1. TÌM X
\(\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=\frac{-13}{12}\)
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a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne=\)
Nên x + 1 = 0 => x = -1
b) \(\frac{x+1}{14}+\frac{x+2}{13}=\frac{x+3}{12}+\frac{x+4}{11}\)
\(\Leftrightarrow\frac{x+1}{14}+1+\frac{x+2}{13}+1=\frac{x+3}{12}+1+\frac{x+4}{11}+1\)
\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}=\frac{x+15}{12}+\frac{x+15}{11}\)
\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}-\frac{x+15}{12}-\frac{x+15}{11}=0\)
\(\Leftrightarrow\left(x+15\right)\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)=0\)
Vì \(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\ne0\)
Nên x +15 = 0 => x = -15
a,\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)-\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)=0\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\frac{1}{10}>\frac{1}{13};\frac{1}{11}>\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}>\frac{1}{13}+\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}>\frac{1}{13}+\frac{1}{14}\)
\(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
b, Bạn cộng thêm 1 vào \(\frac{x+1}{14};\frac{x+1}{13};\frac{x+1}{12};\frac{x+1}{11}\)Mội bên phân số 1 đơn vị rồi áp dụng như bài 1
\(\frac{x+1}{12}+\frac{x+2}{13}=\frac{x+3}{14}+\frac{x+4}{15}\) .Trừ 1 ở mỗi hạng tử của 2 vế ,ta có :
\(\frac{x-11}{12}+\frac{x-11}{13}=\frac{x-11}{14}+\frac{x-11}{15}\Rightarrow\left(\frac{1}{12}+\frac{1}{13}\right)\left(x-11\right)=\left(\frac{1}{14}+\frac{1}{15}\right)\left(x-11\right)\)
\(\Rightarrow\left[\left(\frac{1}{12}+\frac{1}{13}\right)-\left(\frac{1}{14}+\frac{1}{15}\right)\right]\left(x-11\right)=0\)
\(\frac{1}{12}>\frac{1}{14};\frac{1}{13}>\frac{1}{15}\Rightarrow\frac{1}{12}+\frac{1}{13}>\frac{1}{14}+\frac{1}{15}\Rightarrow\left(\frac{1}{12}+\frac{1}{13}\right)-\left(\frac{1}{14}+\frac{1}{15}\right)\ne0\)
\(\Rightarrow x-11=0\Rightarrow x=11\)
\(\frac{x+1}{12}+\frac{x+2}{13}=\frac{x+3}{14}+\frac{x+4}{15}\)
\(\Leftrightarrow\frac{x+1}{12}-1+\frac{x+2}{13}-1=\frac{x+3}{14}-1+\frac{x+4}{15}-1\)
\(\Leftrightarrow\frac{x-11}{12}+\frac{x-11}{13}=\frac{x-11}{14}+\frac{x-11}{15}\)
\(\Leftrightarrow\frac{x-11}{12}+\frac{x-11}{13}-\frac{x-11}{14}-\frac{x-11}{15}=0\)
\(\Leftrightarrow\left(x-11\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
Mà: \(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\ne0\)
\(\Rightarrow x-11=0\Rightarrow x=11\)
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
a) \(\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)
\(\Leftrightarrow\frac{x+2}{12}+\frac{x+2}{13}-\frac{x+2}{14}-\frac{x+2}{15}=0\)
\(\Leftrightarrow\left(x+2\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
Vì \(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}>0\)
\(\Rightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
b) \(\frac{x+4}{2016}+\frac{x+3}{2017}=\frac{x+2}{2018}+\frac{x+1}{2019}\)
\(\Leftrightarrow\frac{x+4}{2016}+1+\frac{x+3}{2017}+1=\frac{x+2}{2018}+1+\frac{x+1}{2019}+1\)
\(\Leftrightarrow\frac{x+2020}{2016}+\frac{x+2020}{2017}-\frac{x+2020}{2018}-\frac{x+2020}{2019}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\)
Vì \(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}>0\)
\(\Rightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
a) \(\left(x+2\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
=>\(x+2=0\)
=>\(x=-2\)
nếu có sai thì mong bn thông cảm nha
\(\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=-\frac{13}{12}\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)=\frac{-13}{12}\)
\(\Rightarrow x.\frac{13}{12}=\frac{-13}{12}\)
\(\Rightarrow x=-1\)
\(\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=-\frac{13}{12}\)
\(\Rightarrow\frac{6x}{12}+\frac{4x}{12}+\frac{3x}{12}=-\frac{13}{12}\)
\(\Leftrightarrow6x+4x+3x=-13\)
\(\Leftrightarrow13x=-13\)
\(\Leftrightarrow x=-1\)