tìm X (-x-6).(x+9)=0 Giúp mình với
.là nhân nha
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a) \(\left(x^2-9\right)\cdot\left(4^x-16\right)=0\)
\(\Rightarrow x^2-9=0\)hoặc \(4^x-16=0\)
\(x^2=9\) \(4^x=16\)
\(x^2=\left(\pm3\right)^2\) \(4^x=4^2\)
\(\Rightarrow x=\pm3\)hoặc \(x=2\)
b) \(5^x+5^{x+2}=650\)
\(\Rightarrow5^x+5^x\cdot25=650\)
\(\Rightarrow5^x\cdot\left(1+25\right)=650\)
\(\Rightarrow5^x\cdot26=650\)
\(\Rightarrow5^x=650\div26=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)Vậy \(x=2\)
c) \(2^{x+2}-2^x=96\)
\(2^x\cdot4-2^x=96\)
\(2^x\cdot\left(4-1\right)=96\)
\(2^x\cdot3=96\)
\(2^x=96\div3=32\)
\(2^x=2^5\)Vậy \(x=5\)
\(9x-5x+3x=56\)
<=> \(7x=56\)
<=> \(x=8\)
Vậy...
\(9x-x+3x=165\)
<=> \(11x=165\)
<=> \(x=15\)
Vậy....
\(7x+8x-x=168\)
<=> \(14x=168\)
<=> \(x=12\)
Vậy...
\(\frac{5}{8}x-\frac{1}{3}x-\frac{1}{6}x=15\)
<=> \(\frac{1}{8}x=15\)
<=> \(x=120\)
Vậy...
Dấu . là dấu nhân nha
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a, 9 . x - 5 . x + 3 . x = 56
x . (9 - 5 + 3) = 56
x . 7 = 56
x = 56 : 7
x = 8
b, 9 . x - x + 3 . x = 165
x . (9 - 1 + 3) = 165
x . 11 = 165
x = 165 : 11
x = 15
c, 7 . x + 8 . x - x = 168
x . (7 + 8 - 1) = 168
x . 14 = 168
x = 168 : 14
x = 12
d, 5/8 . x - 1/3 . x - 1/6 . x = 15
x . (5/8 - 1/3 - 1/6) = 15
x . 1/8 = 15
x = 15 : 1/8
x = 120
\(\left(2.x-4\right).\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2.x-4=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}\)
\(\orbr{\begin{cases}2X-4=0\\X-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}X=2\\X=1\end{cases}}\)
\(1.x^2+11x=0\)
\(\Leftrightarrow x\left(x+11\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+11=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-11\end{cases}}\)
\(2.\left(x^2-1\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+9\right)\left(x-9\right)=0\)
chia thành 4 TH :
\(TH1:X-1=0\)
\(\Leftrightarrow x=1\)
\(TH2:x+1=0\)
\(\Leftrightarrow x=-1\)
\(TH3:X+9=0\)
\(\Leftrightarrow X=-9\)
\(TH4:x-9=0\)
\(\Leftrightarrow x=9\)
Kết luận ....
\(3.\left(\left|x+1\right|-5\right)\left(x^2-9\right)\)
\(\Leftrightarrow\left(\left|x+1\right|-5\right)\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}\left|x+1\right|-5=0\\x-3=0\\x+3=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left|x+1\right|=5\\x=3\\x=-3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+1=+_-5\Leftrightarrow x+1=5,x+1=-5\Leftrightarrow x=4,x=-6\\x=3x\\x=-3\end{cases}}\)
kết luận x=.....
