -4 - 2021 = -2025