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\(a,\left(-5\right).\left|x\right|=-75\)
\(\left|x\right|=\frac{-75}{-5}=15\)
\(\Rightarrow\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)
Vậy....
\(b,\left(-6\right)^3.x^2=-1944\)
\(-216.x^2=-1944\)
\(x^2=9\)
\(\Rightarrow x=\pm3\)
Vậy....
\(d,\left|9-x\right|=-7+64\)
\(\left|9-x\right|=57\)
\(\Rightarrow\orbr{\begin{cases}9-x=57\\9-x=-57\end{cases}\Rightarrow\orbr{\begin{cases}x=-48\\x=66\end{cases}}}\)
Vậy...
\(e,\left|x+101\right|-\left(-16\right)=\left(-43\right).\left(-5\right)\)
\(\left|x+101\right|+16=215\)
\(\left|x+101\right|=199\)
\(\Rightarrow\orbr{\begin{cases}x+101=199\\x+101=-199\end{cases}\Rightarrow\orbr{\begin{cases}x=98\\x=-300\end{cases}}}\)
Vậy..
hok tốt!!
a,\(\left(-5\right).\left|x\right|=-75\)
\(=>\left|x\right|=-75:\left(-5\right)=15\)
\(=>\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)
b,\(\left(-6\right)^3.x^2=-1944\)
\(=>\frac{1944}{216}=x^2\)
\(=>x=\sqrt{\frac{1944}{216}}=3\)
câu1
(3x-1).(1/2x5)=0
=>3x-1=0 hoặc 1/2x5=0
=>x=1/3 =>x=0
câu2
1/4+1/3 :(2x-1)=5
=> 1/3:(2x-1)=19/4
=>2x-1 =57/4
=>2x=61/4
=>x=61/8
còn hai câu sau bn ghi đề mik ko hỉu
1.
a)(3x-1)(1/2x5)=0
=>3x-1=0 hoặc 1/2x5=0
3x=0+1 x=0:1/2:5
x=1/3 x=0
Vậy x=1/3 hoặc x=0
b)1/4+1/3:(2x-1)=5
1/3:(2x-1)=5-1/4=20/4-1/4=19/4
2x-1=1/3:19/4=1/3*4/19=4/57
2x=4/57+1=4/57+57/57=61/57
x=61/57:2=61/57*1/2=61/114
Vậy x=61/114
c)(2x+2/5)2-9/25=0=02-9/25
=>2x+2/5=0
2x=0-2/5
x=-2/5:2=-2/5*1/2
x=-1/5
Vậy x=-1/5
d)(3x-1/2)3+1/9=0=03+1/9
=>3x-1/2=0
3x=0+1/2
x=1/2:3=1/2*1/3
x=1/6
Vậy x=1/6
a) \(x=\dfrac{25}{72}\)
b)\(x=-\dfrac{1}{4}\)
\(x=\dfrac{3}{2}\)
c)\(x=\dfrac{5}{4}\) hoặc
x \(=\dfrac{8}{5}\)
d và e chịu vì mk kg giỏi lắm về mũ
f)\(x=-2\)
G)\(x=-\dfrac{5}{12}\)
a)2^x + 1 . 2^2009 = 2^2010
=> 2^x + 1 + 2009 = 2^2010
=>2^x + 2010 = 2^2010
=>x + 2010 = 2010
=>x = 2010 - 2010 = 0
b)Chắc ý bạn là 6^17 : 6^15 + 44 : 11 đúng không?
Nếu thế thì mình sẽ giải như sau:
7x - 2x = 6^17 : 6^15 + 44 : 11
7x - 2x = 6^17 - 15 + 44 : 11
7x - 2x = 6^2 + 44 : 11
7x - 2x = 6^2 + 4
7x - 2x = 36 + 4
7x - 2x = 40
(7 - 2)x = 40
5x = 40
x = 40 : 5
x = 8
c)0 : x = 0
=>x ϵ N*
3^x = 9
3^x = 3^2
=> x = 2
d) x^4 = 16; 2^x : 2^5 = 1
x^4 = 2^4
x = 2
2^x : 2^5 = 1
2^x : 2^5 = 2^0
2^x - 5 = 2^0
=>x - 5 = 0
=>x = 0 + 5 = 5
e)|x - 2|= 0
<=> x - 2 = 0
<=> x = 0 + 2
<=> x = 2
g)4^x = 64
4^x = 4^3
x = 3
9^x - 1 = 9
9^x - 1 = 9^1
x - 1 = 1
x = 1 + 1
x = 2
a)
Để \(\left(3x-1\right).\left(-\frac{1}{2}x+5\right)=0\)=> 3x-1=0 hoặc \(-\frac{1}{2}x+5=0\)
=> x= \(\frac{1}{3}\) hoăc \(x=10\)
b)
\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=5\) => \(\frac{1}{3}:\left(2x-1\right)=5-\frac{1}{4}=\frac{19}{4}=>2x-1=\frac{1}{3}:\frac{19}{4}=\frac{4}{57}=>x=\frac{61}{114}\)
c) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0=>\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)\(=>2x+\frac{3}{5}\in\left\{\pm\frac{3}{5}\right\}=>2x\in\left\{0;\frac{-6}{5}\right\}=>x\in\left\{0;\frac{-3}{5}\right\}\)
d) Xem lại đề
a) để (3x-1).(\(-\dfrac{1}{2}x+5\))=0
=> 3x-1 hoặc \(-\dfrac{1}{2}x+5\) =0
TH1 : 3x-1=0
3x = 0+1=1
x = 1:3 = \(\dfrac{1}{3}\)
TH2 : \(-\dfrac{1}{2}x+5\)= 0
\(-\dfrac{1}{2}x\)= 0 -5 = -5
x= -5 : \(-\dfrac{1}{2}\)
x= 10
`-3x.(2x+16)=0`
`@TH1:`
`-3x=0`
`x=0:(-3)`
`x=0`
@TH2:`
`2x+16=0`
`2x=0-16`
`2x=-16`
`x=-16:2`
`x=-8`
_________________________
`(x^2 - 9).(x-3)=0`
`@TH1:`
`x^2-9=0`
`x^2=0+9`
`x^2=9`
\(x^2=\left(\pm3\right)^2\)
`@TH2:`
`x-3=0`
`x=0+3`
`x=3`
Vậy `x={-3;3}`