\(4.\left(3x-16\right)⋮\left(x+2\right)\)
\(\Leftrightarrow\left(3x+6\right)-22\)
\(\Leftrightarrow3\left(x+2\right)-22⋮\left(x+2\right)\)
Vì\(\left(x+2\right)⋮\left(x+2\right)\)
\(\Rightarrow\left(3x-16\right)⋮\left(x+2\right)\)
Kết luận x=.....
a, => x=0 hoặc x+5=0
=> x=0 hoặc x=-5
b, => 2x-6=0 hoặc -4x-8=0
=> x=3 hoặc x=-2
Tk mk nha
a.x.(x+5)=0 <=>\(\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
b.(2x-6)(-4x-8)=0 <=> \(\orbr{\begin{cases}2x-6=0\\-4x-8=0\end{cases}}\) <=>\(\orbr{\begin{cases}2x=6\\-4x=8\end{cases}}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
Bài 1:
a) \(x^3-16x=x\left(x-4\right)\left(x+4\right)\)
b) \(3x^2+3y^2-6xy-12=3\left(x^2-2xy+y^2-4\right)=3\left(x-y-2\right)\left(x-y+2\right)\)
c) \(x^2+6x+5=\left(x+1\right)\left(x+5\right)\)
d) \(x^4+x^3+2x^2+x+1=\left(x^2+x+1\right)\left(x^2+1\right)\)
Bài 2:
a) Ta có: \(\left(x+6\right)^2=144\)
\(\Leftrightarrow\left[{}\begin{matrix}x+6=12\\x+6=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-18\end{matrix}\right.\)
b) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
c) Ta có: \(2x^2-x-6=0\)
\(\Leftrightarrow2x^2-4x+3x-6=0\)
\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(6\cdot x-5=613\)
\(6\cdot x=613+5\)
\(6\cdot x=618\)
\(x=618\div6\)
\(x=103\)
Vậy \(x=103\)
\(12\cdot x+3\cdot x=30\)
\(x\cdot\left(12+3\right)=30\)
\(x\cdot15=30\)
\(x=30\div15\)
\(x=2\)
Vậy \(x=2\)
\(125-25\cdot\left(x-1\right)=100\)
\(25\cdot\left(x-1\right)=125-100\)
\(25\cdot\left(x-1\right)=25\)
\(x-1=25\div25\)
\(x-1=1\)
\(x=1+1\)
\(x=2\)
Vậy \(x=2\)
\(\left(x-2\right)\cdot\left(x-14\right)=0\)
\(\Rightarrow\) \(x-2=0\) hoặc \(x-14=0\)
TH1: \(x-2=0\) TH2: \(x-14=0\)
\(x=0+2\) \(x=0+14\)
\(x=2\) \(x=14\)
Vậy \(x=2\) hoặc \(x=14\)
\(128-3\cdot\left(x+4\right)=23\)
\(3\cdot\left(x+4\right)=128-23\)
\(3\cdot\left(x+4\right)=105\)
\(x+4=105\div3\)
\(x+4=35\)
\(x=35-4\)
\(x=31\)
Vậy \(x=31\)
12.x+3.x=30
x.(12+3)=30
x.15=30
x =30:15
x =2
125-25.(x-1)=100
25.(x-1)=125-100
25.(x-1)=25
x-1=25:25
x-1=1
x =1+1
x=2
(x-2).(x-14)=0
x=14
128-3.(x+4)=23
3.(x+4)=128-23
3.(x+4)=105
x+4=105:3
x+4=35
x = 35+4
x =39
\(\text{a ) ( x - 6 ) . ( 7 - x ) = 0 }\)
\(\Leftrightarrow\text{13x-x^2 -42= 0 }\)
\(\Leftrightarrow x\left(13-x\right)-42=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\13-x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=13\end{cases}}\)
\(\left(-x-6\right)\left(x+9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}-6-x=0\\x+9=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-6-0=-6\\x=0-9=-9\end{cases}}\)
Vậy x=-6 hoặc x=-9
Từ biểu thức, suy ra:
-x-6=0 hoặc x+9=0
Mà -x-6=0(vô lý)
Suy ra để (-x-6)(x+9)=0 thì x+9=0
<=> x=-9
Vậy x=-9