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\(\frac{2}{3}.x=\frac{1}{3}\)1) x-\(\frac{10}{3}\)=\(\frac{7}{15}.\frac{3}{5}\)
x-10/3=7/25
x=7/25+10/3
x=\(\frac{271}{75}\)
2)\(\frac{8}{23}.\frac{46}{24}.x=\frac{1}{3}\)
2/3.x=1/3
x=1/3:2/3
x=1/2
Tìm x
a, 5x - 10 . (-7) = 15 . (-3)
5x + 70 = -45
5x = -45 - 70
5x = -115
x = -115 : 5
x = -23
Vậy x = -23.
b, 2x + 12 = -8
2x = -8 - 12
2x = -20
x = -20 : 2
x = -10
Vậy x = -10.
c, 3x - 5 = -20
3x = -20 + 5
3x = -15
x = -15 : 3
x = -5
Vậy x = -5.
d, 2x - (-12) = -8 . 23
2x + 12 = -184
2x = -184 - 12
2x = -196
x = -196 : 2
x = -98
Vậy x = -98.
e, x2 = 49
x2 = 72 = (-7)2
=> x = 7 hoặc x = -7.
Vậy x thuộc {-7 ; 7}
i, (x - 2)2 + 1 = 65
(x - 2)2 = 65 - 1
(x - 2)2 = 64
(x - 2)2 = 82 = (-8)2
=> x - 2 = 8 hoặc x - 2 = -8
=> x = 10 hoăc x = -6
Vậy x thuộc {-6 ; 10}.
g, 2 (x - 3)2 - 5 = 5 . (-3)2
2 (x - 3)2 - 5 = 5 . 9
2 (x - 3)2 - 5 = 45
2 (x - 3)2 = 45 + 5
2 (x - 3)2 = 50
(x - 3)2 = 50 : 2
(x - 3)2 = 25
(x - 3)2 = 52 = (-5)2
=> x - 3 = 5 hoặc x - 3 = -5
=> x = 8 hoặc x = -2
Vậy x thuộc {-2 ; 8}.
a) 5x - 10.(-7)=15.(-3) b) 2x+12=-8 c) 3x-5=-20 d)2x-(-12)=-8.23 e)\(^{x^2}\)=49 i)\(\left(x-2\right)^2+1\)=65
5x+70=-45 2x=-8-12 3x=-20+5 2x+12=-184 x=7 \(\left(x-2\right)^2\) =64
5x=-45-70 2x=-20 3x=-15 2x=-184-12 X-2 =8
5x=-115 x=-20:2 x=-15:3 2x=-186 x =10
x=-23 x=-10 x=-5 x=-93
g)\(2.\left(x-3\right)^2-5=5.\left(-3\right)^2\)
\(2\left(x-3\right)^2-5=30\)
\(2.\left(x-3\right)^2=35\)
\(2.\left(x^2-2x.3+3^2\right)=35\)
\(2x^2-24x+18=35\)
\(2x^2-24x=17\)
\(2x\left(x-12\right)=17\)
\(\orbr{\begin{cases}2x=17\\x-12=17\end{cases}}\orbr{\begin{cases}x=\frac{17}{2}\\x=29\end{cases}}\)
=1,4 x\(\dfrac{15}{49}-\) \(\left(\dfrac{4}{5}+\dfrac{2}{3}\right)\) : 2\(\dfrac{1}{5}\)
= \(\dfrac{3}{7}\) - \(\dfrac{22}{15}\) : \(\dfrac{11}{5}\)
= \(\dfrac{3}{7}\) - \(\dfrac{2}{3}\)
= \(-\dfrac{5}{21}\)
( 2\(\dfrac{1}{5}\) + \(\dfrac{3}{5}\) \(\times\) \(x\)) = \(\dfrac{3}{4}\)
\(\dfrac{11}{5}\) + \(\dfrac{3}{5}\)\(x\) = \(\dfrac{3}{4}\)
\(\dfrac{3}{5}\)\(x\) = \(\dfrac{3}{4}\) - \(\dfrac{11}{5}\)
\(\dfrac{3}{5}\)\(x\) = - \(\dfrac{29}{20}\)
\(x\) = -\(\dfrac{29}{12}\)
x = \(\frac{1}{10}\)
1 x = _ 10