Phương trình bậc nhất một ẩn

Linh Ngô

Giải phương trình:

a) \(\dfrac{1+x}{2017}+\dfrac{2+x}{2016}+\dfrac{3+x}{2015}=-3\)

b) \(\dfrac{x-\dfrac{3x-4}{5}}{15}=\dfrac{5x-\dfrac{3-x}{2}}{5}-x+1\)

 Mashiro Shiina
30 tháng 12 2017 lúc 16:41

a)\(\dfrac{1+x}{2017}+\dfrac{2+x}{2016}+\dfrac{3+x}{2015}=-3\)

\(\Rightarrow\left(\dfrac{1+x}{2017}+1\right)+\left(\dfrac{2+x}{2016}+1\right)+\left(\dfrac{3+x}{2015}+1\right)=0\)

\(\Rightarrow\dfrac{2018+x}{2017}+\dfrac{2018+x}{2016}+\dfrac{2018+x}{2015}=0\)

\(\Rightarrow\left(2018+x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\right)=0\)

\(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\ne0\) nên \(2018+x=0\Leftrightarrow x=-2018\)

b) \(\dfrac{x-\dfrac{3x-4}{5}}{15}=\dfrac{5x-\dfrac{3-x}{2}}{5}-x+1\)

\(\Rightarrow\dfrac{x-\dfrac{3x-4}{5}}{15}=\dfrac{15x-\dfrac{9-3x}{2}}{15}-\dfrac{15x+15}{15}\)

\(\Rightarrow\dfrac{x-\dfrac{3x-4}{5}}{15}=\dfrac{15x-\dfrac{9-3x}{2}-15x-15}{15}\)

\(\Rightarrow\dfrac{x-\dfrac{3x-4}{5}}{15}=\dfrac{-\dfrac{9-3x}{2}-15}{15}\)

\(\Rightarrow\dfrac{x-\dfrac{3x-4}{5}+\dfrac{9-3x}{2}+15}{15}=0\)

\(\Rightarrow x-\dfrac{3x-4}{5}+\dfrac{9-3x}{2}+15=0\)

\(\Rightarrow\dfrac{10x}{10}-\dfrac{6x-8}{10}+\dfrac{45-15x}{10}+\dfrac{150}{10}=0\)

\(\Rightarrow\dfrac{10x-6x+8+45-15x+150}{10}=0\)

\(\Rightarrow10x-6x+8+45-15x+150=0\)

\(\Rightarrow-11x+203=0\)

\(\Rightarrow-11x=-203\Leftrightarrow x=\dfrac{203}{11}\)

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lê thị hương giang
30 tháng 12 2017 lúc 16:45

\(a,\dfrac{1+x}{2017}+\dfrac{2+x}{2016}+\dfrac{3+x}{2015}=-3\)

\(\Leftrightarrow\dfrac{1+x}{2017}+1+\dfrac{2+x}{2016}+1+\dfrac{3+x}{2015}+1=-3+3\)

\(\Leftrightarrow\dfrac{1+x+2017}{2017}+\dfrac{2+x+2016}{2016}+\dfrac{3+x+2015}{2015}=0\)

\(\Leftrightarrow\dfrac{x+2018}{2017}+\dfrac{x+2018}{2016}+\dfrac{x+2018}{2015}=0\)

\(\Leftrightarrow\left(x+2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\right)=0\)

\(\Leftrightarrow x+2018=0\)

\(\Leftrightarrow x=-2018\)

b,\(\dfrac{x-\dfrac{3x-4}{5}}{15}=\dfrac{5x-\dfrac{3-x}{2}}{5}-x+1\)

\(\Leftrightarrow\dfrac{\dfrac{5x-3x+4}{5}}{15}=\dfrac{\dfrac{10x-3+x}{2}}{5}-x+1\)

\(\Leftrightarrow\dfrac{\dfrac{2x+4}{5}}{15}=\dfrac{\dfrac{11x-3}{2}}{5}-\dfrac{5x-5}{5}\)

\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{11x-3}{10}-\dfrac{10x-10}{10}\)

\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{11x-3-10x+10}{10}\)

\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{x+7}{10}\)

\(\Leftrightarrow10\left(2x+4\right)=75\left(x+7\right)\)

\(\Leftrightarrow20x+40=75x+525\)

\(\Leftrightarrow20x-75x=525-40\)

\(\Leftrightarrow-55x=485\)

\(\Leftrightarrow x=-\dfrac{97}{11}\)

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Nguyễn Anh Kim Hân
30 tháng 12 2017 lúc 17:01

a) \(\dfrac{1+x}{2017}+\dfrac{2+x}{2016}+\dfrac{3+x}{2015}=-3\)

\(\Leftrightarrow\dfrac{1+x}{2017}+1+\dfrac{2+x}{2016}+1+\dfrac{3+x}{2015}+1=0\)

\(\Leftrightarrow\dfrac{x+2018}{2017}+\dfrac{x+2018}{2016}+\dfrac{x+2018}{2015}=0\)

\(\Leftrightarrow\left(x+2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\right)=0\)

\(\Rightarrow x+2018=0\)

\(\Leftrightarrow x=-2018\)

b) \(\dfrac{x-\dfrac{3x-4}{5}}{15}=\dfrac{5x-\dfrac{3-x}{2}}{5}-x+1\)

\(\Leftrightarrow\dfrac{\dfrac{5x-3x+4}{5}}{15}=\dfrac{\dfrac{10x-3+x}{2}}{5}-x+1\)

\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{11x-3}{10}-x+1\)

\(\Leftrightarrow\dfrac{4x+8}{150}=\dfrac{165x-45}{150}-\dfrac{150x-150}{150}\)

\(\Leftrightarrow4x+8=165x-45-150x+150\)

\(\Leftrightarrow4x-165x+150x=-45+150-8\)

\(\Leftrightarrow-11x=97\)

\(\Leftrightarrow x=-\dfrac{97}{11}\)

\(S=\left\{-\dfrac{97}{11}\right\}\)

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