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a, \(-19-\left(2x-5\right)^3=-18\)
\(\Leftrightarrow\left(2x-5\right)^3=-1\)
\(\Leftrightarrow2x-5=-1\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
b, \(\frac{x}{5}+\frac{1}{3}.\frac{9}{2}=\frac{-1}{5}\)
\(\Leftrightarrow\frac{x}{5}+\frac{3}{2}=\frac{-1}{5}\)
\(\Leftrightarrow\frac{x}{5}=\frac{-17}{10}\)
\(\Leftrightarrow10x=5.\left(-17\right)\)
\(\Leftrightarrow10x=-85\)
\(\Leftrightarrow x=\frac{-17}{2}\)
a) 3x – 15 = 25 – 5x
=> 3x + 5x = 25 + 15
=> 8x = 40
=> x = 5
b) 3x - 17 = 2x – 7
=> 3x - 2x = -7 + 17
=> x = 10
c) 2x – 17 = – (3x – 18)
=> 2x - 17 = -3x + 18
=> 2x + 3x = 18 + 17
=> 5x = 35
=> x = 7
d) 3x – 14 = 2(x – 9) + 1
=> 3x - 14 = 2x - 18 + 1
=> 3x - 2x = -18 + 1 + 14
=> x = -3
f) (x – 5)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
a) Ta có: \(3x-15=25-5x\)
\(\Leftrightarrow3x-15-25+5x=0\)
\(\Leftrightarrow8x-40=0\)
\(\Leftrightarrow8x=40\)
hay x=5
Vậy: x=5
b) Ta có: \(3x-17=2x-7\)
\(\Leftrightarrow3x-17-2x+7=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
Vậy: x=10
c) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=-3x+18\)
\(\Leftrightarrow2x-17+3x-18=0\)
\(\Leftrightarrow5x-35=0\)
\(\Leftrightarrow5x=35\)
hay x=7
Vậy: x=7
d) Ta có: \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14-2x+18-1=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: x=-3
f) Ta có: \(\left(x-5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)
\(\dfrac{2x+5}{9}=\dfrac{7}{18}=>\dfrac{9\cdot7}{18}=2x+5\)
\(2x+5=3.5\)
\(2x=3.5-5\)
\(2x=\left(-1.5\right)\)
\(x=\left(-1.5\right):2\)
\(x=-0.75\)
=> x = -0.75
~ HT ~
\(\dfrac{2x}{18}\) \(=\) \(\dfrac{5}{9}\) \(+\) \(\dfrac{2}{3}\)
\(\dfrac{2x}{18}\) \(=\) \(\dfrac{11}{9}\)
\(2x . 9\) \(=\) \(18 . 11\)
\(2x . 9\) \(=\) \(198\)
\(2x \) \(=\) \(198 : 9\)
\(2x\) \(=\) \(22\)
\(x\) \(=\) \(22 : 2 = 11\)
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
a, 75 - 5 . ( 15 - 10 ) - ( -60 ) = 75 - 5 . 5 + 60 = 135 - 25 = 110
b, 136 - ( -7 ) + 6 - 23 - 36 = 136 + 7 + 6 - 23 - 36 = ( 136 - 36 ) + ( 7 + 6 - 23 ) = 100 - 10 = 90
c, ( -15 - 17 ) . ( -15 + 7 ) = -32 . ( -8 ) = 256
d, ( -9 - 18 ) : ( -9 ) = -27 : ( -9 ) = 3
Bài 2:
a, 2x - 17 = ( -3x - 18 )
<=> 2x + 3x = 17 - 18
<=> 5x = -1
<=> x = -1/5
b, ( 13 - IxI ) + 15 = 20
<=> 13 - |x| = 5
<=> |x| = 8
<=> x = 8 hoặc x = -8
c, Ix + 5I - ( x + 5 ) = 0
<=> |x + 5| = x + 5
Đk: x + 5 ≥ 0 => x ≥ -5
\(\Leftrightarrow\orbr{\begin{cases}x+5=x+5\\x+5=-x-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}0x=0\\2x=-10\end{cases}\Leftrightarrow}\orbr{\begin{cases}x\ge-5\\x=-5\end{cases}}\)
Vậy x ≥ -5
1.
a, \(x-14=3x+18\)
\(\Rightarrow x-3x=18+14\)
\(\Rightarrow-2x=32\Rightarrow x=\frac{32}{-2}=-16\)
b, \(\left(x+7\right).\left(x-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)
c, \(\left|2x-5\right|-7=22\)
\(\Rightarrow\left|2x-5\right|=22+7\)
\(\Rightarrow\left|2x-5\right|=29\)
\(\Rightarrow\orbr{\begin{cases}2x+5=29\\2x-5=29\end{cases}}\Rightarrow\orbr{\begin{cases}2x=24\\2x=34\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=17\end{cases}}\)
d, \(\left(\left|2x\right|-5\right)-7=22\)
\(\Rightarrow\left(\left|2x\right|-5\right)=29\)
\(\Rightarrow\left|2x\right|=29+5\Rightarrow\left|2x\right|=34\Rightarrow x=\pm17\)
e, \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\)
Vì \(\left|x+3\right|\ge0;\left|x+9\right|\ge0;\left|x+5\right|\ge0;4x\ge0\)
Nên \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\ge0\)
\(\Rightarrow\left|x+3\right|>0\Rightarrow\left|x+3\right|=x+3\)
\(\left|x+9\right|>0\Rightarrow\left|x+9\right|=x+9\)
\(\left|x+5\right|>0\Rightarrow\left|x+5\right|=x+5\)
Ta có :
\(x+3+x+9+x+5=4x\)
\(\Rightarrow3x+\left(3+9+5\right)=4x\)
\(\Rightarrow4x-3x=17\)
\(\Rightarrow x=17\)
2. a , b sai đề bn
c, \(\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(\text{ }Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
x | 0 | -2/5 | 1/5 | -3/5 | 3/5 | -1 |
y | -3 | 5 | -1 | 3 | 0 | 2 |
d, \(5xy-5x+y=5\)
\(\Rightarrow\left(5xy-5x\right)+y=5\)
\(\Rightarrow5x.\left(y-1\right)+y=5\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
x | 0 | -2 | 1/5 | -3/5 | 3/5 | -1 |
y | -3 | 5 | -1 | 3 | 0 | 2 |
18 - (2x + 5) = 9
2x + 5 = 18 - 9
2x + 5 = 9
2x = 9 + 5
2x = 14
x = 14 : 2
x = 7
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