7)
\(=\dfrac{\left(5-\sqrt{5}\right).\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{3\left(\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}+\sqrt{5}\right)}\)
\(=\dfrac{\left(5-\sqrt{5}\right)\left(1+\sqrt{5}\right)}{1-5}+\dfrac{3\left(\sqrt{2}+\sqrt{5}\right)}{4-5}\)
\(=\dfrac{5+5\sqrt{5}-\sqrt{5}-5}{4}+3\left(\sqrt{2}+\sqrt{5}\right)\)
=\(\dfrac{4\sqrt{5}}{4}+3\left(\sqrt{2}+\sqrt{5}\right)\)
\(=\sqrt{5}+3\sqrt{2}+3\sqrt{5}\)
\(=4\sqrt{5}+3\sqrt{2}\)
 

7: Ta có: \(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}+\dfrac{3}{\sqrt{2}-\sqrt{5}}\)

\(=-\sqrt{5}-3\left(\sqrt{5}+\sqrt{2}\right)\)

\(=-\sqrt{5}-3\sqrt{5}-3\sqrt{2}\)

\(=-4\sqrt{5}-3\sqrt{2}\)

a: Ta có: \(\widehat{ABC}=\widehat{ACB}\)(ΔABC cân tại A)

\(\widehat{ACB}=\widehat{ECN}\)(hai góc đối đỉnh)

Do đó: \(\widehat{ABC}=\widehat{ECN}\)

Xét ΔMBD vuông tại D và ΔNCE vuông tại E có

BD=CE

\(\widehat{MBD}=\widehat{NCE}\)

Do đó: ΔMBD=ΔNCE

=>DM=EN

b: Ta có: DM\(\perp\)BC

EN\(\perp\)BC

Do đó: DM//EN

Xét ΔIDM vuông tại D và ΔIEN vuông tại E có

MD=EN

\(\widehat{MDI}=\widehat{ENC}\)(hai góc so le trong, DM//EN)

Do đó: ΔIDM=ΔIEN

=>IM=IN

=>I là trung điểm của MN

bruhhh

1: \(75^3:\left(-25\right)^3=\left(\dfrac{75}{-25}\right)^3=\left(-3\right)^3=-27\)

2: \(\left(-60\right)^2:\left(-5\right)^2=\dfrac{60^2}{5^2}=12^2=144\)

3: \(169^2:\left(-13\right)^2=\dfrac{169^2}{13^2}=\left(\dfrac{169}{13}\right)^2=13^2=169\)

4: \(\left(\dfrac{1}{2}\right)^2:\left(\dfrac{3}{2}\right)^2=\left(\dfrac{1}{2}:\dfrac{3}{2}\right)^2=\left(\dfrac{1}{3}\right)^2=\dfrac{1}{9}\)

5: \(\left(\dfrac{2}{3}\right)^3:\left(\dfrac{8}{27}\right)^3=\left(\dfrac{2}{3}:\dfrac{8}{27}\right)^3=\left(\dfrac{2}{3}\cdot\dfrac{27}{8}\right)^3=\left(\dfrac{9}{4}\right)^3=\dfrac{729}{64}\)

6: \(\left(\dfrac{5}{4}\right)^4:\left(\dfrac{15}{2}\right)^4=\left(\dfrac{5}{4}:\dfrac{15}{2}\right)^4=\left(\dfrac{5}{4}\cdot\dfrac{2}{15}\right)^4=\left(\dfrac{1}{6}\right)^4=\dfrac{1}{1296}\)

7: \(\left(\dfrac{7}{8}\right)^5:\left(\dfrac{21}{16}\right)^5\)

\(=\left(\dfrac{7}{8}:\dfrac{21}{16}\right)^5\)

\(=\left(\dfrac{7}{8}\cdot\dfrac{16}{21}\right)^5=\left(\dfrac{2}{3}\right)^5=\dfrac{32}{243}\)

8: \(\left(\dfrac{5}{6}\right)^4:\left(\dfrac{25}{18}\right)^4=\left(\dfrac{5}{6}:\dfrac{25}{18}\right)^4=\left(\dfrac{5}{6}\cdot\dfrac{18}{25}\right)^4=\left(\dfrac{3}{5}\right)^4=\dfrac{81}{625}\)

9: 

\(\left(-\dfrac{3}{4}\right)^3:\left(\dfrac{9}{8}\right)^3=\left(-\dfrac{3}{4}:\dfrac{9}{8}\right)^3=\left(-\dfrac{3}{4}\cdot\dfrac{8}{9}\right)^3\)

\(=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)

10:

\(\left(\dfrac{9}{10}\right)^6:\left(\dfrac{27}{-20}\right)^6=\left(\dfrac{9}{10}:\dfrac{-27}{20}\right)^6\)

\(=\left(\dfrac{9}{10}\cdot\dfrac{20}{-27}\right)^6=\left(-\dfrac{2}{3}\right)^6=\dfrac{64}{729}\)

Bài 5:

Gọi độ dài quãng đường là x

Thời gian ô tô thứ nhất đi là x/40(h)

Thời gian ô tô thứ hai đi là x/50(h)

Theo đề, ta có: x/40-x/50=1,5

hay x=300

D. Tất cả các đáp án trên

D

đáp án là câu d nhé