Giải phương trình sau
x+1/2018 + x+2/2019= x-2014/3 + x-2013/4
Giúp mình nha mình cảm ơn
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Gợi ý :
Bài 1 : Cộng thêm 1 vào 3 phân thức đầu, trừ cho 3 ở phân thức thứ 4, có nhân tử chung là (x+2020)
Bài 2 : Trừ mỗi phân thức cho 1, chuyển vế và có nhân tử chung là (x-2021)
Bài 3 : Phân thức thứ nhất trừ đi 1, phân thức hai trù đi 2, phân thức ba trừ đi 3, phân thức bốn trừ cho 4, phân thức 5 trừ cho 5. Có nhân tử chung là (x-100)
bài 3
\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15.\)
=>\(\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=0\)
=>\(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
=>\(\left(x-100\right).\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
=>(x-100)=0 do \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
=> x=100
`@` `\text {Ans}`
`\downarrow`
`a)`
`13/50 + 9% + 41/100 + 0,24`
`= 0,26 + 0,09 + 0,41 + 0,24`
`= (0,26 + 0,24) + (0,09 + 0,41)`
`= 0,5 + 0,5`
`= 1`
`b)`
`2018 \times 2020 - 1/2017 + 2018 \times 2019`
`= 2018 \times (2020 + 2019) - 1/2017`
`= 2018 \times 4039 - 1/2017`
`= 8150702`
`c)`
`1/2 + 1/6 + 1/12 + 1/20 +1/30 +1/42`
`=`\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}\)
`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{6}-\dfrac{1}{7}\)
`=`\(1-\dfrac{1}{7}\)
`= 6/7`
\(a,\dfrac{13}{50}+9\%+\dfrac{41}{100}+0,24\\ 0,26+0,09+0,41+0,24\\ =\left(0,26+0,24\right)+\left(0,09+0,41\right)\\ =0,5+0,5\\ =1\\ b,2018\times2020-\dfrac{1}{2017}+2018\times2019\\ =2018\times\left(2020+2019\right)-\dfrac{1}{2017}\\ =2018\times4039-\dfrac{1}{2017}\\ =3150702-\dfrac{1}{2017}\\ c,\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\\ =1-\dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}.........+\dfrac{1}{6}-\dfrac{1}{7}\\ =1-\dfrac{1}{7}\\ =\dfrac{6}{7}\)
1)
\(\dfrac{x-1}{2014}+\dfrac{x-2}{2013}+\dfrac{x-3}{2012}+...+\dfrac{x-2014}{1}=2014\)
\(\Leftrightarrow\left(\dfrac{x-1}{2014}-1\right)+\left(\dfrac{x-2}{2013}-1\right)+...+\left(\dfrac{x-2014}{1}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-2015}{2014}+\dfrac{x-2015}{2013}+...+\dfrac{x-2015}{1}=0\)
\(\Leftrightarrow\left(x-2025\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1}\right)=0\)
\(\Leftrightarrow x=2015\)
Vậy \(S=\left\{2015\right\}\)
\(\frac{x+1}{2018}+\frac{x+2}{2019}=\frac{x-2014}{3}+\frac{x-2013}{4}\)
\(\Leftrightarrow\frac{x+1}{2018}+\frac{x+2}{2019}-\frac{x-2014}{3}-\frac{x-2013}{4}=0\)
\(\Leftrightarrow\frac{x+1}{2018}-1+\frac{x+2}{2019}-1-\frac{x-2014}{3}+1-\frac{x-2013}{4}+1=0\)
\(\Leftrightarrow\frac{x-2017}{2018}+\frac{x-2017}{2019}-\frac{x-2017}{2013}-\frac{x-2017}{2014}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
có : \(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2014}-\frac{1}{2013}\ne0\)
\(\Leftrightarrow x-2017=0\)
=> x = 2017
\(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}+\frac{x+4}{2011}=0\)
\(\Leftrightarrow\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+2}{2013}+1\right)+\left(\frac{x+3}{2012}+1\right)+\left(\frac{x+4}{2011}+1\right)=0+1+1+1+1\)
\(\Leftrightarrow\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}+\frac{x+2015}{2011}=4\)
\(\Leftrightarrow\left(x+2015\right).\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=4\)
\(\Leftrightarrow x+2015=4:\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)\)
\(\Leftrightarrow x+2015=2012,499379\)
\(\Leftrightarrow x=2012,499379-2015\)
\(\Leftrightarrow x=-2,5006118\)
P/s : Số xấu quá !