b) 3.[5^2x+1 - (5²)^x] = 3.100

3( 5^2x+1-5^2x) = 3.100

-> 5^2x+1-5^2x = 100

mà 5³-5²= 100

-> 5^2x+1-5^2x = 5³-5²

^{Vậy x = 1}

c) (4²)^x : 4^x = 4²

<=>4^2x-x =4²

<=> 4^x = 4²

-> x=2

d) 2^x( 0,5 +4) = 72

2^x. 4,5 = 72

2^x = 72:4,5 = 16

2^x = 2⁴ = 16

-> x= 4

Bài 1:

a) \(x^2-3=1\)

\(\Rightarrow x^2=1+3=4\)

\(\Rightarrow x=\pm2\)

b)\(2x^3+12=-4\)

\(\Rightarrow2x^3=-4-12=-16\)

\(\Rightarrow x^3=-8\)

\(\Rightarrow x=-2\)

c)\(\left(2x-3\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=\dfrac{7}{2}\\-\dfrac{1}{2}\end{matrix}\right.\)

a) \(x^2-3=1\Rightarrow x^2=4\Rightarrow x=\pm2\)

b) \(2x^3+12=-4\Rightarrow2x^3=-16\)

\(\Rightarrow x^3=-\dfrac{16}{2}=-8=-2^3\)

\(\Rightarrow x=-2\)

c) \(\left(2x-3\right)^2=16\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

d,h,i,k cững tương tự....

a) x-8.5=4

x=4+8.5

x=12.5

\(\frac{3}{\sqrt{16}}-\left(x+\frac{1}{2}\right)^2=\frac{1}{2}\)

<=>  \(\left(x+\frac{1}{2}\right)^2=\frac{3}{\sqrt{16}}-\frac{1}{2}=\frac{1}{4}\)

<=>  \(\left(x+\frac{1}{2}\right)^2=\frac{1}{4}\)

<=>  \(\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{2}\\x+\frac{1}{2}=-\frac{1}{2}\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Vậy...

b: \(\left(\dfrac{2}{5}-\dfrac{7}{10}x\right):\dfrac{5}{3}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{2}{5}-\dfrac{7}{10}x=\dfrac{-3}{4}\cdot\dfrac{5}{3}=\dfrac{-5}{4}\)

\(\Leftrightarrow x\cdot\dfrac{7}{10}=\dfrac{2}{5}+\dfrac{5}{4}=\dfrac{8+25}{20}=\dfrac{33}{20}\)

\(\Leftrightarrow x=\dfrac{33}{20}:\dfrac{7}{10}=\dfrac{33}{20}\cdot\dfrac{10}{7}=\dfrac{33}{14}\)

c: \(\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)-\dfrac{11}{6}=0\)

\(\Leftrightarrow\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)=\dfrac{11}{6}\)

\(\Leftrightarrow x\cdot\dfrac{1}{4}+\dfrac{9}{2}=\dfrac{11}{6}:\dfrac{7}{16}=\dfrac{88}{21}\)

\(\Leftrightarrow x\cdot\dfrac{1}{4}=\dfrac{88}{21}-\dfrac{9}{2}=-\dfrac{13}{42}\)

hay \(x=-\dfrac{26}{21}\)

1. Tìm x, biết :

a. ( x - \(\frac{3}{4}\)) \(^2\)= 0

=> x - \(\frac{3}{4}\)= 0

=> x = 0 + \(\frac{3}{4}\)

=> x = \(\frac{3}{4}\)

b. ( x + \(\frac{1}{2}\)) \(^2\)= \(\frac{9}{64}\)

=> ( x + \(\frac{1}{2}\)) \(^2\)= ( \(\frac{3}{8}\)) \(^2\)

=> x + \(\frac{1}{2}\)= \(\frac{3}{8}\)

=> x = \(\frac{3}{8}\)- \(\frac{1}{2}\)

=> x = \(\frac{-1}{8}\)

c.  \(\frac{\left(-2\right)^x}{16}=-8\)

=> \(\frac{\left(-2\right)^x}{16}=\frac{-8}{1}=\frac{-128}{16}\)

=> ( -2)\(^x\)= -128

=> ( -2 ) \(^x\)= ( -2) \(^7\)

=> x = 7

a. \(\dfrac{3}{4}-\left(2x-\dfrac{2}{3}\right)=\dfrac{-5}{6}\)

\(\Rightarrow2x-\dfrac{2}{3}=\dfrac{3}{4}-\dfrac{-5}{6}\)

\(\Rightarrow2x-\dfrac{2}{3}=\dfrac{19}{12}\)

\(\Rightarrow2x=\dfrac{19}{12}+\dfrac{2}{3}=\dfrac{9}{4}\)

\(\Rightarrow x=\dfrac{9}{4}:2=\dfrac{9}{8}\)

Vậy............

b. \(1,5-\left(x+\dfrac{7}{2}\right)=2^7:2^5\)

\(\Rightarrow1,5-\left(x+\dfrac{7}{2}\right)=2^2=4\)

\(\Rightarrow x+\dfrac{7}{2}=1,5-4=\dfrac{-5}{2}\)

\(\Rightarrow x=\dfrac{-5}{2}-\dfrac{7}{2}=-6\)

Vậy.............