\(7\left(5-x\right)+5\left(x-2\right)=15\)

\(3x-7x+5x-10=15\)

\(x-10=15\)

\(x=25\)

1, x-(-19)+(-2)⁵=14-(-2)⁴

    x+19+(-32)=14-16

    x+19+(-32)=-2

    x+19=-2-(-32)

    x+19=30

          x=30-19

          x=11

Vậy x=11

2)7.(5-x)+5(x-2)=15

   35-7x+5x-10=15

   -7x+5x=15-35+10

    -2x=-10

     2x=10

       x=10:2

       x=5

Vậy x=5

a) \(9^{21}.9^{33}=9^{21+33}=9^{54}\)

b) \(19^{11}.19.19=19^{11+1+1}=19^{13}\)

c) \(25^2.5^2.125=5^4.5^2.5^3=5^{4+2+3}=5^9\)

d) \(t^{2021}.t^2.\left(t^2\right)^2=t^{2021}.t^2.t^4=t^{2021+2+4}=t^{2027}\)

e) \(123^{14}:123^{13}=123^{14-13}=123\)

f) \(64^2:8^3=\left(8^2\right)^2:8^3=8^4:8^3=8^{4-3}=8=2^3\)

g) \(6^{10}:6^3:36=6^{10}:6^3:6^2=6^{10-3-2}=6^5\)

h) \(m^{20}:m^{10}.m^{10}=m^{20-10+10}=m^{20}\)

oke

Rút gọn hả bạn ?

( 3x - 1 )² - 9( x - 1 )( x + 1 )

= 9x² - 6x + 1 - 9( x² - 1 )

= 9x² - 6x + 1 - 9x² + 9

= 10 - 6x

( 2x + 3 )( 2x - 3 ) - ( 2x - 1 )² - ( x - 1 )

= 4x² - 9 - ( 4x² - 4x + 1 ) - x + 1

= 4x² - x - 8 - 4x² + 4x - 1

= 3x - 9

2( x - 2y )( x + 2y ) + ( x - 2y )² + ( x + 2y )²

= [ ( x + 2y ) + ( x - 2y ) ]²

= [ x + 2y + x - 2y ]²

= ( 2x )² = 4x²

a, \(\left(x+\frac{4}{3}y^2\right)^2\)

\(=x^2+\frac{8}{3}xy^2+\frac{16}{9}y^4\)

b, \(\left(2x-3y\right)^2\)

\(=4x^2-12xy+9y^2\)

c, \(\left(x^2+2x\right)\left(2x-x^2\right)\)

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

\(=4x^2-x^4\)

d, \(\left(x+\frac{1}{2}\right)^3\)

\(=x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)

e, \(\left(2x-\frac{6}{5}y\right)^3\)

\(=8x^3-\frac{72}{5}x^2y+\frac{216}{25}xy^2-\frac{216}{125}y^3\)

\(\left(2x-1\right)^4=16=\left(\pm2\right)^4\\ =>\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\left(2x+1\right)^3=125=5^3\\ =>2x+1=1\\ =>x=2\)

???????

\(\left(\dfrac{1}{2}\right)^{50}=\left[\left(\dfrac{1}{2}\right)^5\right]^{10}=\left(\dfrac{1}{32}\right)^{10}\)

1/12>1/32

=>(1/12)^10>(1/32)^10

=>(1/12)^10>(1/2)^50

Có: \(\left(\dfrac{1}{12}\right)^{10}=\dfrac{1}{12^{10}}\)

\(\left(\dfrac{1}{2}\right)^{50}=\dfrac{1}{2^{50}}=\dfrac{1}{\left(2^5\right)^{10}}=\dfrac{1}{32^{10}}\)

Do \(12< 32\Rightarrow12^{10}< 32^{10}\)

\(\Rightarrow\dfrac{1}{12^{10}}>\dfrac{1}{32^{10}}\) hay \(\left(\dfrac{1}{12}\right)^{10}>\left(\dfrac{1}{2}\right)^{50}\)

2: \(\left(\dfrac{1}{6}\right)^{10}=\dfrac{1}{6^{10}}\)

(1/2)^50=1/2^50

mà 6^10<2^50

nên (1/6)^10<(1/2)^50

3: 30^-50=1/30^50=(1/3)^500=(1/243)^100

5^-30=1/5^30=(1/125)^100

=>30^50<5^-30

Ta có :

\(\frac{1}{243^9}=\frac{1}{\left(81.3\right)^9}=\frac{1}{81^9.27^3}>\frac{1}{81^9.81^3}=\frac{1}{81^{11}}>\frac{1}{8^{12}}>\frac{1}{8^{13}}\)

\(\Rightarrow\frac{1}{243^9}>\frac{1}{83^{13}}\)

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