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17 tháng 6 2020
* \(\left(x-\frac{5}{24}\right)\cdot\frac{18}{7}=-\frac{12}{7}\)
<=> \(x-\frac{5}{24}=-\frac{2}{3}\)
<=> \(x=-\frac{11}{24}\)
* \(\frac{3}{4}+\frac{1}{4}\left(x-1\right)=\frac{1}{2}\)
<=> \(\frac{x-1}{4}=\frac{-1}{4}\)
<=> \(x-1=-1\)
<=> \(x=0\)
* \(\left(4x-\frac{1}{2}\right)\left(\frac{x}{3}-\frac{1}{5}\right)=0\)
<=> \(\orbr{\begin{cases}4x-\frac{1}{2}=0\\\frac{x}{3}-\frac{1}{5}=0\end{cases}}\)<=> \(\orbr{\begin{cases}4x=\frac{1}{2}\\\frac{x}{3}=\frac{1}{5}\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{1}{8}\\x=\frac{3}{5}\end{cases}}\)
1 tháng 6 2016
\(\text{Ta có: }\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+.....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{\left(x+2\right)}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{\left(x+5\right)}=\frac{1}{2}-\frac{3}{20}\)
4 tháng 6 2016
Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
MM
22 tháng 8 2019
1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)
\(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)
\(x=\frac{45}{4}+\frac{9}{4}\)
\(x=\frac{27}{2}\)
9 tháng 6 2017
\(\frac{1}{x+2}-\frac{1}{x+5}+...+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\frac{1}{x+2}-\frac{1}{x+7}=\frac{x}{\left(x+2\right)\left(x+7\right)}\)
\(\Rightarrow x=1\)
7 tháng 7 2019
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_Tần vũ_
7 tháng 7 2019
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Leftrightarrow3x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{1}{18}\)
_Tần Vũ_
\(\frac{1}{3}\left(3+\frac{3}{5}x\right)-4x=20\%x-1\)
=> \(1+\frac{1}{5}x-4x=\frac{1}{5}x-1\)
=> \(1+\frac{1}{5}x-4x-\frac{1}{5}x+1=0\)
=> \(\left(1+1\right)+\left(\frac{1}{5}x-\frac{1}{5}x-4x\right)=0\)
=> \(2-4x=0\)
=> \(4x=2\)
=> \(x=\frac{1}{2}\)
Vậy : ...
P/S : Lớp 6 có phương trình ???
\(\frac{1}{3}\left(3+\frac{3}{5}x\right)-4x=20\%.x-1\)
\(\Leftrightarrow1+\frac{1}{5}x-4x=\frac{1}{5}x-1\)
\(\Leftrightarrow1-4x=-1\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=2\)