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Từ 2(4x-3)-3(x+5)+4(x-10)= 5(x+2)
-> 8x - 6 - 3x - 15+4x - 40= 5x + 10 (mở ngoạc bạn nhé!)
-> 8x - 6x - 4x - 6 -15 -40 = 5x + 10
-> 9x + 61 = 5x +10
-> 9x - 5x = 10 - 61
-> 4x = -51
-> x = -12.75
Vậy x= -12.75 bạn nhé!
Chúc bạn học tôtr. Nhớ tick cho mk nha!
( 4x - 3 ) - ( x + 5 ) = 3 . ( 10 - x )
<=> 4x - 3 - x - 5 = 30 - 3x
=> 3x - 8 = 30 - 3x
=> 3x + 3x = 30 + 8
=> 6x = 38
=> x = 19/3
Vậy x = 19/3
(4x−3)−(x+5)=3(10−x)
⇒4x−3−x−5=30−3x
⇒3x−8=30−3x
⇒6x=38
⇒x=\(\dfrac{19}{3}\)
Vậy x= \(\dfrac{19}{3}\)
⇒ 8x - 6 - 3x - 15 + 4x - 40 = 5x + 10
⇒ 9x - 61 = 5x + 10
⇒ 4x = 71
⇒ x = 17,7
\(\Rightarrow\) 8x - 6 - 3x - 15 + 4x - 40 = 5x + 10
\(\Rightarrow\) 9x - 61 = 5x + 10
\(\Rightarrow\) 4x = 71
\(\Rightarrow\) x = 17,75
4x-3-x-5=30-3x
3x-3-5=30-3x
3x-8=30-3x
3x+3x=30+8
6x=38
x=38:6
x=38/6
(4x-3)-(x+5)=3(10-x)
<=>4x-3-x-5=30-3x
<=>4x-3-x-5-30+3x=0
<=>6x-38=0
<=>6x=38
<=>x=\(\frac{19}{3}\)
\(\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\)
<=> \(4x-3-x-5=30-3x\)
<=>\(3x-8-30+3x=0\)
<=> \(0x=38\)
<=>x thuộc rỗng, không tìm được giá trị x trong trường hợp này
(4x−3)−(x+5)=3(10−x)
⇒4x−3−x−5=30−3x
⇒3x−8=30−3x
⇒6x=38
⇒x=19/3
Vậy x=19/3
\(a,\left(x-8\right)\left(x^3+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
\(b,\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\\ \Leftrightarrow4x-3-x-5=30-3x\\ \Leftrightarrow3x-8-30+3x=0\\ \Leftrightarrow6x-38=0\\ \Leftrightarrow x=\dfrac{19}{3}\)
TK
`a.(x-8)(x+8)=0`
`⇔³{x−8=0x³+8=2 `
`⇔³³{x=8x³=−2³ `
`⇔{x=8x=−2`
Vậy ` x = 8;-2`
`b. ( 4 x − 3 ) − ( x + 5 ) = 3 . ( 10 − x )`
`⇔ 4 x − 3 − x − 5 = 30 − 3 x`
`⇔ 3 x − 8 = 30 − 3 x`
`⇔ 3 x + 3 x = 30 + 8`
`⇔ 6 x = 38`
`⇔ x = 19/ 3`
Vậy ` x = 19/ 3`
\(a.\)
\(\left(x-8\right)\left(x^3+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
\(S=\left\{8,-2\right\}\)
\(b.\)
\(\left(4x-3\right)-\left(x+5\right)=3\cdot\left(10-x\right)\)
\(\Leftrightarrow4x-3-x-5-30+3x=0\)
\(\Leftrightarrow6x-38=0\)
\(\Leftrightarrow x=\dfrac{38}{6}\)
\(S=\left\{\dfrac{38}{6}\right\}\)
a) (x - 8 )( x3 + 8) = 0
\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x^3=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
b)(4x - 3) – ( x + 5) = 3(10 - x)
\(\Leftrightarrow4x-3-x-5=30-3x\)
\(\Leftrightarrow3x-8=30-3x\)
\(\Leftrightarrow3x-8-30+3x=0\)
\(\Leftrightarrow6x-38=0\)
\(\Leftrightarrow x=\dfrac{19}{3}\)
2 PHẦN 7 CUA 491 LÀ bao nhiêu
(4x - 3) - (x + 5) = 3(10-x)
4x - 3 - x - 5 = 30 - 3x
(4x - x) - (3+5) = 30 - 3x
3x - 8 = 30 - 3x
3x + 3x = 30 + 8
6x = 38
x = 38 : 6
x = 19/3
Vậy x = 19/3