K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NT
0
AH
Akai Haruma
Giáo viên
7 tháng 8 2021
Lời giải:
$x+\frac{2}{-15}=\frac{-5}{3}$
$x=\frac{-5}{3}-\frac{2}{-15}=\frac{-5}{3}+\frac{2}{15}$
$x=\frac{-23}{15}$
H9
HT.Phong (9A5)
CTVHS
27 tháng 6 2023
Bài 1:
a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)
\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)
\(=10x^2+10x^2\)
\(=20x^2\)
b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)
\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)
\(=-4x^4+9x^3+4x^2-44x\)
NH
1
H9
HT.Phong (9A5)
CTVHS
22 tháng 4 2023
a) \(\dfrac{3,5}{15}=\dfrac{-2}{x}\)
\(\Rightarrow x=\dfrac{15.-2}{3,5}\)
\(\Rightarrow x=-8,57\)
b) \(2\left(3x-2\right)-3\left(x-2\right)-=-1\)
\(\Rightarrow6x-4-3x+6=-1\)
\(\Rightarrow6x-3x=-1+4-6\)
\(\Rightarrow3x=-3\)
\(\Rightarrow x=-\dfrac{3}{3}=-1\)
OD
1
OY
7 tháng 8 2021
a) \(\dfrac{x.2}{-15}=\dfrac{-5}{3}\)
\(\dfrac{x.2}{-15}=\dfrac{25}{-15}\)
x.2=25
x=12,5
b) \(\dfrac{x-1}{-12}=\dfrac{-3}{x-1}\)
(x-1)2=-3.(-12)
(x-1)2=36
⇒(x-1)2\(\Rightarrow\left[{}\begin{matrix}x-1=6\\x-1=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)
YT
3
H9
HT.Phong (9A5)
CTVHS
9 tháng 8 2023
a) \(1=\left(2x+0,5\right)^{600}\)
\(\Rightarrow1^{600}=\left(2x+0,5\right)^{600}\)
\(\Rightarrow\left[{}\begin{matrix}2x+0,5=1\\2x+0,5=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=0,5\\2x=-1,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0,25\\x=-0,75\end{matrix}\right.\)
b) \(\left(x-0,125\right)^2=0,25\)
\(\Rightarrow\left(x-0,125\right)^2=0,5^2\)
\(\Rightarrow\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)
c) \(\left(x-3\right)^{11}=\left(x-3\right)^{41}\)
\(\Rightarrow\left(x-3\right)^{11}-\left(x-3\right)^{41}=0\)
\(\Rightarrow\left(x-3\right)^{11}\left[1-\left(x-3\right)^{30}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-3=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
9 tháng 8 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
`1 = (2x + 0,5)^600`
`=> (2x+0,5)^600 = (+-1)^600`
`=> \text {TH1: } 2x + 0,5 = 1`
`=> 2x = 1 - 0,5`
`=> 2x = 0,5`
`=> x = 0,5 \div 2`
`=> x = 0,25`
`\text {TH2: } 2x + 0,5 = -1`
`=> 2x = -1 - 0,5`
`=> 2x = -1,5`
`=> x = -1,5 \div 2`
`=> x = -0,75`
Vậy, `x \in {-0,75; 0,25}.`
`b)`
`(x - 0,125)^2 = 0,25`
`=> (x - 0,125)^2 = (+-0,5)^2`
`=> `\(\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0,5+0,125\\x=-0,5+0,125\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)
Vậy, `x \in {-0,375; 0,625}.`
`c)`
`(x - 3)^11 = (x - 3)^41`
`=> (x - 3)^11 - (x - 3)^41 = 0`
`=> (x - 3)^11 * [ 1 - (x - 3)^30] = 0`
`=>`\(\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\1-\left(x-3\right)^{30}=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Vậy, `x \in {3; 4}.`
SV
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
22 tháng 12 2022
a.\(\dfrac{1}{3}\) + x = \(\dfrac{5}{6}\)
x = \(\dfrac{5}{6}\) - \(\dfrac{1}{3}\)
x = \(\dfrac{1}{2}\)
b. | x-1| - \(\dfrac{2}{5}\) = \(\dfrac{11}{10}\)
| x-1| = \(\dfrac{11}{10}\) + \(\dfrac{2}{5}\)
|x-1| = \(\dfrac{3}{2}\)
\(\left[{}\begin{matrix}x-1=\dfrac{3}{2}\\x-1=-\dfrac{3}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{2}+1\\x=-\dfrac{3}{2}+1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
c, \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = 1
\(\dfrac{2}{3}\) (\(\dfrac{x}{2}\) + 3) = 1 - \(\dfrac{1}{3}\)
\(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = \(\dfrac{2}{3}\)
\(\dfrac{x}{2}\) + 3 = 1
\(\dfrac{x}{2}\) = 1 - 3
\(\dfrac{x}{2}\) = -2
\(x\) = -4
d, \(\dfrac{x+2}{3}\) = \(\dfrac{27}{x+2}\)
(x+2)2 = 27.3
(x+2) =92
\(\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)
8 tháng 8 2023
a: =>2^x*4-2^x*3=32
=>2^x=32
=>x=5
b: =>(4x-3)^2-(4x-3)=0
=>(4x-3)(4x-3-1)=0
=>(4x-3)(4x-4)=0
=>x=3/4 hoặc x=1
c: =>7^2x+7^2x*7^3=344
=>7^2x=1
=>2x=0
=>x=0
d: =>(7x-3)^2012-(7x-3)^2010=0
=>(7x-3)^2010*[(7x-3)^2-1]=0
=>(7x-3)^2010*(7x-4)(7x-2)=0
=>x=2/7; x=4/7; x=3/7
e: =>(4x^2-3)^3=-8
=>4x^2-3=-2
=>4x^2=1
=>x^2=1/4
=>x=1/2 hoặc x=-1/2
T
8 tháng 8 2023
a) 2x(22 - 3) = 32
2x.1=25
=> x = 5
b) (4x - 3)2 = 4x -3
=> (4x - 3)2 - (4x - 3) = 0
(4x-3)[(4x - 3) - 1] = 0
(4x-3)(4x - 4)=0
\(\Rightarrow\left[{}\begin{matrix}4x-3=0\\4x-4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=1\end{matrix}\right.\)
c) 72x + 72x+3 = 344
=> 72x(1 + 73) =344
72x . 344 = 344
=> 2x = 0 => x = 0
d) (7x - 3)2012 = (3 - 7x)2010
=> (7x - 3)2012 - (7x - 3)2010 = 0
(7x - 3)2010 [(7x - 3)2 - 1] = 0
\(\Rightarrow\left[{}\begin{matrix}7x-3=0\\\left(7x-3\right)^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{7}\\7x=4\\7x=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{7}\\x=\dfrac{4}{7}\\x=\dfrac{2}{7}\end{matrix}\right.\)
e) (4x2 - 3)3 + 8 = 0
(4x2 - 3)3 = (-2)3
=> 4x2 - 3 = -2
4x2 = 1
x2 = 1/4
=> \(x=\pm\dfrac{1}{2}\)
hhhbnvnhhgfjk,bnbmh
1+1=2 2 nhân 2 bằng 4