TÌm x
5^x+3.5^4x-5^2x.5^3x+2=100
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(575-\left(2x+70\right)=445\)
=>\(2x+70=575-445=130\)
=>\(2x=130-70=60\)
=>x=60/2=30
b: \(575-2\left(x+70\right)=445\)
=>\(2\left(x+70\right)=575-445=130\)
=>x+70=130/2=65
=>x=65-70=-5
c: \(x^5=32\)
=>\(x^5=2^5\)
=>x=2
d: \(\left(3x-1\right)^3=8\)
=>\(\left(3x-1\right)^3=2^3\)
=>3x-1=2
=>3x=3
=>\(x=\dfrac{3}{3}=1\)
e: \(\left(x-2\right)^3=27\)
=>\(\left(x-2\right)^3=3^3\)
=>x-2=3
=>x=5
f: \(\left(2x-3\right)^2=9\)
=>\(\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
g: \(2x+5=3^4:3^2\)
=>\(2x+5=3^2\)
=>2x+5=9
=>2x=9-5=4
=>x=4/2=2
h: \(\left(4x-5^2\right)\cdot7^3=7^4\)
=>\(4x-25=\dfrac{7^4}{7^3}=7\)
=>4x=25+7=32
=>\(x=\dfrac{32}{4}=8\)
Tìm x, biết:
1) 2x ( x - 5) - x ( 2x - 4 ) = 15
<=> 2x2 - 10x - 2x2 + 4x - 15 = 0
<=> -6x - 15 = 0
<=> -6x = 15
<=> x = -15/6
2) ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6
<=> x2 + 2x + x + 2 - x2 - 3x - 4x - 12 - 6 = 0
<=> -4x = -16
<=> x = 4
3) 4x2 - 4x + 5 - x ( 4x - 3) = 1 - 2x
<=> 4x2 - 4x + 5 - 4x2 + 3x - 1 + 2x = 0
<=> x + 4 = 0
<=> x = -4
4) ( x + 3 ) ( 2x + 1 ) - 2x2 = 4x - 5
<=> 2x2 + x + 6x + 3 - 2x2 - 4x + 5 = 0
<=> 3x + 8 = 0
<=> 3x = -8
<=> x = -8/3
5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0
<=> - 8x + 32 + 8x2 + 6x - 4x - 3 = 0
.......
6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)
<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0
<=> -2x + 40 = 0
<=> -2x = -40
<=> x = 20
Còn lại tương tự ....
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
c) x.(1+2+3+4+...+100)=0
x.5050=0
x=0:5050=0
Vậy x=0
d) x.(1+2+3+4+5+...+100)=5050
x.5050=5050
x=1
Vậy x=1
e) x+1+x+2+x+3+x+4+...+x+100=5050
(x+x+x+x+...+x)+(1+2+3+4+...+100)=5050
100 số hạng x
x.100+5050=5050
x.100=0
x=0
Vậy x=0
latex
là sao nhỉ