Tìm x biết (4/5) 2x+7 = 625/ 256
a, (4/5)2x+7= 625/256
b, 7x+2 +7x+1 + 7x / 57 = 52x + 2x+1 + 52x+3 / 131
a: \(\Leftrightarrow2x+7=-4\)
=>2x=-11
hay x=-11/2
b: \(\Leftrightarrow\dfrac{7^x\cdot49+7^x\cdot7+7^x}{57}=\dfrac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)
\(\Leftrightarrow7^x=5^{2x}\)
=>x=0
tìm x biết : \(\left(\frac{4}{5}\right)^{2x+7}=\frac{625}{256}\)
\(\left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\)
Ta có: \(\left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\)
\(\Leftrightarrow\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\)
\(\Leftrightarrow2x+5=4\)
\(\Leftrightarrow2x=-1\)
hay \(x=-\dfrac{1}{2}\)
Vậy: \(x=-\dfrac{1}{2}\)
Tìm X biết :\(\left(\frac{4}{5}\right)^{2x+7}=\frac{625}{256}\)
tìm x
\(\left(\frac{4}{5}\right)^{2x+7}=\frac{256}{625}\)
256/625=(4/5)4
=>2x+7=4
2x=-3
x=-3/2
bài 3
d) \(\left(\frac{4}{5}\right)^{2x+7}=\frac{625}{256}\)
e) \(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+2}}{131}\)
d, \(=>\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4.\)
=> \(2x+7=4\)
=> 2x= -3
=> x=-3/2 . Vậy x=-3/2
e, => \(\frac{7^x.7^2+7^x.7+7^x}{57}=\frac{5^{2x}+5^{2x}.5+5^{2x}.5^2}{131}.\)
=> \(\frac{7^x\left(7^2+7+1\right)}{57}=\frac{5^{2x}\left(1+5+5^2\right)}{131}\)
= > \(\frac{7^x.57}{57}=\frac{5^{2x}.131}{131}\)
=> \(7^x=5^{2x}\)
Đến đoạn này là mik nghĩ không ra nhé
Cô làm tiếp giúp Linh Đan:
\(7^x=5^{2x}\Rightarrow7^x=25^x\Rightarrow\frac{7^x}{25^x}=1\Rightarrow\left(\frac{7}{25}\right)^x=1\Rightarrow x=0\)
Tìm x biết :
\(\left(\frac{4}{5}\right)^{2x+7}=\frac{625}{256}\)
( 5/4 ) ^2x+ 7 = ( 5/4 )^4
=> 2x + 7 = 4
=> 2x = -3
x = -3/2
xem lại 5/4 hay 4/5
Để sai rồi :
\(\left(\frac{5}{4}\right)^{2x+7}=\frac{625}{256}\)
\(\Leftrightarrow\left(\frac{5}{4}\right)^{2x+7}=\left(\frac{5}{4}\right)^4\)
<=> 2x + 7 = 4
<=> 2x = -3
<=> x = -3/2
Vậy x = -3/2
\(\left(\frac{5}{4}\right)^{2x+7}\Leftrightarrow\left(\frac{5}{4}\right)^4\)
suy ra 2x+7=4
2x =-3
x = -3/2
Bài 3 Tìm x biết
a) \(\left(x-4\right)^2=\left(x-4\right)^4\)
b) \(\left(\dfrac{4}{5}\right)^{2x+7}=\dfrac{625}{256}\)
c) \(\dfrac{7^{x+2}+7^{x+1}+7^x}{57}=\dfrac{5^{2x}+5^{2x+1}+5^{2x+2}}{131}\)
a) \(\left(x-4\right)^2=\left(x-4\right)^4\)
\(\Rightarrow\left(x-4\right)^2-\left(x-4^4\right)=0\)
\(\Rightarrow\left(x-4\right)^2.\left[1-\left(x-4\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-4\right)^2=0\\1-\left(x-4\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\\left(x-4\right)^2=1^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-4=1\\x-4=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=5\\x=3\end{matrix}\right.\)
Tìm x, biết:
a) \(\left(\frac{4}{5}\right)^{2x+7}=\frac{625}{256}\)
b) \(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{5^x+5^{2x+1}+5^{2x+2}}{132}\)