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a) | 2x - 5 | = 13
=> 2x - 5 = 13 hoặc 2x - 5 = -13
+ Nếu 2x - 5 = 13
2x = 13 + 5
2x = 18
x = 18 : 2
x = 9
+ Nếu 2x - 5 = -13
2x = ( -13 ) + 5
2x = -8
x = ( -8 ) : 2
x = -4
=> x = { -4 ; 9 }
Tck nha
|7x + 3| = 66
7x + 3 = 66
7x = 66-3
7x = 63
x = 63 : 7
x = 9
a) 14-(7-x+3)=5-{4-(5- |3| ) }
14-(10-x) = 5-{4-(5-3) }
x +14-10=5-(4-2)
x+4 = 5-2
x+4 =3
x =3-4
x =-1 Vậy x= -1
-7 + [ - (-3) + |6| - (544 + |-6 |) ] = 5 - ( 7 - x + 4)
-7+{ 3+6-(544+6) } =5-(11-x)
-7+(9-600) =x+5-11
-7+-591 =x+(-6)
-598 = x+ (-6)
x =-598 - (-6)
x = -592
Vậy x= -592
tick mình nha
\(\left(7-x\right)^3+\left(11-7\right)^2=141\\ \left(7-x\right)^3+4^2=141\\ \left(7-x\right)^3+16=141\\ \left(7-x\right)^3=141-16=125=5^3\\ Nên:7-x=5\\ Vậy:x=7-5=2\)
b)\(\left(x-8\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x-8=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=2\end{cases}}\)
c) \(\left(x+1\right)+\left(x+2\right)+...+\left(x+10\right)=9x+200\)
\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+2+...+10\right)=9x+200\) (10 số hạng x)
\(\Leftrightarrow10x+55=9x+200\Leftrightarrow x+55=200\)
\(\Leftrightarrow x=145\)
-3 + 7 -9-54 + 3 +50 +13
= (-3+3)+ (13 -9)+ 7 + 54 - 50
= 0 + 4 + 7 + 54 - 50
=0 + 7 + 54 - ( 50 +4 )
= 0 + 7 + 54 - 54
= 7 + 0
=7
a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
( x2 - x + 7 ) ⋮ ( x + 1 )
=> ( x2 + x - 2x - 2 + 9 ) ⋮ ( x + 1 )
=> [ x( x + 1 ) - 2( x + 1 ) + 9 ] ⋮ ( x + 1 )
=> [ ( x + 1 )( x - 2 ) + 9 ] ⋮ ( x + 1 )
=> 9 ⋮ ( x + 1 )
=> ( x + 1 ) ∈ Ư(9) = { ±1 ; ±9 }
=> x ∈ { 0 ; -2 ; 8 ; -10 }
|7x+3|=66
<=> \(\orbr{\begin{cases}7x+3=66\\7x+3=-66\end{cases}}\)
<=> \(\orbr{\begin{cases}x=9\\x=-\frac{69}{7}\end{cases}}\)
Vậy ............
|7x + 3 |= 66
=> 7x + 3 = 66
7x = 66 - 3
x = 63 : 7
x = 9
hoặc 7x + 3 = -66
7 x = - 66 - 3
7x = -69
x = -69 : 7
x = -9,9
=> x \(\in\){ 9 }