2x - 10 = 0
7x - 28 =0
3x -7 = 14
44 - 3(x + 2 ) = 2.7
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NH
21 tháng 9 2021
a)2x=0+10
2x=10
x=10:2
x=5
b)7x=0+28
7x=28
x=28:7
x=4
c)3x=7+14
3x=21
x=21:3
x=7
d)3(x+2)=44-14
3(x+2)=30
x+2=30:3
x-2=10
x=10+2
x=12
DT
1
21 tháng 11 2018
/x-3/=7-(-2)
/x-3/=9
x=/9+3/
x=/12/
x=/12/
<=> x= 12; -12
( 3x-24)*73=2*74
(3x-16)*343=4802
3x-16=14
3x=14+16
3x=30
=> x = 10
x-[42+(-28)]=-8
x-14=8
x=8+14
x=22
PT
1
17 tháng 9 2021
\(x^2+3y^2-4x+6y+7=0\\ \Leftrightarrow\left(x^2-4x+4\right)+\left(3y^2+6y+3\right)=0\\ \Leftrightarrow\left(x-2\right)^2+3\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\\ 3x^2+y^2+10x-2xy+26=0\\ \Leftrightarrow\left(x^2-2xy+y^2\right)+2x^2+36=0\\ \Leftrightarrow\left(x-y\right)^2+2x^2+36=0\\ \Leftrightarrow x,y\in\varnothing\left[\left(x-y\right)^2+2x^2+36\ge36>0\right]\\ 3x^2+6y^2-12x-20y+40=0\\ \Leftrightarrow\left(3x^2-12x+12\right)+\left(6y^2-20y+28\right)=0\\ \Leftrightarrow3\left(x-2\right)^2+6\left(y^2-\dfrac{10}{3}y+\dfrac{14}{3}\right)=0\\ \Leftrightarrow3\left(x-2\right)^2+6\left(y^2-2\cdot\dfrac{5}{3}y+\dfrac{25}{9}+\dfrac{17}{9}\right)=0\)
\(\Leftrightarrow3\left(x-2\right)^2+6\left(y-\dfrac{5}{3}\right)^2+\dfrac{34}{3}=0\\ \Leftrightarrow x,y\in\varnothing\)
21 tháng 3 2020
a,
128-3x-12=23
3x=128-12-23
3x=93
x=93:3
= 31
b,
(12x+84+55):5=35
12x+84+55=35.5
12x+84+55=175
12x=175-55-84
12x=36
x=36:12
x=3
TA
1
15 tháng 12 2017
(3x-24 ).73 =2.74
3x-16 =2.74 :73 =2.7=14
3x=14+16=30
x=30:3
x=10
AP
1
14 tháng 3 2023
`x(x+3) - (2x-1) . (x+3) = 0`
`<=>(x+3)(x-2x+1)=0`
`<=>(x+3)(-x+1)=0`
`** x+3=0`
`<=>x=-3`
`** -x+1=0`
`<=>x=1`
`x(x-3) - 5 (x-3) = 0`
`<=>(x-3)(x-5)=0`
`** x-3=0`
`<=>x=3`
`** x-5=0`
`<=>x=5`
`3x + 12 = 0`
`<=>3x=-12`
`<=> x=-4`
`2x (x-2) + 5 (x-2) = 0`
`<=>(x-2)(2x+5)=0`
`** x-2=0`
`<=>x=2`
`** 2x+5=0`
`<=> x= -5/2`
KV
16 tháng 12 2023
Bài 1
a) (3x - 2⁴).7³ = 2.7⁴
3x - 16 = 2.7⁴ : 7³
3x - 16 = 2.7
3x - 16 = 14
3x = 14 + 16
3x =30
x = 30 : 3
x = 10
b) x - [42 + (-28)] = -8
x - 14 = -8
x = -8 + 14
x = 6
c) 4 - 7.x = x - (13 - 4)
x + 7x = 4 + 9
8x = 13
x = 13/8
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
16 tháng 12 2023
Bài 1: a, (3\(x\) - 24 ).73 = 2.74
3\(x\) - 24 = 2.74 : 73
3\(x\) - 16 = 14
3\(x\) = 30
\(x\) = 10
b, \(x\) - (42 + (-28) ] = -8
\(x\) - 14 = -8
\(x\) = 14 - 8
\(x\) = 6
c, 4 - (7\(x\)) = \(x\) - (13 - 4)
4 - 7\(x\) = \(x\) - 9
\(x\) + 7\(x\) = 4 + 9
8\(x\) = 13
\(x\) = \(\dfrac{13}{8}\)
TL
2
24 tháng 1 2021
(4x - 3)2 - (2x + 1)2 = 0
\(\Leftrightarrow\) (4x - 3 - 2x - 1)(4x - 3 + 2x + 1) = 0
\(\Leftrightarrow\) (2x - 4)(6x - 2) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ...
3x - 12 - 5x(x - 4) = 0
\(\Leftrightarrow\) 3x - 12 - 5x2 + 20x = 0
\(\Leftrightarrow\) -5x2 + 23x - 12 = 0
\(\Leftrightarrow\) 5x2 - 23x + 12 = 0
\(\Leftrightarrow\) 5x2 - 20x - 3x + 12 = 0
\(\Leftrightarrow\) 5x(x - 4) - 3(x - 4) = 0
\(\Leftrightarrow\) (x - 4)(5x - 3) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-4=0\\5x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy ...
(8x + 2)(x2 + 5)(x2 - 4) = 0
\(\Leftrightarrow\) (8x + 2)(x2 + 5)(x - 2)(x + 2) = 0
Vì x2 \(\ge\) 0 \(\forall\) x nên x2 + 5 > 0 \(\forall\) x
\(\Rightarrow\) (8x + 2)(x - 2)(x + 2) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}8x+2=0\\x-2=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=2\\x=-2\end{matrix}\right.\)
Vậy ...
Chúc bn học tốt!
24 tháng 1 2021
a) Ta có: \(\left(4x-3\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3-2x-1\right)\left(4x-3+2x+1\right)=0\)
\(\Leftrightarrow\left(2x-4\right)\left(6x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{2;\dfrac{1}{3}\right\}\)
b) Ta có: \(3x-12-5x\left(x-4\right)=0\)
\(\Leftrightarrow3\left(x-4\right)-5x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(3-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{4;\dfrac{3}{5}\right\}\)
c) Ta có: \(\left(8x+2\right)\left(x^2+5\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow2\left(4x+1\right)\left(x^2+5\right)\left(x-2\right)\left(x+2\right)=0\)
mà \(2>0\)
và \(x^2+5>0\forall x\)
nên \(\left(4x+1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+1=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-1\\x=2\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{4};2;-2\right\}\)
2x-10=0
<=> 2x=10
<=> x=5
7x-28=0
<=> 7x=28
<=> x=4
3x-7=14
<=> 3x=21
<=> x=7
44-3(x+2)=2.7
<=> 44-3(x+2)=14
<=> 3(x+2)=30
<=> x+2=10
<=> x=8
2x - 10 = 0
2x = 0 + 10
2x = 10
x = 10 : 2
x = 5
7x - 28 =0
7x = 0 + 28
7x = 28
x = 28 : 7
x = 4
3x -7 = 14
3x = 14 + 7
3x = 21
x = 21 : 3
x = 7
44 - 3(x + 2 ) = 2.7
44 - 3(x + 2 ) =14
3(x+2)=44-14
3(x+2) =30
x+2 = 30 : 3
x+2 = 10
x = 10 - 2
x = 8
tk cj nha