\(\left[{}\begin{matrix}2x-\dfrac{2}{3}+\dfrac{1}{2}x=0\\x^2+5=0\end{matrix}\right.\)

\(x^2+5>0\)

\(\Rightarrow x^2+5=0\) ( vô lý )

x = 4/15

\(\dfrac{8^2.125.9^2-32.5^3.81}{20^3.3^4-6^8.5^4}\)

\(=\dfrac{2^6.5^3.3^4-2^5.5^3.3^4}{4^3.5^3.3^4-2^8.3^8.5^4}\)

\(=\dfrac{2^6.5^3.3^4-2^5.5^3.3^4}{2^6.5^3.3^4-2^8.3^8.5^4}\)

\(=\dfrac{2^5.5^3.3^4\left(2-1\right)}{2^6.5^3.3^4\left(1-2^2.3^4.5\right)}\)

\(=\dfrac{2^5.5^3.3^4.1}{2^6.5^3.3^4\left(1-810\right)}\)

\(=\dfrac{1}{2.\left(-809\right)}\)

\(=-\dfrac{1}{1618}\)

Ta có: \(2003^{2003}+1=2003^{2002+1}+1và2003^{2004}+1=2003^{2003+1}+1\)

\(\Rightarrow A>B\)

A > B

\(\widehat{AMB}=\widehat{AME}+\widehat{EMB}=3\widehat{EMB}+\widehat{EMB}=4\widehat{EMB}=180^o\)

\(\Rightarrow\widehat{EMB}=180^o:4=45^o\) 

Ta có

\(\widehat{AME}+\widehat{EMB}+\widehat{MND}=\widehat{AMB}+\widehat{MND}=225^o\)

\(\Rightarrow180^o+\widehat{MND}=225^o\Rightarrow\widehat{MND}=225^o-180^o=45^o\) 

Gọi O là giao của AB và CD xét tg OMN có

\(\widehat{MON}=180^o-\left(\widehat{EMB}+\widehat{MND}\right)=180^o-\left(45^o+45^o\right)=90^o\)

\(\Rightarrow AB\perp CD\)

cảm ơn minh

Lời giải:

$M=(x^{10}-24x^9)-(x^9-24x^8)+(x^8-24x^7)-(x^7-24x^6)+(x^6-24x^5)-(x^5-24x^4)+(x^4-24x^3)-(x^3-24x^2)+(x^2-24x)-(x-24)+1$

$=x^9(x-24)-x^8(x-24)+x^7(x-24)-.....+x(x-24)-(x-24)+1$

$=(x-24)(x^9-x^8+x^7-...+x-1)+1$

$=0.(x^9-x^8+....+x-1)+1=1$

Do ∠NIB = 1/5 ∠MIB

⇒ ∠MIB = 5 ∠NIB

Ta có:

∠MIB + ∠NIB = 180⁰ (kề bù)

⇒ 5 ∠NIB + ∠NIB = 180⁰

⇒ 6 ∠NIB = 180⁰

⇒ ∠NIB = 180⁰ : 6

= 30⁰

⇒ ∠MIA = ∠NIB = 30⁰ (đối đỉnh)