15(x+9)(x-3)(x+21)=0
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\(15\left(x+9\right)\left(x-3\right)\left(x+21\right)=0\\ \Leftrightarrow\left(x+9\right)\left(x-3\right)\left(x+21\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+9=0\\x-3=0\\x+21=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=3\\x=-21\end{matrix}\right.\)
VẬy...
\(15\left(x+9\right)\left(x-3\right)\left(x+21\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+9=0\\x-3=0\\x+21=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-9\\x=3\\x=-21\end{cases}}\)
Vậy ...................
vì 15(x+9)(x-3)(x+21)=0 nên một trong ba số x+9;x-3,x+21 sẽ bằng 0 CÓ 3TH:TH1: x+9=0 x=0-9=-9 TH2: x-3=0 x=0+3=3 TH3:x+21=0 x=0-21=-21 Vậy x =-9;3;-21. Nếu bạn ko tin thì cứ bấm máy tính nhé
a: =>4/x=y/21=4/7
=>x=7; y=21*4/7=12
b: x/7=9/y
=>xy=63
mà x>y
nên \(\left(x,y\right)\in\left\{\left(63;1\right);\left(21;3\right);\left(9;7\right);\left(-7;-9\right);\left(-3;-21\right);\left(-1;-63\right)\right\}\)
c: x/15=3/y
=>xy=45
mà x<y<0
nên \(\left(x,y\right)\in\left\{\left(-45;-1\right);\left(-15;-3\right);\left(-9;-5\right)\right\}\)
d: x/y=21/28=3/4
=>x/3=y/4=k
=>x=3k; y=4k(k\(\in Z\))
1) x - 2 = -6
x = -6 + 2
x = -4
2) -5 . x - ( -3 ) =13
-5 . x = 13 + ( -3 )
-5 . x = 10
x = 10 : ( -5 )
x = -2
a: =>4/x=y/-21=4/7
=>x=7; y=-12
b: =>xy=63
mà x>y
nên \(\left(x,y\right)\in\left\{\left(9;7\right);\left(21;3\right);\left(63;1\right);\left(-7;-9\right);\left(-3;-21\right);\left(-1;-63\right)\right\}\)
c: =>xy=45
mà x<y<0
nên \(\left(x,y\right)\in\left\{\left(-45;-1\right);\left(-15;-3\right);\left(-9;-5\right)\right\}\)
1) 5.( x - 6 ) - 2.( x + 9 ) = 21
5x - 30 - 2x - 18 = 21
3x - 48 = 21
3x = 21 + 48
3x = 69
x = 23
2) 2.( x + 3 ) + 3.( x + 1 ) = 15 - ( - 9 )
2x + 6 + 3x + 3 = 24
5x + 9 = 24
5x = 24 - 9
5x = 15
x = 3
3) ( - x + 5 ).(3 - x ) = 0
=> - x + 5 = 0 hoặc 3 - x = 0
=> x = 5 hoặc x = 3
4) ( x - 12 ) - 15 = ( 20 - 7 ) - ( 18 + x )
x - 12 - 15 = 13 - 18 - x
x - 27 = - 5 - x
x + x = - 5 + 27
2x = 22
x = 11
5) x - ( 17 - 8 ) = 5 + ( 10 - 3x )
x - 9 = 5 + 10 - 3x
x + 3x = 15 + 9
4x = 24
x = 6
1) (x-17)*17=0
x-17=0
x=17
2) 32(x-11)=32
x-11=1
x=12
3) (x-25)-75=0
x-25=75
x=3
4) 575-(6*x+70)=445
6*x+70=130
6*x=60
x=10
5) 315+(125-x)=435
125-x=120
x=5
6) x-105:21=15
x-5=15
x=20
7) (x-105):21=15
x-105=315
x=420
8) (x-38):19=12
x-38=228
x=266
9) (x-15)*(x-19)=0
=>x-15=0 hoặc x-19=0
x=15 hoặc x=19
10) 96-3(x+1)=42
3(x+1)=54
x+1=18
x=17
a) ( x - 17 ) . 17 = 0
=> x - 17 = 0 : 17
=> x - 17 = 0
=> x = 0 + 17
=> x = 17
b) 32 . ( x - 11 ) = 32
=> x - 11 = 32 : 32
=> x - 11 = 1
=> x = 1 + 11
=> x = 12
c) ( x - 25 ) - 75 = 0
=> x - 25 = 0 + 75
=> x - 25 = 75
=> x = 75 + 25
=> x = 100
d) 575 - ( 6 . x + 70 ) = 445
=> 6 . x + 70 = 575 - 445
=> 6 . x + 70 = 130
=> 6 . x = 130 - 70
=> 6 . x = 60
=> x = 60 : 6
=> x = 10
e) 315 + ( 125 - x ) = 435
=> 125 - x = 435 - 315
=> 125 - x = 120
=> x = 125 - 120
=> x = 5
f) x - 105 : 21 = 15
=> x - 5 = 15
=> x = 15 + 5
=> x = 20
g) ( x - 105 ) : 21 = 15
=> x -105 = 15 . 21
=> x - 105 = 315
=> x = 315 + 105
=> x = 420
h) ( x - 38 ) : 19 = 12
=> x - 38 = 12 . 19
=> x - 38 = 228
=> x = 228 + 38
=> x = 266
i) ( x - 15 ) . ( x - 19 ) = 0
=> x - 15 = 0 => x = 0 + 15 => x = 15
x - 19 = 0 => x = 0 + 19 = x = 19
=> x = 15 ; 19
j) 96 - 3( x + 1 ) = 42
=> 3( x + 1 ) = 96 - 42
=> 3( x + 1 ) = 54
=> x + 1 = 54 : 3
=> x + 1 = 18
=> x = 18 - 1
=> x = 17
a: \(=\dfrac{3}{5}:\dfrac{7}{5}=\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{7}\)
b: \(=\dfrac{9}{17}\left(\dfrac{8}{5}-\dfrac{3}{5}\right)+\dfrac{8}{17}\)
=9/17+8/17=1
c: =>x-3/10=7/15*1/5=7/75
=>x=7/75+3/10=59/150
\(15\left(x+9\right)\left(x-3\right)\left(x+21\right)=0\)
\(\Leftrightarrow\left(15x+135\right)\left(x-3\right)\left(x+21\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}15x+135=0\\x-3=0\\x+21=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=3\\x=-21\end{matrix}\right.\)
Vậy ...................
15(x + 9)(x - 3)(x + 21) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-3=0\\x+21=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=3\\x=-21\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{-9;3;-21\right\}\)
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