Ta có:

(-1/5)³⁰⁰ = (-1)³⁰⁰/5³⁰⁰ = 1/(5³)¹⁰⁰ = 1/125¹⁰⁰

(-1/3)⁵⁰⁰ = (-1)⁵⁰⁰/3⁵⁰⁰ = 1/(3⁵)¹⁰⁰ = 1/243¹⁰⁰

Vì 125¹⁰⁰ < 243¹⁰⁰

=> 1/125¹⁰⁰ > 1/243¹⁰⁰

=> (-1/5)³⁰⁰ > (-1/3)⁵⁰⁰

Ta có : \(\left(-\frac{1}{5}\right)^{300}=\left(-\frac{1}{5}\right)^{3.100}=\left(-\frac{1}{125}\right)^{100}=\left(\frac{1}{125}\right)^{100}\)

            \(\left(-\frac{1}{3}\right)^{500}=\left(-\frac{1}{3}\right)^{5.100}=\left(-\frac{1}{243}\right)^{100}=\left(\frac{1}{243}\right)^{100}\)

Mà \(125< 243\Rightarrow\frac{1}{125}>\frac{1}{243}\Rightarrow\left(\frac{1}{125}\right)^{100}>\left(\frac{1}{243}\right)^{100}\)

\(=>\left(-\frac{1}{5}\right)^{300}>\left(-\frac{1}{3}\right)^{500}\)

x<y nhé bạn :)

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Chắc đấy! Tick nhé!

Bài 1 có nhiều giá trị mà?

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{\left|x-5\right|}{\left|x-3\right|}=\frac{\left|x-1\right|}{\left|x-3\right|}=\frac{\left|x-5\right|-\left|x-1\right|}{\left|x-3\right|-\left|x-3\right|}=\frac{\left|x-5\right|-\left|x-1\right|}{0}\)

Do đó không tồn tại x thỏa mãn.

\(\left(\frac{1}{3}\right)^{500}=\left(\frac{1}{3}^5\right)^{100}=\frac{1}{243}^{100}\)

\(\left(\frac{1}{5}\right)^{300}=\left(\frac{1}{5}^3\right)^{100}=\frac{1}{125}^{100}\)

Vì \(\frac{1}{243}<\frac{1}{125}=>\frac{1}{243}^{100}<\frac{1}{125}^{100}=>\left(\frac{1}{3}\right)^{500}<\left(\frac{1}{5}\right)^{300}\)

3^-500=(3⁵)^-100= 243^-100

5^-300= (5³)^-100 =125^-100

^{243>125 =>}    243^-100<125^-100

Hay 3^-500 <5^-300

a) Ta có :\(\left(\frac{-1}{5}\right)^{300}=\frac{-1^{300}}{5^{300}}=\frac{1}{125^{100}}\)

\(\left(\frac{-1}{3}\right)^{500}=\frac{-1^{500}}{3^{500}}=\frac{1}{243^{100}}\)

Mà \(\frac{1}{125^{100}}>\frac{1}{243^{100}}\)

\(\Rightarrow\left(\frac{-1}{5}\right)^{300}>\left(\frac{-1}{3}\right)^{500}\)

b)Ta có :\(2^{90}=\left(2^{15}\right)^6=32768^6\)

\(5^{36}=\left(5^6\right)^6=15625^6\)

Vì \(32768^6>15625^6\Rightarrow2^{90}>5^{36}\)

a.Ta có: \(\left(\frac{-1}{5}\right)^{300}=\left(\frac{-1}{5}^3\right)^{100}=\left(\frac{-1}{125}\right)^{100}=\left(\frac{1}{125}\right)^{100}\)

\(\left(\frac{-1}{3}\right)^{500}=\left(\frac{-1}{3}^5\right)^{100}=\left(\frac{-1}{243}\right)^{100}=\left(\frac{1}{234}\right)^{100}\)

Mà: \(\frac{1}{125}>\frac{1}{234}\Rightarrow\left(\frac{1}{125}\right)^{100}>\left(\frac{1}{234}\right)^{100}\)

Vậy \(\left(\frac{-1}{5}\right)^{300}>\left(\frac{-1}{3}\right)^{500}\)

b.Ta có: \(2^{90}=\left(2^{10}\right)^9=1024^9\)

\(5^{36}=\left(5^4\right)^9=625^9\)

Mặt khác: \(1024>625\Rightarrow1024^9>625^9\)

 Vậy \(2^{90}>5^{36}\)

post từng câu một thôi bn nhìn mệt quá

a/ Ta có 

\(K^4+\frac{1}{4}=K^4+K^2+\frac{1}{4}-K^2=\left(K^2+\frac{1}{2}\right)^2-K^2=\left(K^2+K+\frac{1}{2}\right)\left(K^2-K+\frac{1}{2}\right)\)

Ta lại có 

\(K^2+K+\frac{1}{2}=\left(K+1\right)^2-\left(K+1\right)+\frac{1}{2}\)

\(\Rightarrow K^4+\frac{1}{4}=\left(K^2-K+\frac{1}{2}\right)\left(\left(K+1\right)^2-\left(K+1\right)+\frac{1}{2}\right)\)

Áp dụng vào bài toán ta được

\(=\frac{101^2-101+0,5}{1^2-1+0,5}=20201\)\(1S=\frac{\left(2^2-2+0,5\right)\left(3^2-3+0,5\right)\left(4^2-4+0,5\right)\left(5^2-5+0,5\right)...\left(100^2-100+0,5\right)\left(101^2-101+0,5\right)}{\left(1^2-1+0,5\right)\left(2^2-2+0,5\right)\left(3^2-3+0,5\right)\left(4^2-4+0,5\right)...\left(99^2-99+0,5\right)\left(100^2-100+0,5\right)}\)

b/

\(\frac{3\left(x+y\right)}{3\sqrt{x\left(4x+5y\right)}+3\sqrt{y\left(4y+5x\right)}}\)

\(\ge\frac{3\left(x+y\right)}{\frac{9x+4x+5y}{2}+\frac{9y+4y+5x}{2}}\)

\(=\frac{1}{3}\)

Dấu = xảy ra khi x = y