Tim x
A 2/3 +1/3 ÷ x=1
b165 - (7÷x + 3) × 15 =15
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a) ( 2x - 5 ) - 3 = 2
2x - 5 = 2 + 3
2x - 5 = 5
2x = 5 + 5
2x = 10
x = 10 : 2
x = 5
Vậy x = 5
a,(2.x-5)-3=2
<=> 2x-5-3=2
<=> 2x-5-3-2=0
<=> 2x-10=0
<=> 2(x-5)=0
<=> x-5=0
<=> x=5
Đs: x=5
b,(3.x+1)+(x+1)=4
<=> 3x+1+x+1=4
<=> 3x+1+x+1-4=0
<=> 4x-2=0
<=> 2(x-1)=0
<=> x-1=0
<=> x=1
Đs:x=1
c,-12.(x-5)+7(3-x)=15
<=> -12x+60+21-3x=15
<=> -12x+60+21-7x-15=0
<=> -19x+66=0
<=> -19x+66=0
<=> -19x=-66
<=> x=66/19
Đs: 66/19
a)|x+15|=7
=>x+15=7 hoặc x+15=-7
x=7-15 x=-7-15
x=-8 x=-22
b)2*|x-5|=8
|x-5|=8/2
|x-5|=4
=>x-5=4 hoặc x-5=-4
x=4+5 x=-4+5
x=9 x=1
c)3*|2x-7|=9
|2x-7|=9/3
|2x-7|=3
=>2x-7=3 hoặc 2x-7=-3
2x=3+7 2x=-3+7
2x=10 2x=4
x=10/2 x=4/2
x=5 x=2
| x+15| =7
| x| = 7-15
=> x = 8 hoặc x= -8
2. |x-5| =8
|x-5| = 4
|x| = 9
=> x=9 hoặc x=-9
3. |2x-7|= 9
|2x| = 16
|x| = 8
=> x= 8 hoặc x= -8
Ta có:
A = 1 + 3 + 5 + 7 +... + 101
A = \(\frac{102.51}{2}=2601\)
M = 16 - 18 + 20 - 22 + 24 - 26 + .. + 64 - 66 + 68
M = ( 16 - 18 ) + ( 20 - 22 ) + ( 24 - 26 ) + ... + ( 64 - 66 ) + 68
M = (- 2 + - 2 + -2 + ... + - 2 ) + 68
M = 25/2 . ( - 2 ) + 68
M = -25 + 68
M = 43
H = ( 1 + 2 + 3 +...+ 99 ) x ( 13 x 15 - 12 x 15 - 15 )
H = ( 1 + 2 + 3 +...+ 99 ) x { (13 - 12 - 1) x 15 }
H = ( 1 + 2 + 3 +...+ 99 ) x 0
H = 0
G = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - 11 - 12 + 13 + 14 - ... + 301 + 302
G = ( 1 + 2 ) + ( -3 - 4 ) + ( 5 + 6 ) + ( -7 - 8 ) + ( 9 + 10 ) + ( - 11 - 12 ) + ( 13 + 14 ) -...+ ( 301 + 302 )
G = ( 3 - 7 ) + ( 11 - 15 ) + ( 19 - 23 ) + 27 - ... + 603
G = -4 + - 4 + -4 + 27 - ... + 603
G = 75 x ( -4 ) + 603
G = -300 + 603
G = 303
2.
a) 1 + 2 + 3 + 4 +...+ 99 + 100 + 2 x X = 5052
= > \(\frac{100.101}{2}\)+ 2 x X = 5052
= > 5050 + 2 x X = 5052
= > 2X = 2
= > X = 1
1.
b) \(3^x+3^{x+2}=2430\)
\(\Rightarrow3^x.1+3^x.3^2=2430\)
\(\Rightarrow3^x.\left(1+3^2\right)=2430\)
\(\Rightarrow3^x.10=2430\)
\(\Rightarrow3^x=2430:10\)
\(\Rightarrow3^x=243\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=15\\2x-15=\pm1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=15:2\\2x-15=1\\2x-15=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\2x=16\\2x=14\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{15}{2};8;7\right\}.\)
Chúc bạn học tốt!
=> \(3x-15=0\)
=> \(3x=0+15\)
=> \(3x=15\)
=> \(x=15:3\)
=> \(x=5\)
\(\left(3x-5\right)^7=0\)
\(\Rightarrow3x-5=0\)
\(\Rightarrow3x=5\)
\(\Rightarrow x=\frac{5}{3}\)
A,
2/3+1/3:x=1
=>1/3:x=1-2/3
=>1/3:x=1/3
=>x=1/3:1/3
=>x=1
B,
165-(7:x+3)x15=15
=>(7:x+3)x15=165-15
=>(7:x+3)x15=150
=>7:x+3=150:15
=>7:x+3=10
=>7:x=10-3
=>7:x=7
=>x=7:7
=>x=1
\(\frac{2}{3}+\frac{1}{3}:x=1\)
\(\Rightarrow\frac{1}{3}:x=\frac{1}{3}\)
\(\Rightarrow x=1\)
\(165-\left(7:x+3\right)\times15=15\)
\(\Rightarrow\left(7:x+3\right)\times15=150\)
\(\Rightarrow7:x+3=10\)
\(\Rightarrow7:x=7\)
\(\Rightarrow x=1\)