\(a)x^{15}=x\times1000^0\)
\(b)x^{15}=x\times x^0\)
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a. 395 x 15 + 85 x 395= 395x ( 15+85 )
= 395 x 100
= 39500
b. 289 x 47 -289 x 17 = 289 x (47-17)
= 289 x 20
= 5780
c. 2051 x ( 15 - 9) = 2051 x 6
= 12306
d. 2912 x 94 - 2912 x 44 = 2912x ( 94-44 )
= 2912 x 50
= 145600
nhé !
a. 395 x 15 + 85 x 195
= 395 x (15 + 85)
= 395 x 100
= 39500
b. 289 x 47 - 289 x 17
= 289 x (47 - 17)
= 289 x 30
= 8670
c. 2051 x (15 - 9)
= 2051 x 6
= 12306
d. 2912 x 94 - 2912 x 44
= 2912 x (94 - 44)
= 2912 x 50
= 145600
k nha!
\(x=\left\{x\in N;2x\left(2x-6\right)\left(3x-15\right)\right\}=0\)
\(\Leftrightarrow2x=0\Rightarrow x=0\)
\(2x-6=0\Rightarrow2x=6\Rightarrow x=3\)
\(3x-15=0\Rightarrow3x=15\Rightarrow x=5\)
\(\Rightarrow x=\left\{0;3;5\right\}\)
a) 0
b) x - 15 =75
x=15+75=
x=90
c) ( x - 7 ) = 114+0
x - 7 = 114
x=114+7
x=121
d) ( x - 15 ) = 0:2015
x - 15 =0
x=15+0
x=15
e)( x - 15 ) =0:13
x - 15 =0
x=15+0
x+15
Bài 3: Tìm x, biết:
a) \(16x^2-\left(4x-5\right)^2=15\)
\(\Leftrightarrow16x^2-16x^2+40x-25-15=0\)
\(\Leftrightarrow40x-40=0\)
\(\Leftrightarrow4x=40\)
\(\Leftrightarrow x=10\)
Vậy x = 10
b) \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)
\(\Leftrightarrow\left(2x+3\right)^2-4\left(x^2-1\right)=49\)
\(\Leftrightarrow4x^2+12x+9-4x^2+4-49=0\)
\(\Leftrightarrow12x-36=0\)
\(\Leftrightarrow12x=36\)
\(\Leftrightarrow x=3\)
Vậy x = 3
c) \(\left(2x+1\right)\left(1-2x\right)+\left(1-2x\right)^2=18\)
\(\Leftrightarrow\left(1-2x\right)\left(2x+1+1-2x\right)=18\)
\(\Leftrightarrow2\left(1-2x\right)=18\)
\(\Leftrightarrow2-4x=18\)
\(\Leftrightarrow4x=-16\)
\(\Leftrightarrow x=-4\)
Vậy x =-4
d) \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)
\(\Leftrightarrow2x^2+4x+2-x^2+9-x^2+8x-16=0\)
\(\Leftrightarrow12x-5=0\)
\(\Leftrightarrow12x=5\)
\(\Leftrightarrow x=\frac{5}{12}\)
Vậy \(x=\frac{5}{12}\)
e) \(\left(x-5\right)^2-x\left(x-4\right)=9\)
\(\Leftrightarrow x^2-10x+25-x^2+4x=9\)
\(\Leftrightarrow25-6x=9\)
\(\Leftrightarrow6x=16\)
\(\Leftrightarrow x=\frac{8}{3}\)
Vậy \(x=\frac{8}{3}\)
f) \(\left(x-5\right)^2+\left(x-4\right)\left(1-x\right)=0\)
\(\Leftrightarrow x^2-10x+25+x-x^2-4+4x=0\)
\(\Leftrightarrow21-5x=0\)
\(\Leftrightarrow5x=21\)
\(\Leftrightarrow x=\frac{21}{5}\)
Vậy \(x=\frac{21}{5}\)
a) x + 123 = 234
=> x = 234 - 123
=> x = 111
b) 205 – x = 15
=> x = 205 - 15
=> x = 19
c) 2x – 10 = 60
=> 2x = 70
=> x = 35
d) x : 34 = 10
=> x = 10 * 34
=> x = 340
e) 12 + (5 + x) = 20
=> 5 + x = 8
=> x = 3
f) 130 – (100 + x) = 25
=> 100 + x = 130 - 25
=> x = 105 - 100
=> x = 5
g) 15x – 133 = 17
=> 15x = 150
=> x = 10
h) 175 + (30 – x) = 200
=> 30 – x = 25
=> x = 5
A.\(\left(x-15\right).15=0\)
\(x-15=0:15\)
\(x-15=0\)
\(x=15+0\)
\(x=15\)
B.\(32\left(x-10\right)=32\)
\(x-10=32:32\)
\(x-10=1\)
\(x=10+1\)
\(x=11\)
`a) `
`(x-15)xx15=0`
`<=> x-15 = 0 : 15`
`<=> x-15 = 0`
`<=> x = 0 + 15`
`<=> x =15`
`b)`
`32.(x-10)=32`
`<=> x - 10 = 32:32`
`<=>x-10=1`
`<=> x = 1+10`
`<=> x =11`
`c)`
`(x-5).(x-7)=0`
`<=>` \(\left[ \begin{array}{l}x-5 = 0\\x-7=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=5\\x=7\end{array} \right.\)
`d)`
`(x-35)xx35=35`
`<=> x - 35 = 35:35`
`<=> x - 35 = 1`
`<=> x = 1+35`
`<=> x = 36`
a) x^15= x.1000^0 b ) x^15 = x.x^0
=>x^15= x . 1 =>x^15 = x.1
=> x^15 = x => x^15 = x
=> x = 1 => x = 1