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1a)1+2+3+...+x=0
\(\frac{x.\left(x+1\right)}{2}\)=0
x.(x+1) =0:2
x.(x+1) =0
x.(x+1) =0.1
vây x=0
tich dung cho minh nha
a/ \(2x+\frac{1}{7}=\frac{1}{3}\)
=> \(2x=\frac{1}{3}-\frac{1}{7}=\frac{7}{21}-\frac{3}{21}\)
=> \(2x=\frac{4}{21}\)
=> \(x=\frac{4}{21}:2=\frac{4}{21}.\frac{1}{2}=\frac{2}{21}\)
b/ \(3\left(x-\frac{1}{2}\right)=\frac{4}{9}\)
=> \(x-\frac{1}{2}=\frac{4}{9}:3=\frac{4}{9}.\frac{1}{3}\)
=> \(x-\frac{1}{2}=\frac{4}{27}\)
=> \(x=\frac{4}{27}+\frac{1}{2}=\frac{8}{54}+\frac{27}{54}=\frac{35}{54}\)
c/ \(\left(x-5\right)^2+4=68\)
=> \(\left(x-5\right)^2=68-4=64\)
=> \(\left[{}\begin{matrix}x-5=8\\x-5=-8\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=8+5=13\\x=-8+5=-3\end{matrix}\right.\)
d/ \(\left(\left|x\right|-\frac{1}{2}\right)\left(2x+\frac{3}{2}\right)=0\)
=> \(\left[{}\begin{matrix}\left|x\right|-\frac{1}{2}=0\\2x+\frac{3}{2}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left|x\right|=0+\frac{1}{2}=\frac{1}{2}\\2x=0-\frac{3}{2}=-\frac{3}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\\x=-\frac{3}{2}:2=-\frac{3}{2}.\frac{1}{2}=-\frac{3}{4}\end{matrix}\right.\)
e) \(5x+2=3x+8\)
=> \(5x-3x=8-2=6\)
=> \(2x=6\)
=> \(x=6:2=3\)
f/ \(26-\left(5-2x\right)=27\)
=> \(5-2x=26-27=-1\)
=> \(2x=5-\left(-1\right)=5+1=6\)
=> \(x=6:2=3\)
g/ \(\left(4x-8\right)-\left(2x-6\right)=4\)
=> \(4x-8-2x+6=4\)
=> \(\left(4x-2x\right)+\left(-8+6\right)=4\)
=> \(2x+-2=4\)
=> \(2x=4+2=6\)
=> \(x=6:2=3\)
h/ \(\left(x+3\right)^3:3-1=-10\)
=> \(\left(x+3\right)^3:3=-10+1=-9\)
=> \(\left(x+3\right)^3=-9.3=-27\)
=> \(x+3=-3\)
=> \(x=-3-3=-6\)
Bài 3:
a: \(\frac13+\frac23:x=-7\)
=>\(\frac23:x=-7-\frac13=-\frac{22}{3}\)
=>\(x=\frac23:\frac{-22}{3}=\frac{-2}{22}=-\frac{1}{11}\)
b: \(3\frac12-\frac12x=\frac23\)
=>\(\frac72-\frac{x}{2}=\frac13\)
=>\(7-x=\frac23\)
=>\(x=7-\frac23=\frac{21}{3}-\frac23=\frac{19}{3}\)
c: \(\left\lbrack\left(x+\frac13\right)\cdot\frac34+5\right\rbrack:2=3\)
=>\(\left(x+\frac13\right)\cdot\frac34+5=3\cdot2=6\)
=>\(\left(x+\frac13\right)\cdot\frac34=1\)
=>\(x+\frac13=1:\frac34=\frac43\)
=>\(x=\frac43-\frac13=\frac33=1\)
d: \(\left(1\frac23-x\right)\cdot0,75-2=\frac12\)
=>\(\left(\frac53-x\right)\cdot\frac34=2+\frac12=\frac52\)
=>\(\frac53-x=\frac52:\frac34=\frac52\cdot\frac43=\frac{5\cdot2}{3}=\frac{10}{3}\)
=>\(x=\frac53-\frac{10}{3}=-\frac53\)
e: \(x+75\%\cdot x=-1,6\)
=>\(x+0,75x=-1,6\)
=>1,75x=-1,6
=>\(x=-\frac{160}{175}=-\frac{32}{35}\)
f: \(\left(x-\frac23\right)\left(2x+1\right)=0\)
=>\(\left[\begin{array}{l}x-\frac23=0\\ 2x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac23\\ x=-\frac12\end{array}\right.\)
g: \(1-\left(2x+\frac12\right)^2=\frac34\)
=>\(\left(2x+\frac12\right)^2=1-\frac34=\frac14\)
=>\(\left[\begin{array}{l}2x+\frac12=\frac12\\ 2x+\frac12=-\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac12-\frac12=0\\ 2x=-\frac12-\frac12=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-\frac12\end{array}\right.\)
h: \(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\cdots+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2020}{2021}\)
=>\(1-\frac16+\frac16-\frac{1}{11}+\cdots+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2020}{2021}\)
=>\(1-\frac{1}{5x+6}=\frac{2020}{2021}\)
=>\(\frac{1}{5x+6}=\frac{1}{2021}\)
=>5x+6=2021
=>5x=2015
=>x=403