Ta có : 9¹⁰ = (3²)¹⁰ = 3²⁰ < 10²⁰

Nên : 9¹⁰ < 10²⁰

a)10²⁰ và 9¹⁰
Vì 10 > 9
20 > 10
=) 10²⁰ > 9¹⁰
b) (-5)³⁰ và (-3)⁵⁰
Có \(\left(-5\right)^{30}=\left[\left(-5\right)^3\right]^{10}=\left(-125\right)^{10}\)
\(\left(-3\right)^{50}=\left[\left(-3\right)^5\right]^{10}=\left(-243\right)^{10}\)
Vì (-125)¹⁰ > (-243)¹⁰ =) (-5)³⁰ > (-3)⁵⁰
c) \(64^8\)vaf \(16^{12}\)
Có \(64^8=\left(4^3\right)^8=4^{24}\)
\(16^{12}=\left(4^2\right)^{12}=4^{24}\)
Vì \(4^{24}=4^{24}\rightarrow64^8=16^{12}\)
d) \(\left(\frac{1}{16}\right)^{10}\)và \(\left(\frac{1}{2}\right)^{50}\)
Có \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1^{10}}{\left(2^4\right)^{10}}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1^{50}}{2^{50}}=\frac{1}{2^{50}}\)
Vì \(\frac{1}{2^{40}}>\frac{1}{2^{50}}\)=) \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

Ta có ; 2⁵⁰⁰ = (2⁵)¹⁰⁰ = 32¹⁰⁰

          5²⁰⁰ = (5²)¹⁰⁰ = 25¹⁰⁰ 

Vậy 2⁵⁰⁰ > 5²⁰⁰ 

Ta có:

\(2^{500}=\left(2^5\right)^{100}=32^{100}\)

\(5^{200}=\left(5^2\right)^{100}=25^{100}\)

\(\Rightarrow32^{100}>25^{100}\Rightarrow2^{500}>5^{200}\)

Ai k mik mik k lại nha

Ta có:\(\frac{x}{4}=\frac{y}{7}\Rightarrow\frac{x^2}{16}=\frac{y^2}{49}\)

\(=\frac{3x^2}{48}=\frac{4y^2}{196}=100\)

\(\Rightarrow\hept{\begin{cases}x^2=1600\\y^2=4900\end{cases}}\)

TH1:x=40,y=70

TH2:x=40,y=-70

TH3:x=-40,y=70

TH4:x=-40,y=-70

thầy ơi bài này làm rồi

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)

Vậy:

\(\frac{a\cdot c}{b\cdot d}=\frac{bk\cdot dk}{b\cdot d}=\frac{k^2\cdot\left[b\cdot d\right]}{b\cdot d}=k^2\)

và

\(\frac{2009a^2+2010c^2}{2009b^2+2010d^2}=\frac{2009\left[bk\right]^2+2010\left[dk\right]^2}{2009b^2+2010d^2}=\frac{2009\cdot b^2k^2+201d^2k^2}{2009b^2+2010d^2}=\frac{k^2\left[2009b^2+2010d^2\right]}{2009b^2+2010d^2}=k^2\)Vậy khi \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{ac}{bd}=\frac{2009a^2+2010c^2}{2009b^2+2010d^2}\)

Ta có:\(\frac{\left[x\left(x-2\right)\right]}{x^2+8x-20}+12x-3=\frac{x\left(x-2\right)}{x^2-2x+10x-20}+12x-3\)

\(=\frac{x\left(x-2\right)}{x\left(x-2\right)+10\left(x-2\right)}+12x-3=\frac{x\left(x-2\right)}{\left(x+10\right)\left(x-2\right)}+12x-3\)

\(=\frac{x}{x+10}+12x-3=\frac{x+\left(12x-3\right).\left(x+10\right)}{x+10}=\frac{x+12x^2+120x-3x-30}{x+10}\)

\(=\frac{12x^2+118x-30}{x+10}\)