5^x+1 - 5^x = 100 . 25²⁹

5^x . 5 - 5^x = 100  .  25²⁹

5^x . ( 5 - 1 ) = 100 . 25²⁹

5^x.: 4 = 100 . 25²⁹

5^x : 25²⁹ = 100 : 4

5^x : 25²⁹  = 25

5^x = 25 . 25²⁹

5^x = 25³⁰

5^x = (5²)³⁰

5^x = 5⁶⁰ 

=> x = 60

\(a,x.x=-5.-5\)

\(x^2=25< =>5^2=25=>x=5\)Vậy x=5

\(b,-2.8=x.-x\)

\(-16=-\left(x^2\right)< =>-16=-\left(4^2\right)=>x=4\)Vậy x=4

\(c,\left(x+1\right)\left(x+1\right)=4.9\)

\(\left(x+1\right)^2=36\)

\(\left(x+1\right)^2=6^2\)

\(< =>x+1=6\)

\(x=5\)

Vậy x=5

bằng em không biết hihi

(-2,5)^2= ?

Trả lời:

(-2,5)²=(-2,5).(-2,5)=6,25

Giải :

( - 2,5 )² = \(\left(\frac{-5}{2}\right)^2\)

\(=\frac{\left(-5\right)^2}{2^2}=\frac{25}{4}\)

\(( 1 - 2x )^4 = \frac{1}{128}\)

\(\Rightarrow1-2x=\frac{1}{\sqrt[4]{128}}=\frac{-1}{\sqrt[4]{128}}\)

\(\Rightarrow\orbr{\begin{cases}1-2x=\frac{1}{\sqrt[4]{128}}\\1-2x=\frac{-1}{\sqrt[4]{128}}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=1-\frac{1}{\sqrt[4]{128}}\\2x=1-\frac{-1}{\sqrt[4]{128}}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1-\frac{1}{\sqrt[4]{128}}}{2}\\x=\frac{1-\frac{-1}{\sqrt[4]{128}}}{2}\end{cases}}\)

\(\left(1-2x\right)^4=\frac{1}{128}\)

\(\Leftrightarrow1-2x=\frac{\sqrt[4]{2}}{4}\)

\(\Leftrightarrow2x=\frac{1-\sqrt[4]{2}}{4}\)

hay \(x=\frac{1-\sqrt[4]{2}}{8}\)

 số 1/128 ko tach dc mũ 4 

\(\frac{a-1}{2}=\frac{b-2}{3}=\frac{c-3}{4}=\frac{2b-4}{6}=\frac{3c-9}{12}\)

\(=\frac{a-1+2b-4+3c-9}{2+6+12}\)

\(=\frac{a+2b+3c-14}{20}\)

\(=\frac{14-14}{20}=0\)

\(\frac{a-1}{2}=0\Rightarrow a=1\)

\(\frac{b-2}{3}=0\Rightarrow b=2\)

\(\frac{c-3}{4}=0\Rightarrow c=3\)

Vậy a = 1; b = 2; c = 3.

\(\frac{a-1}{2}=\frac{b-2}{3}=\frac{c-3}{4}\)\(=\frac{2b-4}{6}\)\(=\frac{3c-9}{12}\)

\(=\)\(\frac{a-1+2b-4+3c-9}{2+6+12}\)

\(=\)\(\frac{a+2b+3c-14}{20}=\)\(\frac{14-14}{20}20\)

\(\frac{a-1}{2}=0\Rightarrow a=1\)

\(\frac{b-2}{3}=0\Rightarrow b=2\)

\(\frac{c-3}{4}=0\Rightarrow c=3\)