Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 6 :
a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)
c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)
d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)
Bài 7 :
a) \(3^x+3^{x+2}=9^{17}+27^{12}\)
\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)
\(\Rightarrow10.3^x=3^{34}+3^{36}\)
\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)
\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)
b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)
\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)
\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)
c) Bài C bạn xem lại đề
d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
a. x = {3;-3}
b. x thuộc rỗng
c. x2-4=0
x2 = 4
x={2;-2}
d. x2+1=82
x2 =83
x thuộc rỗng
e. (2x)2=6
x thuộc rỗng
f. (x-1)2=9
TH1: x-1=3=>x=4
TH2: x-1=-3=>x=-2
Vậy x={4;-2}
g.(2x+3)2=25
TH1: 2x+3=5=> x=1
Th2: 2x+3=-5=>x=-4
VẬY X={1;-4}
a, x^2= 9
=>\(\sqrt{9}=3\)
b,\(x^2=5=>x=\sqrt{5}\)
c, x^2-4=0
=>x^2=4
=>x=2
d, x^2+1=82
=>x^2=81 =>\(\sqrt{81}=9\)
3, 2x^2=6
=>x= \(\sqrt{6}\)
f, {x-1} ^2=9
=> x-1=3
=>x=2
g{ 2x+3}^2=25
=> 2x+3=5
=>2x=2
=>x=1
3^x*5^x-1=224
3^x*5^x/5=224
15^x=224*5
15^x=1120
=>ko tồn tại x thỏa mãn đề bài vị 15^x luôn có tận cùng bằng 5 (x khác 0 ) hoặc 1 ( x=0) ma 1120 co tận cùng bằng 0
Vũ Hồng Linh bạn check lại bài đầu dùm =_="
\(\left[-\frac{1}{3}\right]^3\cdot x=\frac{1}{81}\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{3}\right]^3\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{27}\right]\)
\(\Leftrightarrow x=\frac{1}{81}\cdot(-27)=-\frac{1}{3}\)
\(\left[x-\frac{1}{2}\right]^3=\frac{1}{27}\)
\(\Leftrightarrow\left[x-\frac{1}{2}\right]^3=\left[\frac{1}{3}\right]^3\)
=> Làm nốt
Mấy bài kia cũng làm tương tự
(- \(\dfrac{1}{3}\))3.\(x\) = \(\dfrac{1}{81}\)
\(x=\dfrac{1}{81}\) : (- \(\dfrac{1}{3}\))3
\(x\) = - (\(\dfrac{1}{3}\))4 :(\(\dfrac{1}{3}\))3
\(x=-\dfrac{1}{3}\)
Vậy \(x=-\dfrac{1}{3}\)
a)3^x+1=9^x
3^x+1=3.3^x
3^x+1=3^x+1
=>x thuộc TH Z
b)2^3.x+2=4^x+5
2^3x+2=2^2.(x+5)
2^3x+2=2^2x+10
2^3x=2^2x+8
3x-2x=8
=>x=8
c)3^2x-1=243
3^2x=243.3
3^2x=729
3^2x=3^6
=>2x=6
x=6:2=3
chúc bạn học tốt nha
a) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
<=> \(x-\frac{1}{2}=\frac{1}{3}\)
<=> x = \(\frac{5}{6}\)
b) \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
<=> \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{array}\right.}\)
Vậy...
a)\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\Rightarrow\left(x-\frac{1}{2}\right)=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Rightarrow x=\frac{5}{6}\)
b)\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\pm\left(\frac{2}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\pm\frac{2}{5}\)
\(\Rightarrow\begin{cases}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{cases}\)