3x - 35 =15
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1 ) ( 2x - 15 ) + 17 = 6
=> 2x - 15 = 6 - 17
=> 2x - 15 = - 11
=> 2x = - 11 + 15
=> 2x = 4
=> x = 4: 2
=> x = 2
Vậy x = 2
2) 15 - ( 4 - 3x ) = -4
=> 4 - 3x = 15 - ( -4 )
=> 4 - 3x = 19
=> 3x = 4 - 19
=> 3x = -15
=> x = -15 : 3
=> x = -5
Vậy x = -5
3) -21 + 3 . ( -x + 7 ) = - 18
=> 3 . ( -x + 7 ) = ( -18 ) - ( -21 )
=> 3 . ( -x + 7 ) = 3
=> -x +7 = 3 : 3
=> -x + 7 = 1
=> -x = 1 - 7
=> -x = - 6
=> x = - (- 6 )
Vậy x = - ( -6 )
4 ) 78 : ( 3x - 2 ) = -3
=> 3x - 2 = 78 : ( -3)
=> 3x - 2 = -26
=> 3x = -26 + 2
=> 3x = -24
=> x = -24 : 3
=> x = -8
Vậy x = -8
5) -35 : 5 . ( -3 - 2x ) = 35
=> -7 . ( -3 ) + 2x = 35
=> -3 + 2x = 35 : ( -7)
=> -3 + 2x = -5
=> 2x = (-5) - (-3)
=> 2x = -2
=> x = -2 : 2
=> x = -
Vậy x = -1
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1/ (2x – 15) + 17 = 6
<=> 2x=4
<=> x=2
2/ 15 – (4 – 3x) = - 4
<=> 3x=-15
<=> x=-5
3/ - 21 + 3(-x + 7) = -18
<=> -3x=-18
<=> x=6
4/ 78 : (3x – 2) = -3
<=> 3x-2=-26
<=> 3x=-24
<=> x=-8
5/ -35 : 5.(-3 – 2x) = 35
<=> -3-2x=-7
<=> 2x=4
<=> x=2
a) (x - 1)(5x + 3) = (3x - 8)(x - 1)
\(\Leftrightarrow\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+11\right)=0\)
\(\Leftrightarrow x-1=0\Rightarrow x=1\)
và\(2x+11=0\Rightarrow x=\frac{-11}{2}\)
3x(25x + 15) – 35(5x + 3) = 0
⇔ 15x(5x + 3) – 35(5x + 3) = 0
⇔ (15x – 35)(5x + 3) = 0 ⇔ 15x – 35 = 0 hoặc 5x + 3 = 0
15x – 35 = 0 ⇔ x = 35/15 = 7/3
5x + 3 = 0 ⇔ x = - 3/5
Vậy phương trình có nghiệm x = 7/3 hoặc x = -3/5
a/ x - ( 25 - x) = x - 15
=> x - 25 + x = x - 15
=> 2x - x = -15 + 25
=> x = 10
b/ x - 8 = -13 - 8
=> x = -13 - 8 + 8 = -13
c/ x - (-15) = 5 - (-19)
=> x + 15 = 5 + 19 => x = 5 + 19 - 15 = 9
d/ (x + 35) = 2x + 15 + (13 - 4)
=> x + 35 = 2x + 15 + 13 - 4
=> x - 2x = 15 + 13 - 4 - 35
=> -x = -11 => x = 11
e/ 5 - 2x = 15-(-9) - 3x
=> 5 - 2x = 15 + 9 - 3x
=> -2x + 3x = 15 + 9 - 5
=> x = 19
g/ x + 135 = 3x - 15
=> x - 3x = -15 - 135
=> -2x = -150
=> x = -150/ - 2 = 75
b) PT \(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)
c) PT \(\Leftrightarrow\left(2-3x\right)\left(x-11\right)+\left(2-3x\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(2-3x\right)\left(-9-4x\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{9}{4}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{2}{3};-\dfrac{9}{4}\right\}\)
a)(x-1)(5x+3)=(3x-8)(x-1)
\(\Leftrightarrow\)(x-1)(5x+3)-(3x-8)(x-1)=0
\(\Leftrightarrow\left(x-1\right)\left(5x-3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\)
\(\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{1;\dfrac{5}{2}\right\}\)
\(3x\left(25x+15\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}15x-35=0\\5x+3=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{-3}{5}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{3};\frac{-3}{5}\right\}\)
3x(25x + 15) - 35(5x + 3) = 0
<=> 15x(5x + 3) - 35(5x + 3) = 0
<=> (5x + 3)(15x - 35) = 0
<=> 5(5x + 3)(3x - 7) = 0
<=> 5x + 3 = 0 hay 3x - 7 = 0 (vì 5 \(\ne\)0)
<=> 5x = -3 I <=> 3x = 7
<=> x =\(\frac{-3}{5}\)I <=> x = \(\frac{7}{3}\)
Vậy S = {\(\frac{-3}{5}\); \(\frac{7}{3}\)}
\(3x-35=15\)
\(3x=15+35\)
\(3x=50\)
\(x=\frac{50}{3}\)
3x-35=15
3x = 15 + 35
3x = 50
=> x = 50/3
Vậy x = 50/3