Tìm x
(5x+1)^2n=36/49
Tìm x, biết:
a) ( 5x+1)^2+36/49
b) (x-2/9)^3=(2/3)^6
c) (8x-1)^2n+1=5^2n+1
Tìm x biết
(-3/4)^3x -1= 256/81
(5x+1) ^2 =36/49
(X-2/9)^3= (2/6)^6
(8x -1)^2n+1 = 5^2n +1
Tìm x, biết:
a) 5x.(53)2=625
b) (5x+1)2=36/49
c) (8x-1)2n+1=52n=1 (n thuộc N)
\(5^x.\left(5^3\right)^2=625\)
\(\Rightarrow5^x.5^6=5^4\)
\(\Rightarrow5^x=5^4:5^6\)
\(\Rightarrow5^x=\frac{1}{25}\)
.......................
còn đoạn sau bn tự giải nha
tíc mình nha
Tìm x biết
a) 5x . ( 53 )2 = 625
b) ( 5x + 1 )2 =36/49
c) ( 8x - 1 )2n + 1 = 5 2n + 1
a) \(5^x\cdot5^6=5^4\)
\(5^x=5^{4-6}\)
\(5^x=5^{-2}\)
=> x = -2
b) \(\left(5x+1\right)^2=\left(\pm\frac{6}{7}\right)^2\)
+) 5x + 1 = 6/7
5x = -1/7
x = -1/35
+) 5x + 1 = -6/7
5x = -13/7
x = -13/35
Vậy,.........
c) => 8x - 1 = 5
=> 8x = 6
=> x = 3/4
Vậy,.....
(5x+1)2 = 36/49
(x-2/9)3 = (2/3)6
(8x-1)2n+1= 52n+1
tìm x, biết:
(5x+1)^2=36/49
\(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\left(\frac{6}{9}\right)^2\)
\(5x+1=\frac{6}{9}\)
\(5x=\frac{6}{9}-1\)
\(x=\frac{-1}{3}:5=\frac{-1}{3}.\frac{1}{5}=\frac{-1}{15}\)
(5x+1)2=36/49
(3x-2)5=-243
5x=125
(3x2-51)2n=(-24)2n (n thuoc N*)
(5x+1)2=36/49
(5x+1)2=\(\left(\frac{6}{7}\right)^2\)
5x+1=6/7
5x=-1/7
x=-1/35
5x=125
5x=53
x=3
(3x-2)5=-243
(3x-2)5=(-3)5
3x-2=-3
3x=-1
x=-1/3
tìm x biết: (5x+1)2 = 36/49
\(\left(5x+1\right)^2=\frac{36}{49}\)
(+) TH 1: 5x + 1 = 6/7
5x = 6/7 - 1
5x = -1/7
x = -1/7 : 5
x = -1 /35
(+) TH2 : 5x + 1 = - 6/7
5x = -6/7 - 1
5x = -13/7
x =-13/7 : 5
x = -13/35
Tìm x biết:
(5x + 1)2 = 36/49
( 5x + 1 )2 = 36/49
<=> ( 5x + 1 )2 = ( ±6/7 )2
<=> 5x + 1 = 6/7 hoặc 5x + 1 = -6/7
<=> x = -1/35 hoặc x = -13/35
\(\left(5x+1\right)^2=\frac{36}{49}\)
\(\Rightarrow\orbr{\begin{cases}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{cases}}\)\(\)
\(\Rightarrow\orbr{\begin{cases}5x=\frac{-1}{7}\\5x=\frac{-13}{7}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{cases}}\)
\(\left(5x+1\right)=\pm\sqrt{\frac{36}{49}}=\pm\frac{6}{7}\)
\(\orbr{\begin{cases}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{cases}}\)
\(\orbr{\begin{cases}5x=\frac{-1}{7}\\5x=\frac{-13}{7}\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{cases}}\)