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e)
\(\left(-\frac{3}{4}\right)^{3x-1}=\frac{256}{81}\)
\(\left(-\frac{3}{4}\right)^{3x}=\left(-\frac{4}{3}\right)^4\)
\(\left(-\frac{3}{4}\right)^{3x}=\left(-\frac{3}{4}\right)^{-4}\)
\(3x=-4\)
\(x=-\frac{4}{3}\)
f) \(172x^2-7^9:98^3=2^{-3}\)
\(172x^2-\frac{7^9}{\left(7^2.2\right)^3}=\frac{1}{2^3}\)
\(172x^2-\frac{7^3}{2^3}=\frac{1}{2^3}\)
\(172x^2=\frac{7^3}{2^3}+\frac{1}{2^3}=\frac{344}{8}\)
\(x^2=\frac{344}{8}:172=\frac{1}{4}\)
x=1/2 hoặc x=-1/2
a)\(\left(5x+1\right)^2=\frac{36}{49}\\ \left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\\ \Rightarrow\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{matrix}\right.\)
vậy...
2.
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
⇒ \(5x+1=\pm\frac{6}{7}\)
⇒ \(\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}5x=\frac{6}{7}-1=-\frac{1}{7}\\5x=\left(-\frac{6}{7}\right)-1=-\frac{13}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-\frac{1}{7}\right):5\\x=\left(-\frac{13}{7}\right):5\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{35};-\frac{13}{35}\right\}.\)
Chúc bạn học tốt!
a: =>2x+5=4
=>2x=-1
hay x=-1/2
b: \(\Leftrightarrow\left(3x-4\right)^2\cdot\left[\left(3x-4\right)^2-1\right]=0\)
=>(3x-4)(3x-5)(3x-3)=0
hay \(x\in\left\{1;\dfrac{4}{3};\dfrac{5}{3}\right\}\)
c: \(\Leftrightarrow3^{x+1}=3^{2x}\)
=>2x=x+1
=>x=1
d: \(\Leftrightarrow2^{2x+3}=2^{2x-10}\)
=>2x+3=2x-10
=>0x=-13(vô lý)
\(\left(\dfrac{-3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
\(\Rightarrow\left(\dfrac{-3}{4}\right)^{3x-1}=\left(\dfrac{4}{3}\right)^4\)
Xem lại đề
a, ĐK: \(x\ne24\)
580 :( x -24 ) =329 -150 : 2
<=> 580 :( x -24 ) = 329 - 75
<=> 580 :( x -24 ) = 254
<=> x - 24 = \(\frac{290}{127}\)
<=> x = \(\frac{3338}{127}\left(TM\right)\)
Vậy \(x=\frac{3338}{127}\)
b, 7 (x-1 ) +35= 25 + 279 :9
<=> 7x - 7 + 35 = 25 + 31
<=> 7x +28 = 56
<=> 7x = 28
<=> x = 4
Vậy x =4
c,4 ( 2x+ 7 ) -3 (3x -2 ) =24
<=> 8x + 28 - 9x + 6 = 24
<=> 34 - x = 24
<=> x = 10
Vậy x = 10
d,( x-1 ) (x-2) =3(x-1)
<=> ( x-1 ) (x-2) - 3(x-1) = 0
<=> (x- 1)(x - 2 - 3) = 0
<=> (x -1)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Vậy x ={1; 5}
e, (x + 3) + (x + 7) + (x + 11) + ... + (x + 79) = 860
x + 3 + x + 7 + x + 11 + ... + x + 79 = 860
Có tất cả: (79 - 3) : 4 + 1 = 20 số hạng \(\Rightarrow\) có 20x
hay x + 3 + x + 7 + x + 11 + ... + x + 79 = 860
\(\Rightarrow\) 20x + (3 + 7 + ... + 79) = 860
3 + 7 + ... + 79 = (79 + 3) x 20 : 2 = 820
\(\Rightarrow\) 20x + (3 + 7 + ... + 79) = 860
\(\Rightarrow\) 20x + 820 = 860
\(\Rightarrow\) 20x = 40
\(\Rightarrow\) x = 2
Vậy x = 2
Chúc bn học tốt!
a.(2x +1). (2x+1)=1
Mà chỉ có 1.1=1
Vậy 2x + 1=1
2x=1-1
2x=0
Suy ra: x= 0
Hoàng Khánh Thi thiếu nha.
a) (2x+1)2 = \(\left(\pm1\right)^2\)
=> 2x + 1 = 1 hoặc 2x + 1 = -1
=> 2x = 0 hoặc 2x = -2
=> x = 0 hoặc x = -1.
\(\left(\frac{-3}{4}\right)^{3x+5}=\left(\frac{81}{256}\right)^{-1}\)
\(\Rightarrow\left(\frac{-3}{4}\right)^{3x+5}=\frac{256}{81}\)
\(\Rightarrow\left(\frac{-3}{4}\right)^{3x+5}=\left(\frac{-3}{4}\right)^{-4}\)
\(\Rightarrow3x+5=-4\)
\(\Rightarrow3x=-9\)
\(\Rightarrow x=-3\)
Vậy x= -3
Bài 2:
a: (x+3)/5=5/7
=>x+3=25/7
hay x=4/7
b: ||x-5|-4|=5
=>|x-5|-4=5 hoặc |x-5|-4=-5
=>|x-5|=9
=>x-5=9 hoặc x-5=-9
=>x=14 hoặc x=-4
c: \(\left(-\dfrac{4}{3}\right)^{3x+1}=\dfrac{256}{81}\)
nên 3x+1=4
=>3x=3
hay x=1