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\(1,\\ x+\dfrac{1}{2}=-\dfrac{5}{3}\\ x=-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{13}{6}\\ Vậyx=-\dfrac{13}{6}\)
\(2,\\ \dfrac{1}{3}-x=\dfrac{3}{5}\\ x=\dfrac{1}{3}-\dfrac{3}{5}\\ x=-\dfrac{4}{15}\\ Vậyx=-\dfrac{4}{15}\)
\(3,\\ 3-4+x=\dfrac{7}{2}\\ -1+x=\dfrac{7}{2}\\ x=\dfrac{7}{2}+1\\ x=\dfrac{9}{2}\\ Vậyx=\dfrac{9}{2}\)
\(4,\\ x-\dfrac{4}{3}=-\dfrac{7}{9}\\ x=-\dfrac{7}{9}+\dfrac{4}{3}\\ x=\dfrac{15}{27}\\ Vậyx=\dfrac{15}{27}\)
\(5,\\ x-\left(-\dfrac{7}{3}\right)=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{3}\\ x=-\dfrac{27}{18}\\ Vậyx=-\dfrac{27}{18}\)
\(6,\\ x-\dfrac{1}{5}=\dfrac{9}{10}\\ x=\dfrac{9}{10}+\dfrac{1}{5}\\ x=\dfrac{11}{10}\\ Vậyx=\dfrac{11}{10}\)
\(7,\\ x+\dfrac{5}{12}=\dfrac{3}{8}\\ x=\dfrac{3}{8}-\dfrac{5}{12}\\ x=-\dfrac{1}{24}\\ Vậyx=-\dfrac{1}{24}\)
\(8,\\ x+\dfrac{5}{4}=\dfrac{7}{6}\\ x=\dfrac{7}{6}-\dfrac{5}{4}\\ x=-\dfrac{9}{24}\\ Vậyx=-\dfrac{9}{24}\)
\(9,\\ x-\dfrac{2}{7}=\dfrac{1}{35}\\ x=\dfrac{1}{35}+\dfrac{2}{7}\\ x=\dfrac{11}{35}\\ Vậyx=\dfrac{11}{35}\\ 10,\\ x-\dfrac{1}{5}=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}+\dfrac{1}{5}\\ x=-\dfrac{1}{2}\\ Vậyx=-\dfrac{1}{2}\)
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\(x\div4\frac{1}{3}=-2,5\)
\(\Leftrightarrow x\div\frac{13}{3}=\frac{-5}{2}\)
\(\Leftrightarrow x=\frac{-5}{2}.\frac{13}{3}\)
\(\Leftrightarrow x=\frac{-65}{6}\)
\(x\div\frac{-3}{5}=\frac{-10}{21}\)
\(\Leftrightarrow x=\frac{-10}{21}.\frac{-3}{5}\)
\(\Leftrightarrow x=\frac{30}{105}\)
\(\Leftrightarrow x=\frac{2}{7}\)
\(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{1}{10}+\frac{1}{2}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{6}{10}\)
\(\Leftrightarrow x=\frac{6}{10}\div\frac{2}{3}\)
\(\Leftrightarrow x=\frac{18}{20}\)
\(\Leftrightarrow x=\frac{9}{10}\)
\(\frac{1}{2}x+\frac{1}{2}=\frac{5}{2}\)
\(\Leftrightarrow\frac{1}{2}\left(x+1\right)=\frac{5}{2}\)
\(\Leftrightarrow\left(x+1\right)=\frac{5}{2}\div\frac{1}{2}\)
\(\Leftrightarrow\left(x+1\right)=5\)
\(\Leftrightarrow x=5-1\)
\(\Leftrightarrow x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ \(x+\dfrac{1}{2}=\dfrac{-5}{3}\)
\(x=\dfrac{-5}{3}-\dfrac{1}{2}\)
\(x=\dfrac{-10}{6}-\dfrac{3}{6}\)
Vậy \(x=\dfrac{-13}{6}\)
2/\(\dfrac{1}{3}-x=\dfrac{3}{5}\)
\(-x=\dfrac{3}{5}-\dfrac{1}{3}\)
\(-x=\dfrac{9}{15}-\dfrac{5}{15}\)
\(-x=\dfrac{4}{15}\)
Vậy \(x=\dfrac{-4}{15}\)
3/ \(3-4+x=\dfrac{7}{2}\)
\(-4+x=\dfrac{7}{2}-3\)
\(-4+x=\dfrac{7}{2}-\dfrac{6}{2}\)
\(-4+x=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+4\)
\(x=\dfrac{1}{2}+\dfrac{8}{2}\)
Vậy \(x=\dfrac{9}{2}\)
4/ \(x-\dfrac{4}{3}=\dfrac{-7}{9}\)
\(x=\dfrac{-7}{9}+\dfrac{4}{3}\)
\(x=\dfrac{-7}{9}+\dfrac{12}{9}\)
