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c: Ta có: \(\dfrac{5}{3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{101\cdot103}\)
\(=\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{101\cdot103}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{102}{103}\)
\(=\dfrac{255}{103}\)
a: =>3/5-x+13/20=5/6
=>11/10-x=5/6
=>x=11/10-5/6=4/15
b: =>x=4/9+5/9=1
c: =>3/2-2x=16/3:8/3=2
=>2x=-1/2
=>x=-1/4
d: =>x+1,5=-5/8*8/5=-1
=>x=-2,5
a: \(\Leftrightarrow\left(\dfrac{13}{4}:x\right)\cdot\left(-\dfrac{5}{4}\right)=\dfrac{-10}{6}-\dfrac{5}{6}=\dfrac{-15}{6}=\dfrac{-5}{2}\)
\(\Leftrightarrow\dfrac{13}{4}:x=\dfrac{5}{2}\cdot\dfrac{5}{4}=\dfrac{25}{8}\)
hay \(x=\dfrac{13}{4}:\dfrac{25}{8}=\dfrac{13}{4}\cdot\dfrac{8}{25}=\dfrac{26}{25}\)
b: \(\Leftrightarrow\dfrac{3}{4}:x=\dfrac{11}{36}-\dfrac{1}{4}=\dfrac{2}{36}=\dfrac{1}{18}\)
=>\(x=\dfrac{3}{4}:\dfrac{1}{18}=\dfrac{54}{4}=\dfrac{27}{2}\)
c: \(\Leftrightarrow\left(-\dfrac{6}{5}+x\right):\left(-3.6\right)=-\dfrac{7}{4}+\dfrac{1}{4}\cdot8=\dfrac{1}{4}\)
=>x-6/5=-9/10
=>x=3/10
Bài 1:
a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)
\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)
hay \(x=-\dfrac{1}{3}\)
Vậy: \(x=-\dfrac{1}{3}\)
b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)
hay \(x=\dfrac{50}{9}\)
Vậy: \(x=\dfrac{50}{9}\)
c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)
\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)
hay \(x=\dfrac{22}{15}\)
Vậy: \(x=\dfrac{22}{15}\)
d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)
\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)
\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)
hay \(x=\dfrac{15}{19}\)
Vậy:\(x=\dfrac{15}{19}\)
\(\left(x^2+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+5=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-5\\x=5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x\in\varnothing\\x=5\end{cases}}}\)
Vậy x=5
TK MK ĐÊ RỒI MK LÀM TIẾP!
\(x-\frac{1}{9}=\frac{8}{3}\)
\(\Leftrightarrow x=\frac{8}{3}+\frac{1}{9}\)
\(\Leftrightarrow x=\frac{24+1}{9}=\frac{25}{9}\)
`Answer:`
a. \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\Leftrightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)
\(\Leftrightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{41}{4}+\frac{3}{4}\\2x=-\frac{41}{4}+\frac{3}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=11\\2x=-\frac{19}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=11:2\\x=-\frac{19}{2}:2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=-\frac{19}{4}\end{cases}}\)
b. \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Leftrightarrow\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)
\(\Leftrightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(x+\frac{1}{5}\right)=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}-\frac{1}{5}\\x=-\frac{3}{5}-\frac{1}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{4}{5}\end{cases}}\)
c. \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Leftrightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}-\left(-\frac{24}{27}\right)\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Leftrightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Leftrightarrow3x=\frac{1}{9}\)
\(\Leftrightarrow x=\frac{1}{9}:3\)
\(\Leftrightarrow x=\frac{1}{27}\)
