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\(a,-12\left(x-5\right)+7\left(3-x\right)=5\)
\(-12x+60+21-7x=5\)
\(-12x-7x+81=5\)
\(-19x=5-81\)
\(-19x=-76\)
\(x=-76:\left(-19\right)\)
\(x=4\)
\(Vậyx=4\)
\(b,30\left(x+2\right)-6\left(x-5\right)-24x=100\)
\(30x+60-6x-30-24x=100\)
\(30x-6x-24x+60-30=100\)
\(0x+30=100\)
\(\Rightarrow Vôlý\)
Vậy không có giá trị nào của x thỏa mãn đề bài.
\(c,-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-1-\frac{1}{2}x-\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-\frac{1}{2}x-1-\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-\frac{11}{2}x-\frac{2}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-\frac{2}{3}+\frac{5}{6}=\frac{3}{2}x+\frac{11}{2}x\)
\(-\frac{4}{6}+\frac{5}{6}=\frac{14}{2}x\)
\(\frac{1}{6}=7x\)
\(x=\frac{1}{6}:7\)
\(x=\frac{1}{6}.\frac{1}{7}\)
\(x=\frac{1}{42}\)
\(Vậyx=\frac{1}{42}\)
\(d,-3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
\(-3x+\frac{3}{2}-5x-3=-x+\frac{1}{5}\)
\(-3x-5x+\frac{3}{2}-3=-x+\frac{1}{5}\)
\(-8x+\frac{3}{2}-\frac{6}{2}=-x+\frac{1}{5}\)
\(-8x-\frac{3}{2}=-x+\frac{1}{5}\)
\(-\frac{3}{2}-\frac{1}{5}=-x+8x\)
\(\frac{15}{10}-\frac{2}{10}=7x\)
\(7x=\frac{13}{10}\)
\(x=\frac{13}{10}:7\)
\(x=\frac{13}{10}.\frac{1}{7}\)
\(x=\frac{13}{70}\)
\(Vậyx=\frac{13}{70}\)
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
a) \(\frac{3}{4}x-\frac{1}{4}=2\left(x-3\right)+\frac{1}{4}x\)
\(\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)
\(\frac{3}{4}x-2x-\frac{1}{4}x=\frac{1}{4}-6\)
\(x\left(\frac{3}{4}-2-\frac{1}{4}\right)=-\frac{23}{4}\)
\(-\frac{3}{2}x=-\frac{23}{4}\)
\(x=-\frac{23}{4}\div\left(-\frac{3}{2}\right)\)
\(x=\frac{23}{6}\)
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
c, Ta có: \(2.|\dfrac{1}{2}x - \dfrac{1}{3}| - \dfrac{3}{2} = \dfrac{1}{4}\)
\(\Rightarrow\) \(2.|\dfrac{1}{2}x - \dfrac{1}{3}| = \dfrac{7}{4}\)
\(\Rightarrow\) \(|\dfrac{1}{2}x - \dfrac{1}{3}| = \dfrac{7}{8}\)
\(\Rightarrow\) \(\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{7}{8}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{29}{24}\\\dfrac{1}{2}x=-\dfrac{13}{24}\end{matrix}\right.\)
\(\Rightarrow\) \(\left[{}\begin{matrix}x=\dfrac{29}{24}\\x=-\dfrac{13}{12}\end{matrix}\right.\)