tim x
\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
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\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1+0.\)
\(=\left(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x\right)+1\)
\(=x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1\)
\(=x\left(\frac{1\cdot5}{30}+\frac{1\cdot3}{30}-\frac{4\cdot2}{30}\right)+1\)
\(=x\left(\frac{5}{30}+\frac{3}{30}-\frac{8}{30}\right)+1\)
\(=x\left(\frac{5+3-8}{30}\right)+1\)
\(=x\cdot0+1=1\)
\(\Rightarrow\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1+0=1\)
trả lời:
\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1+0.61x+101x−154x+1+0.
=\left(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x\right)+1=(61x+101x−154x)+1
=x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=x(61+101−154)+1
=x\left(\frac{1\cdot5}{30}+\frac{1\cdot3}{30}-\frac{4\cdot2}{30}\right)+1=x(301⋅5+301⋅3−304⋅2)+1
=x\left(\frac{5}{30}+\frac{3}{30}-\frac{8}{30}\right)+1=x(305+303−308)+1
=x\left(\frac{5+3-8}{30}\right)+1=x(305+3−8)+1
=x\cdot0+1=1=x⋅0+1=1
\Rightarrow\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1+0=1⇒61x+101x−154x+1+0=1
ko chắc chúc bạn học tốt.
\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
\(\frac{5}{30}x+\frac{3}{30}x-\frac{8}{30}x+1=0\)
\(\frac{8}{30}x-\frac{8}{30}x+1=0\)
\(0+1=0\)
\(\text{Mình thấy bài này có gì sai sai sai thì phải bạn à !}\)
Bài làm
\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
\(x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)
\(x\left(\frac{5}{30}+\frac{3}{30}-\frac{8}{30}\right)+1=40\)
\(x.\frac{1}{30}+1=0\)
\(x\frac{1}{3}=-1\)
\(x=-1:\frac{1}{3}\)
\(x=-1.3\)
\(x=-3\)
# Học tốt #
\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x-1\right)}=\frac{2007}{2009}\)
\(2\cdot\left(\frac{1}{6}+\frac{1}{12}+..+\frac{1}{x\left(x-1\right)}\right)=\frac{2007}{2009}\)
\(\frac{1}{6}+\frac{1}{12}+..+\frac{1}{x\left(x-1\right)}=\frac{2007}{2009}:2\)
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(x-1\right)x}=\frac{2007}{4018}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{x-1}-\frac{1}{x}=\frac{2007}{4018}\)
\(\frac{1}{2}-\frac{1}{x}=\frac{2007}{4018}\)
\(\frac{1}{x}=\frac{1}{2}-\frac{2007}{4018}\)
\(\frac{1}{x}=\frac{1}{2009}\)
=> x = 2009
\(a,\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
TH1 : \(\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{x-2}{7}=0\Rightarrow x-2=0\Leftrightarrow x=2\)
TH2 : \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow\frac{-x+3}{5}=0\Rightarrow-x+3=0\Leftrightarrow x=3\)
TH3 : \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{x+4}{3}=0\Rightarrow x+4=0\Leftrightarrow x=-4\)
\(\Rightarrow x\in\left\{2;3;-4\right\}\)
\(b,\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
\(\Rightarrow\frac{5}{30}x+\frac{3}{30}x-\frac{8}{30}x+1=0\)
\(\Rightarrow\frac{5x+3x-8x}{30}+1=0\)
\(\Rightarrow1=0\)( vô lý )\(\Rightarrow x\in\varnothing\)
a, \(\frac{1}{6}x+\frac{1}{10}-\frac{4}{15}x+1=0\)
\(\Leftrightarrow-\frac{1}{10}x=-\frac{11}{10}\)
\(\Leftrightarrow x=11\)
b,\(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Leftrightarrow\frac{1}{7}x-\frac{2}{7}=0\)hoặc \(-\frac{1}{5}x+\frac{3}{5}=0\)hoặc \(\frac{1}{3}x+\frac{4}{3}=0\)
+) \(\frac{1}{7}x-\frac{2}{7}=0\Leftrightarrow\frac{1}{7}x=\frac{2}{7}\Leftrightarrow x=2\)
+)\(-\frac{1}{5}x+\frac{3}{5}=0\Leftrightarrow-\frac{1}{5}x=-\frac{3}{5}\Leftrightarrow x=3\)
+)\(\frac{1}{3}x+\frac{4}{3}=0\Leftrightarrow\frac{1}{3}x=-\frac{4}{3}\Leftrightarrow x=-4\)
c, \(\frac{1}{2}x-\frac{11}{15}:\frac{33}{35}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{9}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}x=\frac{4}{9}\)
\(\Leftrightarrow x=\frac{8}{9}\)
a/ \(\frac{1}{6}x+\frac{1}{10}-\frac{4}{15}x+1=0\)
\(\Rightarrow-\frac{1}{10}x=-\frac{11}{10}\)
\(\Rightarrow x=11\)
b/ \(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Rightarrow\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{1}{7}x=\frac{2}{7}\Rightarrow x=2\)
hoặc \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow-\frac{1}{5}x=-\frac{3}{5}\Rightarrow x=3\)
hoặc \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{1}{3}x=-\frac{4}{3}\Rightarrow x=-4\)
Vậy x = 2, x = 3, x = -4
c/ \(\frac{1}{2}x-\frac{11}{15}:\frac{33}{35}=-\frac{1}{3}\)
\(\Rightarrow\frac{1}{2}x-\frac{7}{9}=-\frac{1}{3}\)
\(\Rightarrow\frac{1}{2}x=\frac{4}{9}\Rightarrow x=\frac{8}{9}\)
Vậy x = 8/9
Ta có : \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
<=> \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x=-1\)
<=> \(\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)x=-1\)
<=> \(0.x=-1\)
=> x thuộc rỗng
\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
\(\Leftrightarrow\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)x+1=0\)
\(\Leftrightarrow\left(\frac{5}{30}+\frac{3}{30}-\frac{8}{30}\right)x+1=0\)
\(\Leftrightarrow\frac{5+3-8}{30}x+1=0\)
\(\Leftrightarrow0x=-1\)(vô lí)
vậy không có giá trị nào của x thỏa mãn.
b, \(x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)
\(0+1=0\)
=> x thuoc rong
\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
\(\Rightarrow x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)
\(\Rightarrow x.0+1=0\)
=> 1=0 ( Vô lý )
Vậy \(x\in\varnothing\)
\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
\(\Rightarrow x.\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)=1\)
\(\Rightarrow x.0=1\Rightarrow x=0\)
Vậy x=0