Bài 6: 

a: \(15=\sqrt{225}>\sqrt{200}\)

b: \(27=9\sqrt{9}>9\sqrt{5}\)

c: \(-24=-\sqrt{576}< -\sqrt{540}=-6\sqrt{15}\)

Câu 1: a, x + 3 = -8 => x = (-8) - 3 = -11. b, (35 + x) - 12 = 27 => 35 + x = 27 + 12 = 39 => x = 39 - 35 = 4. 

a, Ta có: \(\left(\dfrac{1}{2}\right)^{300}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)
\(\left(\dfrac{1}{3}\right)^{200}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)
=> \(\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}\)=> \(\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)
b, Ta có: \(\left(\dfrac{1}{3}\right)^{75}=\left[\left(\dfrac{1}{3}\right)^3\right]^{25}=\left(\dfrac{1}{27}\right)^{25}\)
\(\left(\dfrac{1}{5}\right)^{50}=\left[\left(\dfrac{1}{5}\right)^2\right]^{25}\)\(=\left(\dfrac{1}{25}\right)^{25}\)
Do \(\left(\dfrac{1}{27}\right)^{25}< \left(\dfrac{1}{25}\right)^{25}=>\left(\dfrac{1}{3}\right)^{75}< \left(\dfrac{1}{5}\right)^{50}\)
Kiểm tra lại bài nhé, học tốt!!

2^x+2^x+3=144

2^x+2^x.2^3=144

2^x(1+2^3)=144

2^x.9=144

2^x=16

2^x=2^4=>x=4

tìm x

3^x+1= 27

3^x+1= 3³

=> x = 2

2³⁰⁰ = (2³)¹⁰⁰ = 8¹⁰⁰

3²⁰⁰ = (3²)¹⁰⁰ = 9¹⁰⁰

8¹⁰⁰ < 9¹⁰⁰

nên 2³⁰⁰ < 3²⁰⁰

1.a)

27⁵ và 243³ Ta có: 27⁵ = 14 348 907

                            243³ = 14 348 907

=> 27⁵ = 243³ ( vì 14 348 907 = 14 348 907 )

tôi kok bik

a,A=27⁵ và B = 243³3

Ta có

     27⁵=(3³)⁵=3¹⁵

      243³=(3⁵)³=3¹⁵

Vì 3¹⁵ = 3¹⁵=>27⁵=243³

Vậy A=B

b,A=2³⁰⁰ và B=3³⁰⁰

Vì 2³⁰⁰<3³⁰⁰=>A<B

k mik nhé