a, (-0,2)² \(\times\) 5 - \(\dfrac{2^{13}\times27^3}{4^6\times9^5}\)

= 0,04 \(\times\) 5 - \(\dfrac{2^{13}\times3^9}{2^{12}\times3^{10}}\)

= 0,2 - \(\dfrac{2}{3}\)

= \(\dfrac{2}{10}\) - \(\dfrac{2}{3}\)

=  - \(\dfrac{7}{15}\)

b, \(\dfrac{5^6+2^2.25^3+2^3.125^2}{26.5^6}\)

 = \(\dfrac{5^6+4.5^6+8.5^6}{26.5^6}\)

= \(\dfrac{5^6.\left(1+4+8\right)}{26.5^6}\)

= \(\dfrac{1}{2}\)

a, (-0,2)2 $\times$ 5 - 213×27346×9546×95213×273​

= 0,04 $\times$ 5 - 213×39212×310212×310213×39​

= 0,2 - 2332​

= 210102​ - 2332​

=  - 715157​

b, 56+22.253+23.125226.5626.5656+22.253+23.1252​

 = 56+4.56+8.5626.5626.5656+4.56+8.56​

= 56.(1+4+8)26.5626.5656.(1+4+8)​

= 1221​
 

Bài 2: 

a: \(\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)

=>(x+5)(x-6)=0

=>x=-5 hoặc x=6

b: \(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)

=>-4x+2=0

hay x=1/2

c: \(\Leftrightarrow\left(x^2+4\right)\left(x^2-1\right)=0\)

=>x=1 hoặc x=-1

vip đây

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a,5mũ 36=(5mũ3)mũ12=125 mũ12

11^24=(11^2)12=121^12

vì 121<125 nên 5^36>11^24

cảm ơn nha

bạn ơi sao bài này khó thế?

a. ( 2x + 1 )² = 49

<=> ( 2x + 1 )² = 7²

<=> 2x + 1 = 7

<=> x = 3

b. ( 2x - 1 )⁴ = 81

<=> ( 2x - 1 )⁴ = 3⁴

<=> 2x - 1 = 3

<=> x = 2

c. ( x + 1 )³ = 2x³

<=> x + 1 = 2x

<=> x = 1

d. ( 2x + 1 )³ = 3x³

<=> 2x + 1 = 3x

<=> x = 1

( 2x + 1 )² = 49

<=> ( 2x + 1 )² = ( ±7 )²

<=> \(\orbr{\begin{cases}2x+1=7\\2x+1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)

( 2x - 1 )⁴ = 81

<=> ( 2x - 1 )⁴ = ( ±3 )⁴

<=> \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

( x + 1 )³ = ( 2x )³

<=> x + 1 = 2x

<=> x - 2x = -1

<=> -x = -1

<=> x = 1

( 2x + 1 )³ = ( 3x )³

<=> 2x + 1 = 3x

<=> 2x - 3x = -1

<=> -x = -1

<=>  x = 1

a) \(\left(2x+1\right)^2=49\)  

\(\left(2x+1\right)=\pm7\) 

\(\orbr{\begin{cases}2x+1=7\\2x+1=-7\end{cases}}\)  

\(\orbr{\begin{cases}2x=6\\2x=-8\end{cases}}\)

\(\orbr{\begin{cases}x=3\\x=-4\end{cases}}\) 

b) \(\left(2x-1\right)^4=81\) 

\(\left(2x-1\right)^4=3^4\) 

\(2x-1=\pm3\) 

\(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\) 

\(\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\) 

\(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)    

c) \(\left(x+1\right)^3=\left(2x\right)^3\)    

\(x+1=2x\) 

\(x-2x=-1\) 

\(-x=-1\) 

\(x=1\)      

d) \(\left(2x+1\right)^3=\left(3x\right)^3\)     

\(2x+1=3x\) 

\(1=3x-2x\) 

\(1=x\)

\(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)

\(=\left(3^{60}-3^{57}\right):3^{57}+\left(5^{27}-5^{24}\right):5^{24}\)

\(=3^{57}\left(3^3-1\right):3^{57}+5^{24}\left(5^3-1\right):5^{24}\)

\(=3^3-1+5^3-1\)

\(=27-1+125-1\)

\(=150\)

2 )

\(x^2-25-\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-5\right)-\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-5-1\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)

Vậy ...

b )

\(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)

\(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)

\(\Leftrightarrow2-4x=0\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy ...

c )

\(x^2\left(x^2+4\right)-x^2-4=0\)

\(\Leftrightarrow x^2\left(x^2+4\right)-\left(4+x^2\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x^2+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\x^2+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=1\\x^2=-4\left(L\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy ...