dề bài đâu mà tìm?

Còn nhớ mình không, **** cho mình nhé

Dương ơi tớ tích cho nhé

a)\(2^{3x+2}=4^{x+5}\)

\(2^{3x+2}=2^{2x+10}\)

\(\Rightarrow3x+2=2x+10\)

\(\Rightarrow x=8\)

Vậy \(x=8\)

b) \(3^{x+1}=9^x\)

\(3^{x+1}=3^{2x}\)

\(\Rightarrow x+1=2x\)

\(\Rightarrow x=1\)

Vậy \(x=1\)

c) \(\left(\frac{4}{5}\right)^{2x+7}=\frac{625}{256}\)

\(\left(\frac{4}{5}\right)^{2x+7}=\frac{5^4}{4^4}\)

\(\frac{4^{2x+7}}{5^{2x+7}}=\frac{5^4}{4^4}\)

\(\Rightarrow5^{2x+7}.5^4=4^{2x+7}.4^4\)

\(\Leftrightarrow5^{2x+11}=4^{2x+11}\)

\(\Leftrightarrow5=4\)( vô lý )

\(\Rightarrow\)x không có giá trị

Vậy không tìm được giá trị của x

Fan alibaba nguyễn~

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)

a)=18-5-7

=13-7

=5

b)=(48-2)+(-21)

46+(-21)

=25

c)=-200.(-2)+20

=400+20=420