(2x-15)5=(2x-15)3
giups minh voi
K
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16 tháng 2 2021
1) x = 23-20
=>x= 3
2)32-x=18
=>x=32-18
=>x=14
3)/2x+1/=8
=>2x+1=8
hoặc 2x+1=-8
=>x=7/2
hoặc x=-9/2
16 tháng 2 2021
1/ Ta có: \(x+20=-\left(-23\right)\)
\(\Leftrightarrow x+20=23\)
\(\Leftrightarrow x=3\)
Vậy...
2/ Ta có: \(15-x+17=-\left(-6\right)+\left|-12\right|\)
\(\Leftrightarrow-x+32=18\)
\(\Leftrightarrow x=32-18=14\)
Vậy...
3/ Ta có: \(3-\left|2x+1\right|=\left(-5\right)\)
\(\Leftrightarrow\left|2x+1\right|=3-\left(-5\right)\)
\(\Leftrightarrow\left|2x+1\right|=8\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)
Vậy...
YT
1
5 tháng 4 2020
7) 2x – (–17) = 15
\(2x=-2\)
\(x=-1\)
8) |x – 2| = 3.
\(\Rightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}}\)
9) | x – 3| –7 = 13
\(\text{ | x – 3|}=20\)
\(\Rightarrow\orbr{\begin{cases}x-3=20\\x-3=-20\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=23\\x=-17\end{cases}}\)
học tốt ạ
KN
2
NT
18 tháng 10 2017
\(a,\left(27x+6\right):3-11=9\)
\(\Rightarrow\left(27x-6\right):3=20\)\(\Rightarrow27x-6=60\)\(\Rightarrow27x=66\)\(\Rightarrow x=\frac{22}{9}\)
\(b,\left(15-6x\right).3^5=3^6\)
\(\Rightarrow15-6x=3\)\(\Rightarrow6x=12\)\(\Rightarrow x=2\)
\(c,\left(2x-6\right).4^7=4^9\)
\(\Rightarrow2x-6=16\)\(\Rightarrow2x=22\)\(\Rightarrow x=11\)
ON
18 tháng 10 2017
Bài 1: Tìm x
a,(27.x+6):3-11=9
﴾ 27x + 6﴿ : 3 ‐ 11 = 9
﴾ 27x + 6﴿ : 3 = 20
27x + 6 = 60
27x = 54
x = 54 : 27
x = 2
Vậy x = 2
b,( 15-6x ) . 35=36
( 15 - 6x ) = 36 : 35
15 - 6x = 3
6x = 15 - 3
6x = 12
x = 12 : 6
x = 2
Vậy x = 2
c,( 2x-6 ) . 47=49
2x - 6 = 49 : 47
2x - 6 = 16
2x = 16 + 6
2x = 22
x = 22 : 2
x = 11
Vậy x = 11
HQ
10 tháng 2 2021
\(\Leftrightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-15=\pm1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
Vậy.........
10 tháng 2 2021
Ta có: \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Leftrightarrow\left(2x-15\right)^3\cdot\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(2x-15-1\right)\left(2x-15+1\right)=0\)
\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(2x-16\right)\left(2x-14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=15\\2x=16\\2x=14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{15}{2};8;7\right\}\)
27 tháng 12 2022
Tham khảo :
`(2x - 15)^5 = (2x - 15)^3`
`=> (2x - 15)^5 : (2x - 15)^3 = 1`
`=> (2x - 15)^2 = 1`
`=> (2x - 15)^2 = 1^2`
`=>` $\left[\begin{matrix} 2x-15=1\\ 2x-15=-1\end{matrix}\right.$
`=>` $\left[\begin{matrix} 2x=1 + 15\\ 2x=-1 + 15\end{matrix}\right.$
`=>` $\left[\begin{matrix} 2x=16\\ 2x=14\end{matrix}\right.$
`=>` $\left[\begin{matrix} x=8\\ x=7\end{matrix}\right.$
`=> x in {7;8}`
H
2 tháng 3 2023
`(2x-15)^5 =(2x-15)^3`
`=>(2x-15)^5 -(2x-15)^3=0`
`=> (2x-15)^3 [(2x-15)^2 -1]=0`
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-15=0\\2x-15=1\\2x-15=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=15\\2x=16\\2x=14\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
9 tháng 4 2023
\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(=>\left[{}\begin{matrix}2x-15=0\\2x-15=1\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}2x=15\\2x=16\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\end{matrix}\right.\)
\(=>x\in\left\{\dfrac{15}{2};8\right\}\)
28 tháng 11 2015
http://olm.vn/hoi-dap/question/299944.html
Chưa tích cho tớ đâu nha!
28 tháng 11 2015
\(\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=\frac{x-y}{2-\frac{3}{2}}=\frac{15}{\frac{1}{2}}=30\Leftrightarrow x=2.30=60;y=\frac{3}{2}.30=45;z=\frac{4}{3}.30=40\)
b)\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6}=\frac{2x+1+3y-2-2x-3y+1}{5+7-6}=0\)
2x+1=0=> x =-1/2
3y-2=0 => y=2/3
\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-15\right)^3.\left(2x-16\right).\left(2x-14\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=15\\2x=16\\x=14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
`(2x-15)^5 =(2x-15)^3`
`<=> (2x-15)^5 -(2x-15)^3 =0`
`<=>(2x-15)^3*[(2x-15)^2-1]=0`
`@TH1 : (2x-15)^3=0 => 2x-15 =0`
`=> 2x=15 => x=15/2`
`@TH2 : (2x-15)^2 -1 =0`
`=> [(2x-15=1),(2x-15=-1):}`
`=> [(2x=1+15=16),(2x=-1+15=14):}`
`=>[(x=8),(x=7):}`
Vậy `x in {15/2;8;7}`