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|Câu d:
x = 3/4 + 1/-12
x = 9/12 - 1/12
x = 8/12
x = 2/3
Vậy x = 2/3
Câu e:
x /14 = 1/7 + (-3)/14
x/14 = 1/7 - 3/14
x/14 = 2/14 - 3/14
x/14 = -1/14
x = - 1/14 x 14
x = -1
Vậy x = -1
\(a,\frac{-5}{7}+1+\frac{-7}{30}< x< \frac{-1}{6}+\frac{1}{3}+\frac{5}{6}\)
\(\Leftrightarrow\frac{11}{210}< x< 1\)
Vậy tập hợp các số x thuộc Z là > 11/210 và < 1
câu b, c tương tự
mk ko biết đúng hay sai đâu
sai cũng đừng nói quá đáng nha
a) x + 2 = - 9 - 11
x + 2 = - 20
x = - 20 - 2
x = - 22
Vậy x = - 22
b) x - 2 = - 6 + 17
x - 2 = 11
x = 11 + 2
x = 13
Vậy x = 13
đây bạn
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
a, -5/7+ 1+ 30/-7< x < -1/6+ 1/3 +5/6
<=> -4< x <1
<=> x = -3; -2; -1; 0
a, \(\dfrac{-5}{7}+1+\dfrac{30}{-7}\le x\le\dfrac{-1}{6}+\dfrac{1}{3}+\dfrac{5}{6}\)
<=> -4 \(\le x\le1\)
Do x \(\in Z\Rightarrow x=-4;-3;-2;-1;0;1\)
b, \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
<=> -\(\dfrac{1}{12}< x< \dfrac{1}{8}\)
Do x \(\in Z\Rightarrow x=0;1\)
@Mai Tran
a ) 2 - x = 5 - 6
2 - x = - 1
x = 2 - ( - 1 )
x = 3
Vậy x = 3
b ) 7 - x = 17 - ( - 5 )
7 - x = 22
x = 7 - 22
x = - 15
Vậy x = - 15
Ta có :
a) 2 - x = 5 - 6
=> 2 - x = - 1
=> x = 3
b ) 7 - x = 17 - ( - 5 )
=> 7 - x = 22
=> x = - 15
c ) | x + 3 | = 7
=> x + 3 = 7 hoặc x + 3 = - 7
=> x = 4 hoặc x = - 10
a) 2 - x = 5 - 6 c ) |x + 3| = 7
2 - x = -1 => \(\orbr{\begin{cases}x+3=7\\x+3=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-10\end{cases}}\)
x = 2 + 1
x = 3
b) 7 - x = 17 - (-5)
7 - x = 22
x = 7 - 22
x = -15
c ) l x + 3 l = 7
=> \(\orbr{\begin{cases}x+3=7\\x+3=-7\end{cases}}\)=> \(\orbr{\begin{cases}x=7-3\\x=-7-3\end{cases}}\)=> \(\orbr{\begin{cases}x=4\\x=-10\end{cases}}\)
Vậy x thuộc { - 10 ; 4 }
d ) x - ( 17 - x ) = x - 7
x - 17 + x = x - 7
( x + x ) - 17 = x - 7
2x - 17 = x - 7
2x - x = - 7 + 17
x = 10
Vậy x = 10
\(2-x=5-6\)
\(2-x=-1\)
\(x=2-\left(-1\right)\)
\(x=3\)
a) 2 - x = 5 - 6
=> 2 - x = -1
=> x = 2 - ( -1 )
=> x = 3
Vậy : x = 3
b) 7 - x = 17 - ( - 5 )
=> 7 - x = 22
=> x = 7 - 22
=> x = -15
Vậy : x = -15
c) | x + 3 | = 7
=> \(\orbr{\begin{cases}x+3=7\\x+3=-7\end{cases}}\) => \(\orbr{\begin{cases}x=4\\x=-10\end{cases}}\)
Vậy x = 4 ; x = -10
d) x - ( 17 - x ) = x - 7
=> x - 17 + x = x - 7
=> x + x - x = 17 - 7
=> x = 10
Vậy : x = 10
e) ( x - 2 ) . ( 5 - x ) = 0
=> \(\orbr{\begin{cases}x-2=0\\5-x=0\end{cases}}\)=> \(\orbr{\begin{cases}x=2\\x=5\end{cases}}\)
Vậy : x = 2 ; x = 5
a, 2 - x = 5 - 6
2 - x = -1
x = 2 - ( -1 )
x = 3
b, 7 - x = 17 - ( -5 )
7 - x = 22
x = 7 - 22
x = -15
c, | x + 3 | = 7
=> TH1: x + 3 = 7
x = 7 - 3
x = 4
=> TH2 : x + 3 = -7
x = -7 - 3
x = -10
Vậy x=4 hoặc x = -10
d, x - ( 17 - x ) = x - 7
x - 17 + x = x - 7
x + x - 17 = x - 7
x + x - x = -7 + 17
x = 10
e, ( x - 2 ) . ( 5 - x ) = 0
=> TH1 : x - 2 = 0
x = 0 + 2
x = 2
=> TH2: 5 - x = 0
x = 5 -0
x = 5
Vậy x = 2 hoặc x = 5