a: 5x-9=5+3x

=>5x-3x=5+9

=>2x=14

=>\(x=\frac{14}{2}=7\)

b: \(2^3+0,5x=1,5\)

=>0,5x+8=1,5

=>0,5x=1,5-8=-6,5

=>\(x=-6,5:0,5=-13\)

c: \(\left(5x+1\right)^2=\frac{36}{49}\)

=>\(\left[\begin{array}{l}5x+1=\frac67\\ 5x+1=-\frac67\end{array}\right.\Rightarrow\left[\begin{array}{l}5x=\frac67-1=\frac17\\ 5x=-\frac67-1=-\frac{13}{7}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac17:5=\frac{1}{35}\\ x=-\frac{13}{7}:5=-\frac{13}{35}\end{array}\right.\)

d: \(\left(-\frac{3}{81}\right)^{x}=-27\)

=>\(\left(-\frac{1}{27}\right)^{x}=-27\)

=>\(x=-1\)

e: \(2^{x-1}=16\)

=>\(2^{x-1}=2^4\)

=>x-1=4

=>x=4+1=5(nhận)

\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)

a: =>5x+1=6/7 hoặc 5x+1=-6/7

=>5x=-1/7 hoặc 5x=-13/7

=>x=-1/35 hoặc x=-13/35

b: =>x-2/9=4/9

=>x=6/9=2/3

c: =>8x+1=5

=>8x=4

hay x=1/2

a) \(\left(5x+1\right)^2=\dfrac{36}{49}\)

\(\left(5x+1\right)^2=\left(\pm\dfrac{6}{9}\right)\)\(^2\)

\(5x+1=\pm\dfrac{6}{9}\)

+) \(5x+1=\dfrac{6}{9}\)

\(5x=\dfrac{6}{9}-1=\dfrac{6}{9}-\dfrac{9}{9}\)

\(5x=\dfrac{-5}{9}\)

\(x=\dfrac{-5}{9}:5=\dfrac{-1}{45}\)

+) \(5x+1=\dfrac{-6}{9}\)

\(5x=\dfrac{-6}{9}-1=\dfrac{-6}{9}-\dfrac{9}{9}\)

\(5x=\dfrac{-5}{3}\)

\(x=\dfrac{-5}{3}:5=\dfrac{-5}{15}\)

vậy \(x\in\left\{\dfrac{-5}{15};\dfrac{-1}{45}\right\}\)

b, \(\left(5x+1\right)^2=\dfrac{36}{49}\)

\(\left(5x+1\right)^2=\left(\pm\dfrac{6}{7}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=-\dfrac{1}{7}\\5x=\dfrac{-13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)

Vậy .....

Nguyễn Thanh Hằng ;Hồng Phúc Nguyễn ;Mới vô; ... các bn giúp mik vs mik đang cần gấp !

ko ghi lại đề bài lm luôn:

\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)

       \(5x+1=\frac{6}{7}\)

                \(5x=\frac{-1}{7}\)

                  \(x=\frac{-1}{35}\)

tôi lm đại thôi nhưng chắc đ đấy

Bài 1:

\(\dfrac{\left(\dfrac{2}{5}\right)^7\cdot5^7+\left(\dfrac{9}{4}\right)^3:\left(\dfrac{3}{16}\right)^3}{2^7\cdot5^2+512}\)

\(=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}:\dfrac{3}{16}\right)^3}{2^7\cdot5^2+512}\)

\(=\dfrac{2^7+12^3}{2^7\cdot5^2+512}\)

\(=\dfrac{1856}{3712}\)

\(=0,5\)

Bài 2:

\(\left(5x+1\right)^2=\dfrac{36}{49}\)

\(\Rightarrow5x+1=\dfrac{6}{7}\)

\(\Rightarrow5x=\dfrac{-1}{7}\)

\(\Rightarrow x=\dfrac{-1}{35}\)