Vậy \(x=\dfrac{5}{9}\)
5/ \(x-\dfrac{-7}{2}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}-\dfrac{7}{2}\)
\(x=\dfrac{5}{6}-\dfrac{21}{6}\)
Vậy \(x=\dfrac{-16}{6}=\dfrac{-8}{3}\)
6/ \(x-\dfrac{1}{5}=\dfrac{9}{10}\)
\(x=\dfrac{9}{10}+\dfrac{1}{5}\)
\(x=\dfrac{9}{10}+\dfrac{2}{10}\)
Vậy \(x=\dfrac{11}{10}\)
7/ \(x+\dfrac{5}{12}=\dfrac{3}{8}\)
\(x=\dfrac{3}{8}-\dfrac{5}{12}\)
\(x=\dfrac{9}{24}-\dfrac{10}{24}\)
Vậy \(x=\dfrac{-1}{24}\)
8/ \(x+\dfrac{5}{4}=\dfrac{7}{6}\)
\(x=\dfrac{7}{6}-\dfrac{5}{4}\)
\(x=\dfrac{14}{12}-\dfrac{15}{12}\)
Vậy \(x=\dfrac{-1}{12}\)
9/ \(x-\dfrac{2}{7}=\dfrac{1}{35}\)
\(x=\dfrac{1}{35}+\dfrac{2}{7}\)
\(x=\dfrac{1}{35}+\dfrac{10}{35}\)
Vậy \(x=\dfrac{11}{35}\)
10 /\(x-\dfrac{1}{5}=\dfrac{-7}{10}\)
\(x=\dfrac{-7}{10}+\dfrac{1}{5}\)
\(x=\dfrac{-7}{10}+\dfrac{2}{10}\)
Vậy \(x=\dfrac{-5}{10}=\dfrac{-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2. \(7\left(-3\right)-5\left(3-x\right)=11-5\)
\(\Leftrightarrow-21-5\left(3-x\right)=6\)
\(\Leftrightarrow-5\left(3-x\right)=27\)
\(\Leftrightarrow3-x=-\frac{27}{5}\)
\(\Leftrightarrow x=-\frac{42}{5}\)
Vậy \(x=-\frac{42}{5}\)
Khó quá :((
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có : \(10+3\left(x-1\right)=10+6x\)
=> \(10+3x-3-10-6x=0\)
=> \(-3x-3=0\)
=> \(x=-1\)
b, Ta có : \(2\left(x-2\right)+3\left(3-x\right)=-4\)
=> \(2x-4+9-3x=-4\)
=> \(-x=-9\)
=> \(x=9\)
c, \(\left(\left|x-2\right|+1\right)\left(x-3\right)=0\)
TH1 : \(x-2\ge0\left(x\ge2\right)\)
=> \(\left|x-2\right|=x-2\)
Nên ta có phương trình : \(\left(x-2+1\right)\left(x-3\right)=0\)
=> \(\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=1\left(kTM\right)\\x=3\end{matrix}\right.\)
=> \(x=3\)
TH2 : \(x-2< 0\left(x< 2\right)\)
=> \(\left|x-2\right|=2-x\)
Nên ta có phương trình : \(\left(2-x+1\right)\left(x-3\right)=0\)
=> \(\left[{}\begin{matrix}3-x=0\\x-3=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=3\\x=3\end{matrix}\right.\) ( ktm )
d, Ta có : \(\left(x+5\right)\left(-3x-15\right)=0\)
=> \(\left[{}\begin{matrix}x+5=0\\-3x-15=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-5\\x=-5\end{matrix}\right.\)
1) 10 + 3 . (x - 1) = 10 + 6x
10 + 3x - 3 = 10 + 6x
3x - 6x = 10 - 10 + 3
-3x = 0 + 3
-3x = 3
x = 3 : (-3)
x = -1
Vậy x = -1
2) 2 . (x - 2) + 3 . (3 - x) = -4
2x - 4 + 9 - 3x = -4
2x - 3x = -4 + 4 - 9
-x = 0 - 9
-x = -9
=> x = 9
Vậy x = 9
3) (Ix - 2I + 1) . (x - 3) = 0
=> (Ix - 2I + 1) = 0 hoặc (x - 3) = 0
Ix - 2I = 0-1 x = 0 + 3
Ix - 2I = -1( loại ) x = 3( thỏa mãn )
Vậy x = 3
4) (x + 5) . (-3x - 15) = 0
=> (x + 5) = 0 hoặc (-3x - 15) = 0
x = 0 - 5 -3x = 0 + 15
x = -5 -3x = 15
x = 15 : (-3)
x = -5
Vậy x = -5
Tick cho mk nha
13 + 23 + 33 + ... + 103 = (1+2+3+...+10)2 = 552 = (x + 10)2
=> |x+10| = 55
PT có 2 nghiệm x = -65 và x = 45.